AP Inter 2nd Year

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7h Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7h

I.

Question 1.
Evaluate \(\int_{-1}^1(x+1) d x\)
Solution:
Let I = \(\int_{-1}^1(x+1) d x=\left[\frac{x^2}{2}+x\right]_{-1}^1=\left[\frac{1}{2}+1\right]-\left[\frac{1}{2}-1\right]=\frac{1}{2}+1-\frac{1}{2}+1=2\)

Question 2.
Evaluate \(\int_2^3 \frac{1}{x} d x\)
Solution:
Let I = \(\int_2^3 \frac{1}{x} d x=[\log |x|]_2^3=\log |3|-\log |2|=\log \frac{3}{2}\)

Question 3.
Evaluate \(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
Solution:
\(\int_1^2\left[4 x^3-5 x^2+6 x+9\right] d x=\left[4 \frac{x^4}{4}-5 \frac{x^3}{3}+6 \frac{x^2}{2}+9 x\right]_1^2\)
= \(\left[x^4-\frac{5}{3} x^3+3 x^2+9 x\right]_1^2=\left[2^4-\frac{5}{3}(2)^3+3(2)^2+9(2)\right]-\left[1-\frac{5}{3}+3+9\right]\)
= \(\left[16-\frac{40}{3}+12+18\right]-\left[13-\frac{5}{3}\right]=\left[46-\frac{40}{3}\right]-\left[13-\frac{5}{3}\right]\)
= \(46-\frac{40}{3}-13+\frac{5}{3}=33-\frac{40}{3}+\frac{5}{3}=\frac{99-40+5}{3}=\frac{104-40}{3}=\frac{64}{3}\).

Question 4.
Evaluate \(\int_0^\pi 4 \sin 2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{4}} \sin 2 x d x=\left[\frac{-\cos 2 x}{2}\right]_0^{\frac{\pi}{4}}=\left[\frac{-\cos 2 \frac{\pi}{4}}{2}\right]-\left[\frac{-\cos 0}{2}\right]=0-\left(\frac{-1}{2}\right)=0+\frac{1}{2}=\frac{1}{2}\) [∵ \(\cos \frac{\pi}{2}\) = 0]

Question 5.
Evaluate \(\int_0^\pi 2 \cos 2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{2}} \cos 2 x d x=\left[\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}=\frac{\sin \pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}=0\) [∵ sin π = sin 180° = 0]

Question 6.
Evaluate \(\int_4^5 e^x d x\)
Solution:
Let I = \(\int_4^5 e^x d x=\left[e^x\right]_4^5\) = e⁵ – e⁴ = e⁴(e – 1)

Question 7.
Evaluate \(\int_0^\pi 4 \tan x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \tan x d x=[-\log \cos x]_0^{\pi / 4}=-\log \left|\cos \frac{\pi}{4}\right|+\log |\cos 0|\)
= \(-\log \left|\frac{1}{\sqrt{2}}\right|+\log |1|=-\log (2)^{\frac{-1}{2}}=\frac{1}{2} \log 2\)

Question 8.
Evaluate \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x\)
Solution:
I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x=[\log |{cosec} x-\cot x|]_{\pi / 6}^{\pi / 4}=\log \left|{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|{cosec} \frac{\pi}{6}-\cot \frac{\pi}{6}\right|\)
= \(\log |\sqrt{2}-1|-\log |2-\sqrt{3}|=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Question 9.
Evaluate \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1=\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{\pi}{2}-0=\frac{\pi}{2}\)

Question 10.
Evaluate \(\int_0^1 \frac{d x}{1+x^2}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1=\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}\)

Question 11.
Evaluate \(\int_2^3 \frac{d x}{x^2-1}\)
Solution:
Let I = \(\int_2^3 \frac{\mathrm{dx}}{\mathrm{x}^2-1}=\left[\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|\right]_2^3\)
= \(\frac{1}{2}\left[\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|\right]=\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]=\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{3}\right]=\frac{1}{2}\left[\log \frac{3}{2}\right]\)

Question 12.
Evaluate \(\int_2^3 \frac{x d x}{x^2+1}\)
Solution:
Let I = \(\int_2^3 \frac{x}{x^2+1} d x=\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x=\frac{1}{2}\left[\log \left(1+x^2\right)\right]_2^3\)
= \(\frac{1}{2}\left[\log \left(1+3^2\right)-\log \left(1+2^2\right)\right]=\frac{1}{2}[\log (10)-\log (5)]=\frac{1}{2} \log \left(\frac{10}{5}\right)=\frac{1}{2} \log 2\)

Question 13.
Evaluate \(\int_0^1 x e^{x^2} d x\)
Solution:
Let I = \(\int_0^1 x e^{x^2} d x\) put x² = t ⇒ 2x dx = dt
As x → 0, t → 0 and as x → 1, t → 1
∴ I = \(\frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1=\frac{1}{2} e-\frac{1}{2} e^0=\frac{1}{2}(e-1)\)

Question 14.
Evaluate \(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x\)
Solution:
\(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x=\left[2 \tan x+\frac{x^4}{4}+2 x\right]_0^{\pi / 4}\)
= \(\left[\left(2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^4+2\left(\frac{\pi}{4}\right)\right)-(2 \tan 0+0+0)\right]=2+\frac{\pi^4}{4^5}+\frac{\pi}{2}=2+\frac{\pi}{2}+\frac{\pi^4}{1024}\)

Question 15.
Evaluate \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x\)
Solution:
Let I = \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x=-\int_0^\pi\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right) d x\)
= \(-\int_0^\pi \cos x d x=-[\sin x]_0^\pi\) = -(sin π – sin 0) = 0 – 0 = 0

Question 16.
Evaluate \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x=\int_0^{\pi / 2}\left(\frac{1+\cos 2 x}{2}\right) d x=\left[\frac{x}{2}+\frac{\sin 2 x}{4}\right]_0^{\pi / 2}=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^{\pi / 2}\)
= \(\frac{1}{2}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(0+\frac{\sin 0}{2}\right)\right]=\frac{1}{2}\left[\frac{\pi}{2}+0-0-0\right]=\frac{\pi}{4}\)

Question 17.
Find the integral of \(\int_0^1 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{x}}}\)
Solution:

Question 18.
Evaluate \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
I = \(\int_0^{\frac{\pi}{2}} \sin ^2 x \cdot \sin x d x=\int_0^{\frac{\pi}{2}}\left(1-\cos ^2 x\right) \sin x d x=\int_0^{\frac{\pi}{2}} \sin x d x-\int_0^{\frac{\pi}{2}} \cos ^2 x \cdot \sin x d x\)
= \([-\cos x]_0^{\frac{\pi}{2}}+\left[\frac{\cos ^3 x}{3}\right]_0^{\frac{\pi}{2}}=\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[\cos ^3 \frac{\pi}{2}-\cos ^3 0\right]=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}\)

II.

Question 1.
Evaluate \(\int_0^1 \frac{2 x+3}{5 x^2+1} d x\)
Solution:

Question 2.
Evaluate \(\int_0^2 \frac{6 x+3}{x^2+4} d x\)
Solution:

Question 3.
Evaluate \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\)
Solution:
Let I = \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
Dividing the Nr. & Dr. By x² + 4x + 3 and simplifying, we get
Dividing 5x² by x² + 4x + 3, we get
I = \(\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right] d x=\int_1^2 5 d x-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x=[5 x]_1^2-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x\) …(1)
Let I₁ = \(\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x\) …………(1)
Let 20x + 15 = A\(\frac{d}{d x}\)(x² + 4x + 3) + B = 2Ax + (4A + B)
Equating the coefficients of x and constant term, we get
A = 10 and B = -25
∴ 20x + 15 = 10(2x + 4) – 25 [∵ \(\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\)]
⇒ I₁ = \(10 \int_1^2 \frac{2 x+4}{x^2+4 x+3} d x-25 \int_1^2 \frac{d x}{(x+2)^2-1^2}=10\left[\log \left|x^2+4 x+3\right|\right]_1^2-25\left[\frac{1}{2} \log \left(\frac{x+2-1}{x+2+1}\right)\right]_1^2\)
= \(\left[10 \log \left(x^2+4 x+3\right)\right]_1^2-25\left[\frac{1}{2} \log \left(\frac{x+1}{x+3}\right)\right]_1^2\)
= [10log15 – 10log8] – 25 \(\left[\frac{1}{2} \log \frac{3}{5}-\frac{1}{2} \log \frac{2}{4}\right]\)
= [10 log(5 × 3) – 10 log(4 × 2)] – \(\frac{25}{2}\)[log 3 – log 5 – log 2 + log 4]
= [10 log5 + 10log3 – 10log4 – 10log2] – \(\frac{25}{2}\)[log3 – log5 – log2 + log4]
= \(\left[10+\frac{25}{2}\right] \log 5+\left[-10-\frac{25}{2}\right] \log 4+\left[10-\frac{25}{2}\right] \log 3+\left[-10+\frac{25}{2}\right] \log 2\)
= \(\frac{45}{2} \log 5-\frac{45}{2} \log 4-\frac{5}{2} \log 3+\frac{5}{2} \log 2=\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\)
Substituting the value I₁, in (1), we get
I = \(5-\left[\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\right]=5-\frac{5}{2}\left[9 \log \frac{5}{4}-\log \frac{3}{2}\right]\)

Question 4.
Evaluate \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right] d x\)
Solution:
Let I = \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right] d x\)

Question 5.
Evaluate \(\int_1^3 \frac{d x}{x^2(x+1)}\)
Solution:
Let, \(\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \Rightarrow \frac{A x(x+1)+B(x+1)+C\left(x^2\right)}{x\left(x^2\right)(x+1)}\)
⇒ 1 = Ax(x + 1) + B(x + 1) + C(x²) ⇒ 1 = Ax² + Ax + Bx + B + Cx²
Equatingthe coefficients of x², x and constant terms, we get A + C = 0, A + B = 0, B = 1
On solving these equations, we get A = -1, C = 1, B = 1
∴ \(\frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\)
⇒ I = \(\int_1^3\left[-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\right] d x=\left[-\log x-\frac{1}{x}+\log (x+1)\right]_1^3\)
= \(\left[\log \left(\frac{x+1}{x}\right)-\frac{1}{x}\right]_1^3=\log \left(\frac{4}{3}\right)-\frac{1}{3}-\log \left(\frac{2}{1}\right)+1\)
= log 4 – log 3 – log 2 + \(\frac{2}{3}\) = log 2 – log 3 + \(\frac{2}{3}\) = \(\log \left(\frac{2}{3}\right)+\frac{2}{3}\), Hence proved.

Question 6.
Evaluate \(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}} d x\)
Solution:
Put x^1/4 = t ⇒ x = t⁴ ⇒ dx = 4t³ dt.
Also, when x = 0 we get t = 0 and when x = 1 we get t = 1^1/4 = 1

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5d

Question 1.
Differentiate \(\frac{e^x}{\sin x}\) w.r.t. x
Solution:

Question 2.
Differentiate e^sin-1x w.r.t. x
Solution:
Let y = e^sin-1x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^sin-1x \(\frac{\mathrm{d}}{\mathrm{dx}}\)sin^-1x [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= e^sin-1x \(\frac{1}{\sqrt{1-x^2}}\), x ∈ (-1, 1)

Question 3.
Differentiate e^x3 w.r.t. x.
Solution:
Let y = e^x3 = e^(x3)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^(x3) \(\frac{\mathrm{d}}{\mathrm{dx}}\)x³ [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= e^(x3)3x² = 3x²e^(x3)

Question 4.
Differentiate sin(tan^-1e^-x) w.r.t. x:
Solution:
Let y = sin(tan^-1e^-x)

Question 5.
Differentiate log(cos e^x) w.r.t. x.
Solution:
Let y = log(cos e^x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{\cos e^x} \frac{d}{d x}\) (cos e^x)
= \(\frac{1}{\cos \mathrm{e}^{\mathrm{x}}}\)(-sin e^x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^x
= -(tan e^x)e^x = -e^x (tan e^x)

Question 6.
Differentiate e^x + e^x2+ ……….. + e^x5 w.r.t. x.
Solution:
Let y = e^x + e^x2+ e^x3 + e^x4 + e^x5
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x2+ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^x3 + \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^x4 + ……….. + \(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x5
= e^x + e^x2\(\frac{\mathrm{d}}{\mathrm{dx}}\) x² + e^x3 \(\frac{\mathrm{d}}{\mathrm{dx}}\) x³ + e^x4\(\frac{\mathrm{d}}{\mathrm{dx}}\) x⁴ + ……….. + e^x5 \(\frac{\mathrm{d}}{\mathrm{dx}}\) x⁵ [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= e^x + e^x2 2x + e^x3 3x² + e^x4 4x³ + e^x5 5x⁴
= e^x + 2x e^x2+ 3x² e^x3 + 4x³ e^x4 + 5x⁴ e^x5

Question 7.
Differentiate \(\sqrt{e^{\sqrt{x}}}\), x > 0 w.r.t. x.
Solution:
Let y = \(\sqrt{e^{\sqrt{x}}}\) = \(\left(e^{\sqrt{x}}\right)^{1 / 2}\)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{2}\left(e^{\sqrt{x}}\right)^{-1 / 2} \frac{d}{d x} e^{\sqrt{x}}\)
= \(\frac{1}{2 \sqrt{e^{\sqrt{x}}}} e^{\sqrt{x}} \frac{d}{d x} \sqrt{x}\) = \(\frac{1}{2 \sqrt{e^{\sqrt{x}}}} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}\)
= \(\frac{e^{\sqrt{x}}}{4 \sqrt{x} \sqrt{e^{\sqrt{x}}}}\)=\(\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}\)

Question 8.
Differentiate log (log x), x > 1 w.r.t. x
Solution:
Let y = log (log x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{\log x} \frac{d}{d x}(\log x)\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) logf(x) = \(\frac{1}{f(x)} \frac{d}{d x}\) f(x)]
= \(\frac{1}{\log x} \frac{d}{d x}(\log x)\)
= \(\frac{1}{x \log x}\)

Question 9.
Differentiate \(\frac{\cos x}{\log x}\), x > 0 w.r.t. x
Solution:

Question 10.
Differentiate cos(logx + e^x), x > 0 w.r.t x.
Solution:
Let y = cos(logx + e^x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin(log x + e^x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(log x + e^x)
= -sin(log x + e^x) . (\(\frac{1}{x}\) + e^x) = -(\(\frac{1}{x}\) + e^x)sin(log x + e^x)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7g Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7g

I.

Question 1.
Find the integral of \(\sqrt{4-x^2}\)
Solution:
Let I = \(\int \sqrt{4-x^2} d x=\int \sqrt{(2)^2-(x)^2} d x\) [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]
∴ I = \(\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}+C=\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1} \frac{x}{2}+C\)

Question 2.
Find the integral of \(\sqrt{1-4 x^2}\)
Solution:
Let I = \(\int \sqrt{1-4 \mathrm{x}^2} \mathrm{dx}=\int \sqrt{1^2-(2 \mathrm{x})^2} \mathrm{dx}\) dx Put 2x = t ⇒ 2 dx = dt
∴ I = \(\frac{1}{2} \int \sqrt{1^2-t^2}=\frac{1}{2}\left[\frac{t}{2} \sqrt{1-t^2}+\frac{1}{2} \sin ^{-1} t\right]+C=\frac{t}{4} \sqrt{1-t^2}+\frac{1}{4} \sin ^{-1} t+C\)
= \(\frac{2 x}{4} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+C=\frac{x}{2} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+C\)

Question 3.
Find the integral of \(\sqrt{x^2+4 x+6}\)
Solution:

Question 4.
Find the integral of \(\sqrt{x^2+4 x+1}\)
Solution:
Let I = \(\int \sqrt{x^2+4 x+1} d x=\int \sqrt{\left(x^2+4 x+4\right)-3} d x\)
= \(\int \sqrt{(x+2)^2-(\sqrt{3})^2} d x\) [∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C\)]
∴ I = \(\frac{(x+2)}{2} \sqrt{x^2+4 x+1}-\frac{3}{2} \log \left|(x-2)+\sqrt{x^2+4 x+1}\right|+C\)

Question 5.
Find the integral of \(\sqrt{1-4 x-x^2}\)
Solution:
Let I = \(\int \sqrt{1-4 x-x^2} d x=\int \sqrt{1-\left(x^2+4 x+4-4\right)} d x=\int \sqrt{1+4-(x+2)^2} d x\)
= \(\int \sqrt{(\sqrt{5})^2-(x+2)^2} d x\) [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]
∴ I = \(\frac{(x+2)}{2} \sqrt{1-4 x-x^2}+\frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C\)

Question 6.
Find the integral of \(\sqrt{x^2+4 x-5}\)
Solution:
Let I = \(\int \sqrt{x^2+4 x-5} d x\)
= \(\int \sqrt{\left(x^2+4 x+4\right)-9} d x=\int \sqrt{(x+2)^2-3^2} d x\)
∴ I = \(\frac{(x+2)}{2} \sqrt{x^2+4 x-5}-\frac{9}{2} \log \left|(x+2)+\sqrt{x^2+4 x-5}\right|+C\)

Question 7.
Find the integral of \(\sqrt{1+3 x-x^2}\)
Solution:

Question 8.
Find the integral of \(\sqrt{x^2+3 x}\)
Solution:

Question 9.
Find the integral of \(\sqrt{1+\frac{x^2}{9}}\)
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5c

I.

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of 2x + 3y = sin x
Solution:
Given that 2x + 3y = sin x.
Differentiating both sides w.r.t x, we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (3y) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x
∴ 2 + 3 \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\cos x-2}{3}\)
⇒ 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x – 2

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of 2x + 3y = sin y
Solution:
Given that 2x + 3y = sin y ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (3y)
= \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin y
∴ 2 + 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) (cos y – ) = 2
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(=\frac{2}{\cos y-3}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of ax + by² = cos y
Solution:
Given that ax + by² = cos y ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (ax) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (by²) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos y)
∴ a + b.2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin y\(\frac{\mathrm{d}}{\mathrm{dx}}\)
⇒ 2by \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + sin y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -a
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) (2by + sin y) = -a ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-a}{2 b y+\sin y}\)

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of xy + y² = tan x + y
Solution:
Given that xy + y² = tan x + y. ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(xy) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y² = \(\frac{\mathrm{d}}{\mathrm{dx}}\)tan x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y
(x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y + y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec²x + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) [∵ From Product Rule]
⇒ x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec² x + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) ⇒ x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec² x – y
⇒ (x + 2y – 1)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec²x – y
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sec ^2 x-y}{x+2 y-1}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x² + xy + y² = 100
Solution:
Given that ² + xy + y² = 100
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)x² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)xy + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y² = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(100)
∴ 2x + (x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y + y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ 2x + x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ (x + 2y)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2x – y
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(2 x+y)}{x+2 y}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x³ + x²y + xy² + y³ = 81
Solution:
Given that x³ + x²y + xy² + y³ = 81
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)x³ + \(\frac{\mathrm{d}}{\mathrm{dx}}\)x²y + \(\frac{\mathrm{d}}{\mathrm{dx}}\)xy² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y³ = 81
∴ 3x² + x² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y.\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²) + x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y² + y² \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x) = 3y²\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ 3x² + x² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y2x + x2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y² . 1 + 3y² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² +2xy + 3y²) = -3x² – 2xy – y²
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\left(3 x^2+2 x y+y^2\right)}{x^2+2 x y+3 y^2}\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of sin²y + cos xy = k
Solution:
Given that sin² y + cos xy = k
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin y)² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)cos xy = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(k)
∴ 2(sin.y)\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin y – sin xy\(\frac{\mathrm{d}}{\mathrm{dx}}\)(xy) = 0
⇒ 2sin y cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – sin xy(x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y.1) = 0
⇒ sin2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – x sin xy\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y sin xy = 0
⇒ (sin2y – x sin xy)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y sin xy
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y \sin x y}{\sin 2 y-x \sin x y}\)

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of sin²x + cos² y = 1
Solution:
Given that sin² x + cos² y = 1
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x)² + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos y)² = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (1)
∴ 2(sin x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x + 2(cos y)\(\frac{\mathrm{d}}{\mathrm{dx}}\) cos y = 0
⇒ 2 sin x cos x + 2cos y(-sin y\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = 0
⇒ 2 sin x cos x – 2 sin y cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ sin 2x – sin 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = o
⇒ sin 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin 2x
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin 2 x}{\sin 2 y}\)

II.

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Solution:
Given that y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
To simplify the given inverse t-function, we put x = tanθ.
∴ y = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(sin 2θ) = 2θ
⇒ y = 2tan^-1x (∵ x tanθ ⇒ θ = tan^-1x)
∴ \(\frac{d y}{d x}=2 \frac{1}{1+x^2}=\frac{2}{1+x^2}\)

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\), \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Solution:
Given y = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\), \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Put x = tanθ
∴ y = tan^-1\(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\) = tan^-1(tan 3θ) = 3θ
⇒ y = 3 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3 . \(\frac{1}{1+x^2}=\frac{3}{1+x^2}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Solution:
Given that y = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Put x = tanθ
∴ y = cos^-1\(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) = cos^-1(cos 2θ)
y = 2θ = 2tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2 . \(\frac{1}{1+x^2}=\frac{2}{1+x^2}\)

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sin^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Solution:
Given that y = sin^-1\(\left(\frac{1-x^2}{1+x^2}\right)\)
Put x = tanθ
∴ y = sin^-1\(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(cos 2θ)
= sin^-1 sin(\(\frac{\pi}{2}\) – 2θ) = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 – 2\(\frac{1}{1+x^2}=\frac{-2}{1+x^2}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = cos^-1\(\left(\frac{2 x}{1+x^2}\right)\), -1 < x < 1
Solution:
Given that y = cos^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Put x = tanθ
∴ y = cos^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = cos^-1(sin 2θ)
= cos^-1 sin(\(\frac{\pi}{2}\) – 2θ) = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 – 2\(\frac{1}{1+x^2}=\frac{-2}{1+x^2}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sin^-1(2x\(\sqrt{1-\mathrm{x}^2}\)), \(-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
Solution:
Given that y = sin^-1(2x\(\sqrt{1-\mathrm{x}^2}\)). Put x = sinθ
∴ y = sin^-1(2 sinθ\(1-\sin ^2 \theta\)) = sin^-1(2sinθ\(\cos ^2 \theta\)) = sin^-1(2 sinθ cosθ)
⇒ y = sin^-1 (sin 2θ) = 2θ = 2 sin^-1 x [∵ x = sin θ ⇒ θ = sin^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sec^-1\(\left(\frac{1}{2 x^2-1}\right),\), 0 < x < \(\frac{1}{\sqrt{2}}\)
Solution:
Put x = cosθ then 2x² – 1 = 2cos²θ – 1 = cos 2θ) = 2θ
∴ y = sec^-1\(\left(\frac{1}{\cos 2 \theta}\right)\) = sec^-1 (sec 2θ) = 2θ = 2 cos^-1
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2cos^-1x)
= 2\(\left(\frac{-1}{\sqrt{1-x^2}}\right)=\frac{-2}{\sqrt{1-x^2}}\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7f Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7f

I.

Question 1.
Find the integral of x sin x
Solution:
Let I = ∫x sin x dx
Taking u = x, v = sinx and integrating by parts, we have
I = \(x \int \sin x d x-\int\left[\left(\frac{d}{d x}(x)\right) \int \sin x d x\right] d x=x(-\cos x)-\int 1 \cdot(-\cos x) d x\) = -x cosx + sinx + C

Question 2.
Find the integral of x sin 3x
Solution:
Let I = ∫x sin 3x dx
Taking u = x, v = sin3x and integrating by parts, we have
I = \(x \int \sin 3 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sin 3 x d x\right] d x\)
= \(x\left(\frac{-\cos 3 x}{3}\right)-\int 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x=\frac{-x \cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x\)
= \(\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x+C=\frac{-x}{3} \cos 3 x+\frac{1}{9} \sin 3 x+C\)

Question 3.
Find the integral of x log x
Solution:
Let I = ∫x log x dx
Taking u = log x, v = sin3x and integrating by parts, we have
I = \(\log x \int x d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x d x\right] d x\)
= \(\log x \cdot \frac{x^2}{2}-\int \frac{1}{x} \cdot \frac{x^2}{2} d x=\frac{x^2 \log x}{2}-\int \frac{x}{2} d x=\frac{x^2 \log x}{2}-\frac{x^2}{4}+C\)

Question 4.
Find the integral of x log 2x
Solution:
Let I = ∫x log 2x dx
Taking u = log 2x, v = sin3x and integrating by parts, we have
I = \(\log 2 x \int x d x-\int\left[\left(\frac{d}{d x} \log 2 x\right) \int x d x\right] d x\)
= \(\log 2 x \cdot \frac{x^2}{2}-\int \frac{2}{2 x} \cdot \frac{x^2}{2} d x=\frac{x^2 \log 2 x}{2}-\int \frac{x}{2} d x=\frac{x^2 \log 2 x}{2}-\frac{x^2}{4}+C\)

Question 5.
Find the integral of x² log x
Solution:
Let I = ∫x² log x dx
Taking u = log x, v = x² and integrating by parts, we have
I = \(\log x \int x^2 d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x^2 d x\right] d x\)
= \(\log x .\left(\frac{x^3}{3}\right)-\int \frac{1}{x} . \frac{x^3}{3} d x=\frac{x^3 \log x}{3}-\int \frac{x^2}{3} d x=\frac{x^3 \log x}{3}-\frac{x^3}{9}+C\)

Question 6.
Find the integral of x sec² x
Solution:
Let I = ∫x sec² x dx
Taking u = x, v = sec²x and integrating by parts, we have
I = \(x \int \sec ^2 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sec ^2 x d x\right] d x\)
= x tan x – ∫1. tan xdx = x tanx + log|cos x| + C

Question 7.
Find the integral of tan^-1 x
Solution:
Let I = ∫1.tan^-1 x dx
Taking u = tan^-1 x, v = 1 and integrating by parts, we have
I = \(\tan ^{-1} x \int 1 d x-\int\left[\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 . d x\right] d x\)
= \(\tan ^{-1} x . x-\int \frac{1}{1+x^2} x d x=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^2} d x\)
= \(x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^2\right|+C=x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^2\right)+C\)

Question 8.
Find the integral of (x² + 1)log x
Solution:
Let I = ∫(x² + 1)log x dx = ∫x² logxdx + ∫logx dx
Let I = I₁ + I₂ ……….(1)
Where, I₁ = ∫x² logxdx and I₁ = ∫logx dx
I₁ = ∫x²log xdx
Taking u = log x, v = x² and integrating by parts, we have
I₁ = \(\log x \int x^2 d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x^2 d x\right] d x=\log x \cdot \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x\)
= \(\frac{x^3}{3} \log x-\frac{1}{3}\left(\int x^2 d x\right)=\frac{x^3}{3} \log x-\frac{x^3}{9}+C_1\) …(2)
I₂ = ∫log xdx
Taking u = log x, v = 1 and integrating by parts, we have
I₂ = \(\log x \int 1 . d x-\int\left[\left(\frac{d}{d x} \log x\right) \int 1 . d x\right]\)
= \(\log x \cdot x-\int \frac{1}{x} \cdot x d x=x \log x-\int 1 \cdot d x=x \log x-x+C_2\) …………(3)
Using equation (2) and (3) in (1), we get

Question 9.
Find the integral of e^x(sinx + cosx)
Solution:
We know that ∫e^x{f(x) + f(x)} dx = e^xf(x) + C
Here f(x) = sin x and f'(x) = cos x
∴ I = ∫e^x(sin x + cos x) dx = e^xsin x + C

Question 10.
Find the integral of e^x\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)
Solution:
Let I = ∫e^x\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)dx
We know that ∫e^x[f(x) + f'(x)]dx = e^xf(x) + C
Here, f(x) = \(\frac{1}{x}\) & f'(x) = \(\frac{-1}{x^2}\)
∴ I = \(e^x\left(\frac{1}{x}\right)+C=\frac{e^x}{x}+C\)

II.

Question 1.
Find the integral of x²e^x
Solution:
Let I = ∫ x²e^x dx
Taking u = x², v = e^x and integrating by parts, we have
I = \(x^2 \int e^x d x-\int\left[\left(\frac{d}{d x} x^2\right) \int e^x d x\right] d x=x^2 e^x-\int 2 x e^x d x=x^2 e^x-2 \int x e^x d x\)
Again using integration by parts, we have
I = \(x^2 e^x-2\left[x \int e^x d x-\int\left(\frac{d}{d x} x\right) \int e^x d x\right] d x\)
= x²e^x – 2[xe^x – e^x dx] = x²e^x – 2[xe^x – ∫e^x]
= x²e^x – 2xe^x + 2e^x + C = e^x(x² – 2x + 2) + C

Question 2.
Find the integral of x sin^-1 x
Solution:
Let I = ∫ xsin^-1x dx
Taking u = sin^-1 x, v = x and integrating by parts we have
I = \(\sin ^{-1} x \int x d x-\int\left[\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right] d x=\sin ^{-1} x\left(\frac{x^2}{2}\right)-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} d x\)
= \(\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} d x\)

Question 3.
Find the integral of x tan^-1 x
Solution:
Let I = ∫ x tan^-1 xdx
Taking u = tan^-1 x, v = x and integrating by parts we have

Question 4.
Find the integral of x cos^-1 x
Solution:
Let I = ∫ x cos^-1 xdx
Taking u = cos^-1 x, v = x and integrating by parts we have

Question 5.
Find the integral of \(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} \cdot \cos ^{-1} x d x\)
Taking u = cos^-1 x, v = \(\left(\frac{-2 x}{\sqrt{1-x^2}}\right)\) and integrating by parts we have

Question 6.
Find the integral of x(log x)²
Solution:
Let I = ∫ x(log x)² dx
Taking u = (log x)², v = x and integrating by parts, we have
I = \((\log x)^2 \int x d x-\int\left[\left(\frac{d}{d x}(\log x)^2\right) \int x d x\right] d x\)
= \(\frac{x^2}{2}(\log x)^2-\left[\int 2 \log x \frac{1}{x} \frac{x^2}{2} d x\right]=\frac{x^2}{2}(\log x)^2-\int x \log x d x\)
Again, using integrated by parts, we have
I = \(\frac{x^2}{2}(\log x)^2-\left[\log x \int x d x-\int\left(\left(\frac{d}{d x} \log x\right) \int x d x\right) d x\right]\)
= \(\frac{x^2}{2}(\log x)^2-\left[\frac{x^2}{2} \log x-\int \frac{1}{x} \cdot \frac{x^2}{2} d x\right]=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2} \log x+\frac{1}{2} \int x d x\)
= \(\frac{x^2}{2}(\log x)^2-\frac{x^2}{2} \log x+\frac{x^2}{4}+C\)

Question 7.
Find the integral of \(\frac{x e^x}{(1+x)^2}\)
Solution:
We know that ∫e^x[f(x) + f'(x)] dx = e^xf(x) + C
Here, f(x) = \(\frac{1}{1+x}\) and f'(x) = \(\frac{-1}{(1+x)^2}\)
∴ I = \(\int \frac{x e^x}{(1+x)^2} d x=\int e^x\left[\frac{x}{(1+x)^2}\right] d x=\int e^x\left[\frac{1+x-1}{(1+x)^2}\right] d x=\int e^x\left[\frac{1}{1+x}-\frac{1}{(1+x)^2}\right] d x\)
∴ I = \(e^x \frac{1}{1+x}+C=\frac{e^x}{1+x}+C\)

Question 8.
Find the integral of \(e^x\left(\frac{1+\sin x}{1+\cos x}\right)\)
Solution:

[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x) + C Here \(\tan \frac{x}{2}\) = f(x) & f'(x) = \(\frac{1}{2}\)sec²\(\frac{x}{2}\)]

Question 9.
Find the integral of \(\frac{(x-3) e^x}{(x-1)^3}\)
Solution:
Let I = \(\int e^x\left[\frac{x-3}{(x-1)^3}\right] d x=\int e^x\left[\frac{x-1-2}{(x-1)^3}\right] d x=\int e^x\left[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\right] d x\)
Here f(x) = \(\frac{1}{(x-1)^2}\) and f'(x) = \(\frac{-2}{(x-1)^3}\)
We know that ∫ e^x[f(x) + f'(x)] dx = e^xf(x) + C ∵ I = \(\frac{\mathrm{e}^{\mathrm{x}}}{(\mathrm{x}-1)^2}\) + C

Question 10.
Find the integral of e^2x sin x
Solution:
Let I = e^2x sin x dx …….(1)
Taking u = sin x, v = e^2x and integrating by parts, we have

Question 11.
Find the integral of \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Solution:
Let x = tan θ ⇒ dx = sec² θdθ
∴ sin^-1\(\left(\frac{2 x}{1+x^2}\right)\) = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(sin 2θ) = 2θ
∫ sin^-1\(\left(\frac{2 x}{1+x^2}\right)\) dx = ∫ 2θ.sec² θdθ = 2∫θ. sec² θdθ
Using integration by parts, we get
I = \(\left[\theta \cdot \int \sec ^2 \theta \mathrm{~d} \theta-\int\left[\left(\left(\frac{\mathrm{d}}{\mathrm{~d} \theta} \theta\right) \int \sec ^2 \theta \mathrm{~d} \theta\right)\right] \mathrm{d} \theta=2\left[\theta \cdot \tan \theta-\int \tan \theta \mathrm{d} \theta\right]\right.\)
= 2[θ. tanθ + log |cos θ|] + C = \(2\left[x \tan ^{-1} x+\log \left|\frac{1}{\sqrt{1+x^2}}\right|\right]+C\)
= 2x tan^-1 x + 2log(1 + x²)^{\(\frac{-1}{2}\)} + C = \(2 x \tan ^{-1} x+2\left[\frac{-1}{2} \log \left(1+x^2\right)\right]+C\)
= 2x tan^-1 x – log(1 + x²) + C

Question 12.
Find the integral of \(\frac{2+\sin 2 x}{1+\cos 2 x} e^x\)
Solution:
I = \(\int\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) e^x=\int\left(\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right) e^x\)
= \(\int\left(\frac{1+\sin x \cos x}{\cos ^2 x}\right) e^x\) = ∫(sec² x + tan x)e^x
Let f(x) = tan x ⇒ f'(x) = sec² x
∴ I = ∫[f(x) + f'(x)] e^x dx = e^xf(x) + C = e^x tan x + C

III.

Question 1.
Find the integral of \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Solution:
Given integral is \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\) dx
Let x = cosθ ⇒ dx = -sin θ dθ

Question 2.
Find the integral of \(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}\)
Solution:
Given integral of \(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}=\frac{\sqrt{x^2+1}}{x^4}\left[\log \left(x^2+1\right)-\log x^2\right]\)

Question 3.
Find the integral of (sin^-1 x)²
Solution:
Let I = ∫(sin^-1 x)² . 1 dx
Taking u = (sin^-1 x)², v = 1 and integrating by parts, we have

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5b

I.

Question 1.
Differentiate the function sin(x² + 5) with respect to x.
Solution:
Let y = sin(x² + 5)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin(x² + 5) = cos(x² + 5) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x² + 5) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin f(x) = cos f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= cos(x² + 5) (2x + 0) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) xⁿ = nx^n-1 and \(\frac{\mathrm{d}}{\mathrm{dx}}\) (c) = 0]
= 2x cos(x² + 5)

Question 2.
Differentiate the function cos(sin x) with respect to x.
Solution:
Let y = cos(sin x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos(sin x) = -sin(sin x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos f(x) = -sin f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= -sin(sin x) cos x = -cos x sin(sin x)

Question 3.
Differentiate the function sin(ax + b) with respect to x.
Solution:
Let y = sin(ax + b)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)sin(ax + b) = cos(ax + b) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(ax + b)
= cos(ax + b)[a\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)(b)]
= cos(ax + b)[a(1) + 0] = a cos(ax + b)

Question 4.
Differentiate the function sec(tan(\(\sqrt{x}\))) with respect to x.
Solution:
Let y = sec(tan(\(\sqrt{x}\)))

Question 5.
Differentiate the function \(\frac{\sin (a x+b)}{\cos (c x+d)}\) with respect to x.
Solution:

Question 6.
Differentiate the function cosx³ . sin²(x⁵) with respect to x.
Solution:
Let y = cosx³ . sin²(x⁵) = cos x³ (sin x⁵)² [∵ sin²f(x) = [sin f(x)]²]
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x³ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x⁵)² + (sin x⁵)² \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos x³ [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uv) = u\(\frac{\mathrm{dv}}{\mathrm{dx}}\) + v\(\frac{\mathrm{du}}{\mathrm{dx}}\)]
= cos x³ 2(sin x⁵)\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin x⁵ +(sin x⁵)² (-sin x³ )\(\frac{\mathrm{d}}{\mathrm{dx}}\)x³
= cos x³2(sin x⁵)cos x⁵ (5x⁴ + (sin x⁵)²(-sin x³)(3x²) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x⁵ = cos x⁵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) x⁵ = cos x⁵(5x⁴)]
= 10x⁴ cos x³ sin x⁵ cos x⁵ – 3x² sin² x⁵ sin x³
= x² sin x⁵[10x² cos x³ cos x⁵ – 3 sin x⁵ sin x³].

Question 7.
Differentiate the function \(2 \sqrt{\cot \left(x^2\right)}\) with respect to x.
Solution:

Question 8.
Differentiate the function cos(\(\sqrt{x}\)) with respect to x.
Solution:
Let y = cos(\(\sqrt{x}\))
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos(\(\sqrt{x}\))
= -sin \(\sqrt{x}\) \(\frac{\mathrm{d}}{\mathrm{dx}}\) \(\sqrt{x}\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos f(x) = -sin f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) f(x)]
= -sin \(\sqrt{x} \frac{1}{2 \sqrt{x}}\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{x}=\frac{1}{2 \sqrt{x}}\)

II.

Question 1.
Prove that the function f given by f(x) = |x – 1|, x ∈ R k is not differentiable at x = 1.
Solution:
Given f(x) = |x – 1| , x ∈ R ……….. (i)
To prove: f(x) is not differentiable at x = 1
Putting x = 1 in (i), f'(1) = |1 – 1| = |0| = 0

Here, L.H.L of f'(1) ≠ R.H.L of f'(1)
∴ f(x) is not differentiable at x = 1

Question 2.
Prove (hat the greatest nleger function defined by f(x) = |x|, 0 < x < 3 is not differentiable at x = 1 and x = 2.
Solution:
Given f(x) = [x], 0 < x < 3 ……….(i)
(a) Differentiability at x = 1
Putting x = 1 in (i), f(1) = [1] = 1 .
Left Hand derivative of f'(1) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) \(\frac{f(x)-f(1)}{x-1}\) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\)\(\frac{[x]-1}{x-1}\)
Put x = 1 – h, h → 0+
\(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{[1-\mathrm{h}]-1}{1-\mathrm{h}-1}\) = \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{0-1}{-h}\)
= \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{1}{h}\)
We know that as h → 0+, [c – h] = c – 1 if c is an integer
∴ [1 – h] = 1 – 1 = 0
Put h = 0, \(\frac{1}{h}\) = \(\frac{1}{0}\) = ∞ does not exist
∴ f(x) is not differentiable at x = 1
(We need not find R f'(1) as L f'(1) does not exist).

(b) Differentiability at x = 2
Putting x = 2 in (i), f(2) = [2] = 2 .
Left Hand derivative of f'(2) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) \(\frac{f(x)-f(2)}{x-2}\) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\)\(\frac{[x]-2}{x-2}\)
Put x = 2 – h, h → 0+
\(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{[2-\mathrm{h}]-2}{2-\mathrm{h}-2}\) = \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{1-2}{-h}\)
= \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{-1}{-h}\)
(For h → 0+, [2 – h] = 2 – 1 = 1
\(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{1}{h}\) = \(\frac{1}{0}\) = ∞ does not exist
∴ f(x) is not differentiable at x = 2
Note. For h → 0+, [c + h] = c if c is an integer

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7e Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7e

I.

Question 1.
Find the integral of \(\frac{x}{(x+1)(x+2)}\)
Solution:
Let \(\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}=\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}+2)}=\frac{\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)}{(\mathrm{x}+1)(\mathrm{x}+2)}\)
⇒ A(x + 2) + B(x + 1) = x …………(1)
Put x = -1 in (1) ⇒ A(1) + 0 = -1 ⇒ A = -1
Put x = -2 in (1) ⇒ 0 + B(-1) = -2 ⇒ B = 2
\(\frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}\)
⇒ \(\int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x\) = -log |x + 1| + 2 log|x + 2| + C
= log(x + 2)² – log(x + 1) + C = \(\log \frac{(x+2)^2}{|x+1|}+C\) [∵ loga – logb = \({log}\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)\)]

Question 2.
Find the integral of \(\frac{1}{x^2-9}\)
Solution:
Let \(\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}=\frac{A(x-3)+B(x+3)}{(x+3)(x-3)}\)
⇒ A(x – 3) + B(x + 3) = 1 …………..(1)
Put x = 3 in (1) ⇒ 1 = A(3 – 3) + B(3 + 3) ⇒ 6B = 1 ⇒ B = 1/6
Put x = -3 in (1) ⇒ 1 = A(-3 – 3) + B(-3 + 3) ⇒ -6A = 1 ⇒ A = -1/6
∴ \(\frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\)
⇒ \(\int \frac{1}{\left(x^2-9\right)} d x=\int\left(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\right) d x\)
= \(-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+C=\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+C\)

Question 3.
Find the integral of \(\frac{1-x^2}{x(1-2 x)}\)
Solution:
The given integral is an improper rational function.
So by dividing the numerator 1 – x² by the denominator x(1 – 2x) = x – 2x² we get
\(\frac{1-x^2}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)\)
Now let \(\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{1-2 x}=\frac{A(1-2 x)+B x}{(x)(1-2 x)}\)
A(1 – 2x) + Bx = (2 – x) ……..(1)
Pit x = 0 in (1) ⇒ A(1 – 0) + B(0) = 2 – 0 ⇒ A = 2
Put x = 1/2 in (1) ⇒ A(0) + B(1/2) = 1 – 0 ⇒ 2 – \(\frac{1}{2}=\frac{3}{2}\)
∴ \(\frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{1-2 x}\)
∴ \(\frac{1-x^2}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left[\frac{2}{x}+\frac{3}{1-2 x}\right] \Rightarrow \int \frac{1-x^2}{x(1-2 x)} d x=\int\left[\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{1-2 x}\right)\right] d x\)
= \(\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C\) [∵ \(\int \frac{1}{a x+b} d x=\frac{1}{a} \log |a x+b|+C\)]
= \(\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C\)

II.

Question 1.
Find the integral of \(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
Solution:
Let \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\)
= \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
⇒ A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2) = 3x – 1 …..(1)
Put x = 1 in (1) ⇒ A(-1)(-3) + 0 + 0 = 3 – 1 ⇒ 2A = 2 ⇒ A = 1
Put x = 2 in (1) ⇒ 0 + B1(-1) + 0 = 6 – 1 ⇒ -B = 5 ⇒ B = -5
Put x = 3 in (1) ⇒ 0 + 0 + C(2)1 = 9 – 1 ⇒ 2C = 8 ⇒ C = 4
∴ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\)
⇒ \(\int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int\left[\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\right] d x\)
= log |x – 1| – 5log |x – 2| + 4log |x – 3| + C

Question 2.
Find the integral of \(\frac{x}{(x-1)(x-2)(x-3)}\)
Solution:
Let \(\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\)
= \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x -2) = x ……..(1)
Put x = 1 in (1) ⇒ A(-1)(-2) + 0 + 0 = 1 ⇒ 2A = 1 ⇒ A = 1/2
Put x = 2 in (1) ⇒ 0 + B1(-1) + 0 = 2 ⇒ -B = 2 ⇒ B = -2
Put x = 3 in (1) ⇒ 0 + 0 + C(2)1 = 3 ⇒ 2C = 3 ⇒ C = 3/2
∴ \(\frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\)
⇒ \(\begin{aligned}
\int \frac{x}{(x-1)(x-2)(x-3)} d x=\int & {\left[\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\right] d x } \\
& =\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+C
\end{aligned}\)

Question 3.
Find the integral of \(\frac{2 x}{x^2+3 x+2}\)
Solution:
Let \(\frac{2 x}{x^2+3 x+2}=\frac{2 x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\)
⇒ A(x + 2) + B(x + 1) = 2x …….(1)
Put x = -1 ⇒ A(-1 + 2) + B(0)2(-1) ⇒ A = -2
Put x = -2 ⇒ A(-2 + 2) + B(-2 + 1) = 2(-2) ⇒ B = -4
∴ \(\frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)} \Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int\left[\frac{4}{(x+2)}-\frac{2}{(x+1)}\right] d x\)
= 4log |x + 2| – 2log |x + 1| + C

Question 4.
Find the integral of \(\frac{x}{\left(x^2+1\right)(x-1)}\)
Solution:
Let \(\frac{x}{\left(x^2+1\right)(x-1)}=\frac{A x+B}{\left(x^2+1\right)}+\frac{C}{(x-1)}=\frac{(A x+B)(x-1)+C\left(x^2+1\right)}{\left(x^2+1\right)(x-1)}\)
⇒ (Ax + B)(x – 1) + C(x² + 1) = x ………….(1)
Put x = 1 in (1) ⇒ 0 +C(1 + 1) = 1 ⇒ 2C = 2 ⇒ C = 1/2
Put x = 0 in (1) ⇒ (0 + B)( 0 – 1) + C(0 + 1) = 0 ⇒ -B + C = 0 ⇒ B = C = 1/2
Equating the coefficients of x²
⇒ A + C = 0 ⇒ A = -C = -1/2 A = \(-\frac{1}{2}\), B = \(\frac{1}{2}\) and C = \(\frac{1}{2}\)

Question 5.
Find the integral of \(\frac{2}{(1-x)\left(1+x^2\right)}\)
Solution:
Let \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{A}{(1-x)}+\frac{B x+C}{\left(1+x^2\right)}=\frac{A\left(1+x^2\right)+(B x+C)(1-x)}{(1-x)\left(1+x^2\right)}\)
⇒ A(1 + x²) + (Bx + C)(1 – x) = 2 ……..(1)
Put x = 1 in (1) ⇒ A(1 + 1) + 0 = 2 ⇒ 2A = 2 ⇒ A = 1
Equating the coefficients of x² we get A – B = 0 ⇒ B = A = 1
Equating the coefficients of x we get B – C = 0 ⇒ C = A = 1
∴ \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}\) [\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]
⇒ \(\int \frac{2}{(1-x)\left(1+x^2\right)} d x=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^2} d x+\int \frac{1}{1+x^2} d x\)
= \(-\int \frac{-1}{x-1} d x+\frac{1}{2} \int \frac{2 x}{1+x^2} d x+\int \frac{1}{1+x^2} d x=-\log |x-1|+\frac{1}{2} \log \left(1+x^2\right)+\tan ^{-1} x+C\)

Question 6.
Find the integral of \(\frac{3 x-1}{(x+2)^2}\)
Solution:
Let \(\frac{3 x-1}{(x+2)^2}\)
Consider \(\frac{3 x-1}{(x+2)^2}\)
Put x + 2 = y ⇒ x = y – 2
\(\begin{aligned}
\frac{3 x-1}{(x+2)^2} & =\frac{3(y-2)-1}{(y)^2} \\
& =\frac{3 y-6-1}{y^2}=\frac{3 y-7}{y^2} \\
& =\frac{3}{y}-\frac{7}{y^2}
\end{aligned}\)
⇒ A(x + 2) + B = 3x – 1 …………….(1)
Put x = -2 in (1) ⇒ A(0) + B = 3(-2) – 1 ⇒ B = 7
Comparing the coefficient of x we get A = 3

Question 7.
Find the integral of \(\frac{1}{x\left(x^n+1\right)}\)
[Hint: Multiply numerator and denominator by x^n-1 and put xⁿ = t]
Solution:
Multiplying numerator and denominator of the integral by x^n-1, we get
\(\frac{1}{x\left(x^n+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^n+1\right)}=\frac{x^{n-1}}{x^n\left(x^n+1\right)}\)
Put xⁿ = t ⇒ nx^n-1dx = dt
∴ \(\int \frac{1}{x\left(x^n+1\right)} d x=\int \frac{x^{n-1}}{x^n\left(x^n+1\right)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t\)
Let \(\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}=\frac{A(1+t)+B t}{t(t+1)}\)
A(1 + t) + Bt = 1 …………(1)
Put t = 0 in (1) ⇒ A(1 + 0) + 0 = 1 ⇒ A = 1
Put t = -1 in (1) ⇒ 0 – B = 1 ⇒ B = -1

Question 8.
Find the integral of \(\frac{1}{\left(e^x-1\right)}\) [Hint: Put e^x = t]
Solution:
Put e^x = t ⇒ e^x dx = dt
∴ \(\int \frac{1}{\left(e^x-1\right)} d x=\int\left(\frac{1}{t-1}\right) \frac{d t}{t}=\int \frac{1}{t(t-1)} d t\)
Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}=\frac{A(t-1)+B t}{t(t-1)}\) ⇒ A(t – 1) + Bt = 1 ………..(1)
Put t = 0 in (1) ⇒ A(0 – 1) + 0 = 1 ⇒ A = -1
Put t = 1 in (1) ⇒ 0 – B = 1 ⇒ B = 1
∴ \(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\) [∵ log a – log b = \(\log \left(\frac{a}{b}\right)\)]
⇒ \(\int \frac{1}{t(t-1)} d t=-\log t+\log |t-1|=\log \left|\frac{t-1}{t}\right|+C=\log \left(\frac{e^x-1}{e^x}\right)+C\)

Question 9.
Find the integral of \(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\)
Solution:
Put e^x = t ⇒ e^x dx = dt
∴ \(\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x=\int \frac{d t}{(t+1)(t+2)}=\int \frac{(t+2)-(t+1)}{(t+1)(t+2)} d t=\int\left[\frac{1}{(t+1)}-\frac{1}{(t+2)}\right] d t\)
= log |t + 1| – log |t + 2| + C = \(\log \left|\frac{t+1}{t+2}\right|+C=\log \left|\frac{1+e^x}{2+e^x}\right|+C\)

Question 10.
Find the integral of \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\)
Solution:
Put x² = t.
Then \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{(t+1)(t+4)}=\frac{A}{(t+1)}+\frac{B}{(t+4)}=\frac{A(t+4)+B(t+1)}{(t+1)(t+4)}\)
⇒ A(t + 4) + B(t + 1) = 1 ………(1)
Put t = -1 in (1) ⇒ A(-1 + 4) + B(0) = 1 ⇒ 3A = 1 ⇒ 3A = 1/3
Put t = -4 in (1) ⇒ A(0) + B(-4 + 1) = 1 ⇒ -3B = 1 ⇒ B = -1/3

III.

Question 1.
Find the integral of \(\frac{x}{(x-1)^2(x+2)}\)
Solution:
Let \(\frac{x}{(x-1)^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}=\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}\)
A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² = x …….(1)
Put x = 1 in (1) ⇒ 0 + 3B + 0 = 1 ⇒ 3B = 1 ⇒ B = 1/3
Put x = -2 in (1) ⇒ 0 + 0 + C(-2 – 1)² = -2 ⇒ 9C = -2 ⇒ C = -2/9
Equating the coefficient of x² we get
A + C = 0 ⇒ A = -C = 2/9 A = \(\frac{2}{9}\), B = \(\frac{1}{3}\) and C = \(-\frac{2}{9}\)

Question 2.
Find the integral of \(\frac{3 x+5}{x^3-x^2-x+1}\)
Solution:
We have \(\frac{3 x+5}{x^3-x^2-x+1}=\frac{3 x+5}{(x-1)^2(x+1)}\)
Let \(\frac{3 x+5}{(x-1)^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}=\frac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+2)}\)
⇒ A(x – 1)(x + 1) + B(x + 1) + C(x – 1)² = 3x + 5 ….(1)
Put x = 1 in (1) ⇒ 0 + 2B + 0 = 3 + 5 ⇒ B = 4
Put x = -1 in (1) ⇒ 0 + 0 + 4C = -3 + 5 ⇒ 4C = 2 ⇒ C = 1/2
Equating the coefficient of x² we get
A + C = 0 ⇒ A = -C = -1/2 ∴ A = \(-\frac{1}{2}\), B = 4 and C = \(\frac{1}{9}\)

Question 3.
Find the integral of \(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\)
Solution:
We have \(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}\)
Let \(\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2 x+3)}\)
= \(\frac{A(x-1)(2 x-3)+B(x+1)(2 x+3)+C(x+1)(x-1)}{(x+1)(x-1)(2 x-3)}\)
⇒ A(x – 1)(2x – 3) + B(x + 1)(2x + 3) + C(x + 1)(x – 1) = 2x – 3 …..(1)
Put x = -1 in (1) ⇒ A(-1 – 1)(-2 + 3) + 0 + 0 = -2 – 3 ⇒ -2A = -5 ⇒ A = 5/2
Put x = 1 in (1) ⇒ 0 + B(2)(5) + 0 = 2 – 3 ⇒ 10B = -1 ⇒ B = -1/10
Put x = -3/2 in (1) ⇒ 0 + 0 + C\(\left(\frac{-3}{2}+1\right)\left(\frac{-3}{2}-1\right)=2\left(-\frac{3}{2}\right)-3 \Rightarrow-6=C\left(\frac{-1}{2}\right)\left(\frac{-5}{2}\right)\)
⇒ 5C = -24 ⇒ C = \(-\frac{24}{5}\)
∴ A = \(\frac{5}{2}\), B = \(-\frac{1}{10}\) and C = \(-\frac{24}{5}\)

Question 4.
Find the integral of \(\frac{5 x}{(x+1)\left(x^2-4\right)}\)
Solution:
We have \(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{5 x}{(x+1)(x+2)(x-2)}\)
Let \(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\)
= \(\frac{A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)}{(x+1)(x+2)(x-2)}\)
A(x + 2)(x – 2) + B(x + 1)(x – 2) + C(x+ 1)(x + 2) = 5x ……..(1)
Put x = -1 in (1) ⇒ A(-1 + 2)(-1 – 2) + 0 + 0 = -5 ⇒ -3A = -5 ⇒ A = 5/3
Put x = -2 in (1) ⇒ 0 + B(-2 + 1)(-2 – 2)(5) + 0 = -10 ⇒ 4B = -10 ⇒ B = -5/2
Put x = 2 in (1) ⇒ 0 + 0 + C(2 + 1)(2 + 2) = 10 ⇒ 12C = 10 ⇒ C = 5/6
A = \(\frac{5}{3}\), B = \(-\frac{5}{2}\) and C = \(\frac{5}{6}\)

Question 5.
Find the integral of \(\frac{x^3+x+1}{x^2-1}\)
Solution:
The given integral is an improper rational function
On dividing (x³ + x + 1) by x² – 1, we get \(\frac{x^3+x+1}{x^2-1}=x+\frac{2 x+1}{x^2-1}\)
Let \(\frac{2 x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}=\frac{A(x-1)+B(x+1)}{x^2-1}\)
⇒ A(x – 1) + B(x + 1) = 2x + 1 ………..(1)
Put x = -1 in (1) ⇒ A(-1 – 1) + 0 = -2 + 1 ⇒ -2A = -1 ⇒ A = 1/2
Put x = 1 in (1) ⇒ 0 + B(1 + 1) = 2 + 1 ⇒ 2B = 3 ⇒ B = 3/2

Question 6.
Find the integral of \(\frac{1}{x^4-1}\)
Solution:
GE = \(\frac{1}{\left(x^4-1\right)}=\frac{1}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{1}{(x+1)(x-1)\left(x^2+1\right)}\)
Let \(\frac{1}{(x+1)(x-1)\left(x^2+1\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^2+1\right)}\)
= \(\frac{A(x-1)\left(1+x^2\right)+B(x+1)\left(1+x^2\right)+(C x+D)\left(x^2-1\right)}{(x+1)(x-1)\left(x^2+1\right)}\)
A(x – 1)(1 + x²)
⇒ A(x³ + x – x² – 1) + B(x³ + x + x² + 1) + Cx³ + Dx² – Cx – D = 1
⇒ (A + B + C)x³ + (-A + B + D)x² + (A + B – C)x + (-A + B – D) = 1
Put x = -1 in(1) ⇒ A(-1 – 1)(1 + 1) + 0 + 0 = 1 ⇒ -4A = 1 ⇒ A = -1/4
Put x = 1 in (1) ⇒ 0 + B(1 + 1)(1 + 1) + 0 = 1 ⇒ 4B = 1 ⇒ B = 1/4
Equating the coefficient of x³ we get
A + B + C = 0 ⇒ C = -A – B = \(\frac{1}{4}-\frac{1}{4}\) = 0
Equating the coefficient of x² we get
-A + B – D = 1 ⇒ D = A – B = \(-\frac{1}{4}-\frac{1}{4}=\frac{-2}{4}=-\frac{1}{2}\)
∴ A = \(-\frac{1}{4}\), B = \(\frac{1}{4}\), C = 0, D = \(-\frac{1}{2}\)

Question 7.
Find the integral of \(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) [Hint: Put sin x = t]
Solution:
Put sin x = t ⇒ cos xdx = dt
∴ I = \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}\)
Let \(\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}=\frac{A(2-t)+B(1-t)}{(2-t)(1-t)}\)
A(2 – t) + B(1 – t) = 1 …….(1)
Put t = 1 in (1) ⇒ A(2 – 1) + 0 = 1 ⇒ A = 1
put t = 2 in (1) ⇒ 0 + B(1 – 2) = 1 ⇒ B = -1

Question 8.
Find the integral of \(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}\)
Solution:
We have \(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{x^4+3 x^2+2}{x^4+7 x^2+12}=\frac{x^4+7 x^2+12-4 x^2+10}{x^4+7 x^2+12}=1-\frac{\left(4 x^2+10\right)}{\left(x^2+3\right)\left(x^2+4\right)}\) …………(a)
Put x = t then
Let \(\frac{(4 t+10)}{(t+3)(t+4)}=\frac{A}{(t+3)}+\frac{B}{(t+4)}=\frac{A(t+4)+B(t+3)}{(t+3)(t+4)}\)
⇒ A(t + 4) + B(t + 3) = (4t + 10) ..(1)
Put t = -3 in (1) ⇒ A(-3 + 4) + B(0) = -12 + 10 = -2 ⇒ A = -2
Put t = -4 in (1) ⇒ A(0) + B(-4 + 3) = -16 + 10 = -6 ⇒ B = 6
From (1),

Question 9.
Find the integral of \(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)
Solution:
Given integral is \(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\) Put x² = t ⇒ 2x dx = dt
∴ \(\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x=\int \frac{d t}{(t+1)(t+3)}\) ……(1)
Let \(\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}\)
⇒ A(t + 3) + B(t + 1) = 1 …..(1)
Put t = -1 in (1) ⇒ A(-1 + 3) + 0 = 1 ⇒ A = 1/2
Put t = -3 in (1) ⇒ 0 + B(-3 + 1) = 1 ⇒ B = -1/2

Question 10.
Find the integral of \(\frac{1}{x\left(x^4-1\right)}\)
Solution:
Given integrand is \(\frac{1}{x\left(x^4-1\right)}\)
Multiplying Nr and Dr by x³, we get \(\frac{1}{x\left(x^4-1\right)}=\frac{x^3}{x^4\left(x^4-1\right)}\)
∴ \(\int \frac{1}{x\left(x^4-1\right)} d x=\int \frac{x^3}{x^4\left(x^4-1\right)} d x \)
Put x = t ⇒ 4x³ = dt
∴ \(\int \frac{1}{x\left(x^4-1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}\)
Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}=\frac{A(t-1)+B(t)}{(t-1) t}\) ⇒ A(t – 1) + Bt = 1 ………….(1)
Put t = 0 in (1) ⇒ A(0 – 1) + 0 = 1 ⇒ A = -1
Put t = 1 in(1) ⇒ 0 + B = 1 ⇒ B = 1

Question 11.
Find the integral of \(\frac{1}{x-x^3}\)
Solution:
We have \(\frac{1}{x-x^3}=\frac{1}{x\left(1-x^2\right)}=\frac{1}{x(1-x)(1+x)}\)
Let \(\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{(1+x)}=\frac{A\left(1-x^2\right)+B x(1+x)+C x(1-x)}{x(1-x)(1+x)}\)
⇒ A(1 – x²) + Bx(1 + x) + Cx(1 – x) = 1 ……….(1)
Put x = 1 in (1) ⇒ A(0) + B(1 + 1) + C(0) = 1 ⇒ 2B = 1 ⇒ B = 1/2
Put x = 0 in (1) ⇒ A(1) + B(0) + C(0) = 1 ⇒ A = 1
Comparingthe coefficient of x, B + C = 0 ⇒ C = -B = \(-\frac{1}{2}\)
A = 1, B = \(\frac{1}{2}\) , C = \(-\frac{1}{2}\)

Question 12.
Find the integral of \(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\) [Hint: \(\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}\), put x = t⁶]
Solution:
Given that \(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}\)
Put x = t⁶ ⇒ dx = 6t⁵ dt

Question 13.
Find the integral of \(\frac{5 x}{(x+1)\left(x^2+9\right)}\)
Solution:
Let \(\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^2+9\right)}=\frac{A\left(x^2+9\right)+(B x+C)(x+1)}{(x+1)\left(x^2+9\right)}\)
⇒ A(x² + 9) + (Bx + C)(x + 1) = 5x …….(1)
Put x = -1 in (1) ⇒ A(1 + 9) + (Bx + C)(0) = 5(-1) ⇒ 10A = -5 ⇒ A = -1/2
Comparingthe coefficientof x² we have B + A = 0 ⇒ B = -A = 1/2
Comparingthe coefficientof x² we have C + 9A = 0 ⇒ C = -9A = 9/2
A = \(-\frac{1}{2}\), B = \(\frac{1}{2}\), C = \(\frac{9}{2}\)

Question 14.
Find the integral of \(\frac{x^2+x+1}{(x+1)^2(x+2)}\)
Solution:
Let \(\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+2)}=\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^2}{(x+1)^2(x+2)}\)
⇒ A(x + 1)(x + 2) + B(x + 2) + C(x + 1)² = x + x + 1 ……….(1)
= A(x² + 3x + 2) + B(x + 2) + C(x² + 2x + 1)
⇒ x² + x + 1 = (A + C)x² + (3A + B + 2C)x + (2A + 2B + C)
Put x = -1 in (1) ⇒ A(0) + B(-1 + 2) + C(0) = 1 – 1 + 1 ⇒ B = 1
Put x = -2 in (1) ⇒ A(0) + B(0) + C(-2 + 1)² = (-2)² – 2 + 1 ⇒ C = 4 – 2 + 1 = 3
Equating the coefficient os x², we get A + C = 1 ⇒ A = 1 – C = 1 – 3 = -2
A = -2, B = 1, C = 3

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5a

Question 1.
Prove that the function f(x) = 5x – 3 is continuous at x = 0.
Solution:
Given function is f(x) = 5x – 3; At x = 0
(i) f(x) = 5x – 3 = f(0) = 5(0) – 3 – 3 ………..(1)
(ii) \(\underset{\mathrm{x} \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0}{\mathrm{Lt}}\) (5x – 3) = 5(0) – 3 = -3 …………….. (2)
(iii) From (1) & (2), \(\underset{\mathrm{x} \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
So, f(x) is continuous at x = 0

Question 2.
Prove that the function f(x) = 5x – 3 is continuous at x = -3
Solution:
Given function is f(x) = 5x – 3; At x = -3
(i) f(x) = 5x – 3 ⇒ f(-3) = 5(-3) – 3 = -18 …………….. (1)
(ii) \(\underset{x \rightarrow-3}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow-3}{\mathrm{Lt}}\) (5x – 3) = 5(-3) – 3 = -18 …………. (2)
(iii) From (1) & (2), \(\underset{x \rightarrow-3}{\mathrm{Lt}}\) f(x) = f(3).
So, f(x) is continuous at x = -3

Question 3.
Prove that the function f(x) = 5x – 3 is continuous at x = 5.
Solution:
Given function is f(x) = 5x – 3; At x = 5
(i) f(x) = 5x – 3 ⇒ f(5) = 5(5) – 3 = 22 ………….. (1)
(ii) \(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) (5x – 3) = 5(5) – 3 = 22 …………….. (2)
(iii) From (1) & (2), \(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) f(x) = f(5)
So, f(x) is continuous at x = 5

Question 4.
Prove that the function f(x) = 2x² – 1 is continuous at x = 3
Solution:
Given function is f(x) = 2x² – 1; At x = 3
(i) f(x) = 2x² – 1 ⇒ f(3) = 2(3)² – 1 = 17 …………….. (1)
(ii) \(\underset{x \rightarrow3}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow3}{\mathrm{Lt}}\) (2x² – 1) = 2(3)² – 1 = 17 …………. (2)
(iii) From (1) & (2), \(\underset{x \rightarrow3}{\mathrm{Lt}}\) f(x) = f(3).
So, f(x) is continuous at x = 3

Question 5.
Examine f(x) = x – 5 for continuity.
Solution:
The given function is f(x) = x – 5
For a real k, f(k) = k – 5.
\(\underset{\mathrm{x} \rightarrow k}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow k}{\mathrm{Lt}}\) (x – 5) = k – 5 = f(k)
∴ \(\underset{\mathrm{x} \rightarrow k}{\mathrm{Lt}}\) f(x) = f(k)
Thus f is continuous at every real number and hence it is a continuous function.

Question 6.
Examine f(x) = \(\frac{1}{x-5}\), x ≠ 5 for continuity.
Solution:
The given function is f(x) = \(\frac{1}{x-5}\), x ≠ 5
For any real number k ≠ 5 ,we have f(k) = \(\frac{1}{k-5}\)
\(\underset{\mathrm{x} \rightarrow k}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow k}{\mathrm{Lt}}\) \(\frac{1}{x-5}\) = \(\frac{1}{k-5}\)
∴ \(\underset{\mathrm{x} \rightarrow k}{\mathrm{Lt}}\) f(x) = f(k)
Thus f is continuous at every point in the domain of f and hence it is a continuous function.

Question 7.
Examine f(x) = \(\frac{x^2-25}{x+5}\), x ≠ -5 for continuity.
Solution:
The given function is f(x) = \(\frac{x^2-25}{x+5}\), x ≠ -5
For any real number c≠5, we have f(c) = \(\frac{c^2-25}{c+5}\) = \(\frac{(c+5)(c-5)}{c+5}\) = (c – 5)
\(\underset{x \rightarrow c} {\mathrm{Lt}} f(x)=\underset{x \rightarrow c}{\mathrm{Lt}} \frac{x^2-25}{x+5}\)
\(\underset{x \rightarrow c} {\mathrm{Lt}}\frac{(x+5)(x-5)}{x+5}\)
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x – 5) = (c – 5)
∴ \(\underset{\mathrm{x} \rightarrow c}{\mathrm{Lt}}\) f(x) = f(c)
Thus f is continuous at every point in the domain of f and hence it is a continuous function.

Question 8.
Examine f(x) = |x – 5| for continuity.
Solution:
The given function is

This function f is defined at all points on the real line.
Let c be a point on a real line. Then, c < 5, c – 5 or c > 5.
Case i:
c < 5
Here, f(c) = 5 – c \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (5 – x) = 5 – c
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all real numbers less than 5.

Case ii:
c = 5
Here, f(c) = f(5) = (5 – 5) = 0
\(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\)(5 – x) = (5 – 5) = 0
\(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) (x – 5) = 0
\(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 5}{\mathrm{Lt}}\) f(x) = f(c) ∴ f is continuous at x = 5

Case iii:
c > 5
Here f(c) = f(5) = c – 5
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x – 5) = c – 5
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all real numbers greater than 5.
Thus f is continuous at every real number and hence it is a continuous function.

Question 9.
Prove that the function f(x) = xⁿ Is continuous at x = n, where n is a positive Integer.
Solution:
Given function is f(x) = xⁿ
for all positive integers n, we have f(n) = nⁿ
\(\underset{\mathrm{x} \rightarrow \mathrm{n}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{n}}{\mathrm{Lt}}\) (xⁿ) = nⁿ
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{n}}{\mathrm{Lt}}\) f(x) = f(n)
Thus f(x) is continuous at n, where n is a positive integer.

Question 10.
Discuss the continuity of the function f(x) sin x + cos x
Solution:
We know that if g and h are two continuous functions, then g+h is continuous.
Let g(x) = sinx and h(x) = cosx. These two are continuous functions.
g(x) = sinx is defined for every real number.
Let c be a real number.
Put x = c+h. If x → c, then h → 0
Now g(c) = sinc .
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) g(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) sin x = \(\underset{\mathrm{h} \rightarrow \mathrm{0}}{\mathrm{Lt}}\) sin(c + h) = sin(c + 0) = sin c = g (c)
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) g (x ) = g (c)
∴ g(x) = sin x is a continuous function.

II.

Question 1.
Is the function f defined by \(f(x)= \begin{cases}x, & \text { if } x \leq 1 \\ 5, & \text { if } x>1\end{cases}\) continuous at x = 0? At x = 1? At x = 2?
Solution:
Given function is \(f(x)= \begin{cases}x, & \text { if } x \leq 1 \\ 5, & \text { if } x>1\end{cases}\)
(a) At x = 0,
It is clear that f is defined at 0 and its value at 0 is f(0) = 0.
Then, \(\underset{\mathrm{x} \rightarrow \mathrm{0}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{0}}{\mathrm{Lt}}\) (x) = 0
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{0}}{\mathrm{Lt}}\) f(x) = f(0)
∴ f is continuous at x = 0

(b) At x = 1,
It is clear that f is defined at 1 and its value at 1 is f(1) = 1.
L.H.L = \(\underset{\mathrm{x} \rightarrow \mathrm{1-}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{1-}}{\mathrm{Lt}}\) (x) = 1 = 1.
R.H.L = \(\underset{\mathrm{x} \rightarrow \mathrm{1+}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{1+}}{\mathrm{Lt}}\) (5) = 5
Here L.H.L ≠ R.H.L
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{1}}{\mathrm{Lt}}\) f(x) ≠ f(1)
∴ f(x) is not continuous at x = 1

(c) At x = 2
It is clear that f is defined at 2 and its value at 2 is f(2) = 5.
\(\underset{\mathrm{x} \rightarrow \mathrm{2}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{2}}{\mathrm{Lt}}\) 5 = 5
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{1}}{\mathrm{Lt}}\) f(x) = f(2)
∴ f(x) is continuous at x = 2

Question 2.
Find all points of discontinuity of f, where f is defined by f(x) = \(\left\{\begin{array}{l}
2 x+3, \text { if } x \leq 2 \\
2 x-3, \text { if } x>2
\end{array}\right.\)
Solution:
Given function is f(x) = \(\left\{\begin{array}{l}
2 x+3, \text { if } x \leq 2 \\
2 x-3, \text { if } x>2
\end{array}\right.\)
It is clear that the given function is defined at all the points of the real line.
Let c be a point on the real line. Then, three cases arise. c < 2, c > 2, c = 2

Case i: c < 2
Here f(c) = 2c + 3.
Then, \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (2x + 3) = 2c + 3
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x < 2. Case ii: c > 2
Here f(c) = 2c – 3.
Then, \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (2x – 3) = 2c – 3
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 2.

Case iii: c = 2
L.H.L = \(\underset{\mathrm{x} \rightarrow \mathrm{2-}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{2-}}{\mathrm{Lt}}\) (2x + 3) = 2(2) + 3 = 7
R.H.L = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) (2x – 3) = 2(2) – 3 = 1.
Here L.H.L. ≠ R.H.L
∴ f is not continuous at x = 2
Thus, x = 2 is the only point of discontinuity of f.

Question 3.
Find all points of discontinuity of f, where f is defined by f(x) = \(\begin{cases}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3<x<3 \\ 6 x+2, & \text { if } x \geq 3\end{cases}\)
Solution:
The given function is f(x)= \(\begin{cases}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3<x<3 \\ 6 x+2, & \text { if } x \geq 3\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case i:
If c < -3, then f(c) = -c+3,
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (-x + 3) = -c + 3
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x < -3.

Case ii:
If c = -3, then f(-3) = -(-3) + 3 = 6
L.H.L = \(\underset{\mathrm{x} \rightarrow 3-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 3-}{\mathrm{Lt}}\) (-x + 3) = -(3) + 3 = 6;
R.H.L = \(\underset{\mathrm{x} \rightarrow 3+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 3+}{\mathrm{Lt}}\) (-2x)= -2(-3) = 6
Here L.H.L = R.H.L
∴ \(\underset{\mathrm{x} \rightarrow -3}{\mathrm{Lt}}\) f(x) = 6 = f(-3)
∴ f is continuous at x = -3

Case iii:
If -3 < c < 3, then f(c) = -2c \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (-2x) = -2c
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous in (-3, 3)

Case iv:
If c = 3, then
L.H.L = \(\underset{\mathrm{x} \rightarrow 3-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 3-}{\mathrm{Lt}}\) (-2x) = -2(3) = -6
R.H.L = \(\underset{\mathrm{x} \rightarrow 3+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 3+}{\mathrm{Lt}}\) (6x + 2) = 6(3) + 2 = 20
Here L.H.L ≠ R.H.L
∴ f is continuous at x = 3

Case v.
If c > 3, then f(c) = 6c + 2
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (6x + 2) = 6c + 2
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x , such that x > 3.
Hence, x = 3 is the only point of discontinuity of f.

Question 4.
Find all points of discontinuity of f, where f is defined by f(x) = \(\begin{cases}\frac{|x|}{x}, & \text { if } x \neq 0 \\ x & \text { if } x=0\end{cases}\)
Solution:
The given function is f(x) = \(\begin{cases}\frac{|x|}{x}, & \text { if } x \neq 0 \\ x & \text { if } x=0\end{cases}\)
We know that, |x| = -x when x < 0 and |x| = x when x > 0
∴ the given function can be rewritten f (x) = \(\begin{cases}\frac{|x|}{x}=\frac{-x}{x}=-1, & \text { if } x<0 \\ 0, & \text { if } x=0 \\ \frac{|x|}{x}=\frac{x}{x}=1, & \text { if } x>0\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case i:
if c < 0, then f(c) = -1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (-1) = -1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x < 0

Case ii:
If c = 0, then L.H.L = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) (-1) = -1
R.H.L = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) (1) = 1
Here L.H.L ≠ R.H.L
∴ f is not continuous at x = 0.

Case iii.
If c > 0, then f(c) = 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (1) = 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 0.
Hence, x = 0 is the only point of discontinuity of f.

Question 5.
Find all points of discontinuity of f, where f is defined by f(x) = \(\begin{cases}\frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0\end{cases}\)
Solution:
Given function is f(x) = \(\begin{cases}\frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0\end{cases}\)
We know that |x| = -x when x < 0
∴ the given function can be rewritten as f(x) = \(\begin{cases}\frac{x}{|x|}=\frac{x}{-x}=-1, & \text { if } x<0 \\ -1, & \text { if } x \geq 0\end{cases}\)
Let c be any real number.
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (-1) = -1
Also, f(c) = -1 = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x)
∴ the given function is a continuous function.
Hence, the given function f(x) has no point of discontinuity.

Question 6.
Find all points of discontinuity of f, where f is defined by f(x) = \(\begin{cases}x+1, & \text { if } x \geq 1 \\ x^2+1, & \text { if } x<1\end{cases}\)
Solution:
Given function is f (x) = \(\begin{cases}x+1, & \text { if } x \geq 1 \\ x^2+1, & \text { if } x<1\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case i:
If c < 1, then f(c) = c² + 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x² + 1) = c² + 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x < 1.

Case ii.
If c = 1, then f(c) = f(1) = 1 + 1 = 2
L.H.L = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) (x² + 1) = 1² + 1 = 2
R.H.L = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) (x + 1) = 1 + = 2
Here L.H.L = R.H.L
∴ \(\underset{\mathrm{x} \rightarrow 1}{\mathrm{Lt}}\) f(x) = 2 = f(1)
∴ f is continuous at x = 1.

Case iii.
If c > 1, then f(c) = c + 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x + 1) = c + 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 1.
Hence, the given function f(x) has no point of discontinuity.

Question 7.
Find all points of discontinuity of f, where f is defined by f(x) = \(\begin{cases}x^3-3, & \text { if } x \leq 2 \\ x^2+1, & \text { if } x>2\end{cases}\)
Solution:
The given function is f(x) = \(\begin{cases}x^3-3, & \text { if } x \leq 2 \\ x^2+1, & \text { if } x>2\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case i:
If c < 2, then f(c) = c³ – 3
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x³ – 3) = c³ – 3
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x< 2.

Case ii:
If c = 2, then f(c) = f(2) = 2³ – 3 = 5
L.H.L = \(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) (x³ – 3) = 2 – 3 = 5
R.H.L = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) (x² + 1) = 2² + 1 = 5
Here L.H.L = R.H.L
∴ \(\underset{\mathrm{x} \rightarrow 2}{\mathrm{Lt}}\) f(x)= 5 = f(2)
∴ f is continuous at x = 2.

Case iii:
If c > 2, then f(c) = c² + 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x² + 1) = c² + 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 2.
Thus, the given function f is continuous at every point on the real line.
Hence, the given function f(x) has no point of discontinuity.

Question 8.
Find all points of discontinuity off, where f is defined by f(x) = \(\begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1\end{cases}\)
Solution:
The given function is f(x) = \(\begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1\end{cases}\)
The given function fis defmed at all the points of the real line.
Let c be a point on the real line.

Case i:
If c < 1,then f(c) = c¹⁰ – 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x¹⁰ – 1) = c¹⁰ – 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x < 1.

Case ii:
L.H.L = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) (x¹⁰ – 1) = 1¹⁰ – 1 = 1 – 1 = 0
R.H.L = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) (x²) = 1² = 1
Here L.H.L ≠ R.H.L
∴ f is not continuous at x = 1.

Case iii
If c > 1 then f(c) = c²
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x²) = c²
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 1.
We observe that x = 1 is the only point of discontinuity of f.

Question 9.
Is the function defined by f(x) = \(\begin{cases}x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1\end{cases}\) a continuous function?
Solution:
Given function is f(x) = \(\begin{cases}x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case i:
If c < 1, then f(c) = c + 5 \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x + 5) = c + 5
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)

Case ii:
If c = 1, then f(1) = 1 + 5 = 6
L.H.L = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) (x + 5) = 1 + 5 = 6
R.H.L = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) (x – 5) = 1 – 5 = -4
Here L.H.L ≠ R.H.L
∴ f is not continuous at x = 1.

Case iii:
If c >1, then f(c) = c – 5
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x – 5) = c – 5
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 1.
We observe that, x = 1 is the only point of discontinuity of f.

Question 10.
Discuss the continuity of the function f, where f is defined by f(x) = \(\left\{\begin{array}{l}
3, \text { if } 0 \leq x \leq 1 \\
4, \text { if } 1<x<3 \\
5, \text { if } 3 \leq x \leq 10
\end{array}\right.\)
Solution:
The given function is f(x) = \(\left\{\begin{array}{l}
3, \text { if } 0 \leq x \leq 1 \\
4, \text { if } 1<x<3 \\
5, \text { if } 3 \leq x \leq 10
\end{array}\right.\)
The given function is defined at all the points of the interval [0, 10].
Let c be a point in the interval [0, 10].

Case i:
If 0 ≤ c < 1, then f(c) = 3
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (3) = 3
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous in the interval [0, 1)

Case ii:
If c = 1, then f(3) = 3
L.H.L = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) (3) = 3
R.H.L = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) (4) = 4
Here L.H.L≠ R.H.L
∴ f is not continuous at x = 1.

Case iii:
If 1 < c < 3, then f(c) = 4
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (4) = 4
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous in the interval (1, 3).

Case iv:
If c = 3, then f(c) = 5
L.H.L = \(\underset{\mathrm{x} \rightarrow 3-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 3-}{\mathrm{Lt}}\) (4) = 4
R.H.L = \(\underset{\mathrm{x} \rightarrow 3+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 3+}{\mathrm{Lt}}\) (5) = 5
Here L.H.L≠ R.H.L
∴ f is not continuous at x = 3.

Case v:
If 3 < c ≤ 10, then f(c) = 5
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (5) = 5
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points of the interval (3, 10]
Also f is discontinuous at x = 1 and x = 3.

Question 11.
Discuss the continuity of the function f, where f is defined by f(x) = \(\begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}\)
Solution:
Given function is f(x) = \(\begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point in the interval (-∞, 0).

Case i:
If c < 0, then f(c) = 2c
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\)(2x) = 2c
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x < 0.

Case ii:
If c = 0, then f(c) = f(0) = 0
L.H.L = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) (2x) = 2(0) = 0
R.H.L = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) (0) = 0
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(0)
∴ f is continuous at x = 0.

Case iii:
If 0 < c < 1, then f(x) = 0
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (0) = 0
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous in the interval (0, 1)

Case iv:
If c = 1, then f(c) = f(1) = 0
L.H.L = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) (0) = 0
R.H.L = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) (4x) = 4(1) = 4
Here L.H.L ≠ R.H.L
∴ f is not continuous at x = 1.

Case v:
If c > 1, then f(c) = 4c
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (4x) = 4c
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 1.

Question 12.
Discuss the continuity of the function f. where f is defined by f(x) = \(\begin{cases}-2, & \text { if } x \leq-1 \\ 2 x, & \text { if }-11\end{cases}\)
Solution:
Given function is f(x) = \(\begin{cases}-2, & \text { if } x \leq-1 \\ 2 x, & \text { if }-11\end{cases}\)
The given function f is defined at all the points.
Let c be a point on the real line.

Case i:
If c < -1, then f(c) = -2
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (-2) = -2
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c) .
∴ f is continuous at all points x. such that x < -1.

Case ii:
If c = -1, then f(c) = f (-1) = -2 .
L.H.L = \(\underset{x \rightarrow 1-}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1-}{\mathrm{Lt}}\) (-2) = -2
R.H.L = \(\underset{x \rightarrow 1+}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1+}{\mathrm{Lt}}\) (2x) = 2(-1) = -2
Here L.H.L = R.H.L
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = -2 = f(-1)
∴ f is continuous at x = -1.

Case iii:
If -1 < ç < 1, then f(ç) = 2c \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\)
f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (2x) = 2c
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous in the interval (-1, 1).

Case iv:
If c = 1, then f(ç) = f(1) = 2(1) = 2
L.H.L = \(\underset{x \rightarrow 1-}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1-}{\mathrm{Lt}}\) (2x) = 2(1) = 2
R.H.L = \(\underset{x \rightarrow 1+}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1+}{\mathrm{Lt}}\) (2) = 2
Here L.H.L = R.H.L
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at x = 2.

Case v:
If c > 1, then f(c) = 2
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\)(2) = 2
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that x > 1.
Thus f is continuous at all points of all real line.

Question 13.
Find the relationship between a and b so that the function f defined by f(x) = \(\left\{\begin{array}{l}
a x+1, \text { if } x \leq 3 \\
b x+3, \text { if } x>3
\end{array}\right.\) is continuous at x = 3.
Solution:
Given function is f(x) = \(\left\{\begin{array}{l}
a x+1, \text { if } x \leq 3 \\
b x+3, \text { if } x>3
\end{array}\right.\)
For f to be continuous at x = 3, then \(\underset{x \rightarrow 3-}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3+}{\mathrm{Lt}}\) f(x)= f(3)
L.H.L = \(\underset{x \rightarrow 3-}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3-}{\mathrm{Lt}}\) (ax + 1) = 3a + 1
R.H.L = \(\underset{x \rightarrow 3+}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3+}{\mathrm{Lt}}\) (bx + 3) = 3b + 3
Also f(3) = 3a + 1
When L.H.L = R.H.L then 3a + 1 = 3b + 3 = 3a = 3b + 2 = a = b + \(\frac{2}{3}\)
This is the required relation.

Question 14.
For what value of λ is the function defined by f(x) = \(\begin{cases}\lambda\left(x^2-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}\) continuous at x = 0? What about continuity at x = 1?
Solution:
Given function is f(x) = \(\begin{cases}\lambda\left(x^2-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}\)
If f is continuous at x = 0,then \(\underset{x \rightarrow 0-}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0+}{\mathrm{Lt}}\) f(x) = f(0)
∴ \(\underset{x \rightarrow 0-}{\mathrm{Lt}}\) λ(x² – 2x) = \(\underset{x \rightarrow 0+}{\mathrm{Lt}}\) (4x + 1) = X(0² – 2 × 0)
⇒ λ(² – 2 × 0) = 4(0) + 1 = 0
⇒ 0 = 1 = 0 [which is not possible]
∴ f is continuous at x = 0 for no value of λ.
At x = 1, f(1) = 4x + 1 = 4(1) + 1 = 5
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) (4x + 1) = 4(1) + 1 = 5
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x)=f(l)
∴ f, is continuous at x = 1 for any value of λ.

Question 15.
Show that the function defined by g(x) = x – |x| is discontinuous at all integral points. Here |x| denotes (he greatest integer less than or equal to x.
Solution:
The given function ¡s g(x) = x – [x] .
It is clear that g is defined at all integral points.
Let n be an integer
Then, g(n) n – [n] = n – n = 0
LHL = \(\underset{\mathrm{x} \rightarrow \mathrm{n}-}{\mathrm{Lt}}\) g(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{n}-}{\mathrm{Lt}}\) (x – [x]) = \(\underset{\mathrm{x} \rightarrow \mathrm{n}-}{\mathrm{Lt}}\) (x) – \(\underset{\mathrm{x} \rightarrow \mathrm{n}-}{\mathrm{Lt}}\) [x] = n – (n – 1) = 1
R.H.L = \(\underset{\mathrm{x} \rightarrow \mathrm{n}+}{\mathrm{Lt}}\) g(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{n}+}{\mathrm{Lt}}\) (x – [x]) = \(\underset{\mathrm{x} \rightarrow \mathrm{n}+}{\mathrm{Lt}}\) (x) – \(\underset{\mathrm{x} \rightarrow \mathrm{n}+}{\mathrm{Lt}}\) [x] = n – n = 0
Here L.H.L ≠ R.H.L
∴ g is not continuous at x = n.
Hence, g is discontinuous at all integral points.

Question 16.
Is the function defined by f(x) = x² – sin x + 5 continuous at x = π?
Solution:
The given function is f(x) = x² – sin x + 5
It is clear that f is defined at x = π.
At x = π, f(x) = f(π) = π² – sin π + 5 = π² – 0 + 5 = π² + 5
Now \(\underset{\mathrm{x} \rightarrow \pi}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \pi}{\mathrm{Lt}}\) (x² – sin x + 5)
∴ \(\underset{\mathrm{x} \rightarrow \pi}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \pi}{\mathrm{Lt}}\) (x² – sin x) + 5
= (π + 0)² – sin(π + 0) + 5 = π² – sin π + 5
= π² – 0(1) – (-1)0 + 5 = π² + 5 = f(π)
∴ the given function f is continuous at x = π.

Question 17.
Find all points of discontinuity of f, where f(x) = \(\begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \\ x+1, & \text { if } x \geq 0\end{cases}\)
Solution:
The given function is f(x) = \(\begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \\ x+1, & \text { if } x \geq 0\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case i:
If c < 0, then f(c) = \(\frac{\sin c}{c}\)
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) \(\left(\frac{\sin x}{x}\right)=\frac{\sin c}{c}\)
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x, such that, x < 0.

Case ii:
If c > 0, then f(c) = c + 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x + 1) = c + 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x . such that x > 0.

Case iii:
If c = 0, then f(c) = f(0) = 0 + 1 = 1
L.H.L = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) \(\left(\frac{\sin x}{x}\right)\) = 1
R.H.L = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) (x + 1) = 1
\(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) f(x) = f(0)
∴ f is continuous at x = 0
Thus, f is continuous at all points of the real line.
Thus, f has no point of discontinuity.

Question 18.
Determine if f defined by f(x) = \(\begin{cases}x^2 \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}\) is a continuous function?
Solution:
The given function is f(x) = \(\begin{cases}x^2 \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}\)
The given function is defined at all the points of the real line.
Let c be a point on the real line.

Case i:
If c ≠ 0, then f(c) = c²sin\(\frac{1}{c}\)

∴ f is continuous at x = 0
Hence f is continuous at every point of the real line.
Thus, f is a continuous function on R.

Question 19.
Examine the continuity of f, where f ¡s defined by f(x) = \(\begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1, & \text { if } x=0\end{cases}\)
Solution:
The given function is f(x) = \(\begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1, & \text { if } x=0\end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case i:
If c ≠ 0, then f(c) = sin c – cos c
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (sinx – cosx) = sinc – cosc
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous at all points x , such that x ≠ 0.

Case ii:
If c = 0, then f(0) = -1.
L H L = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) (sin x – cos x) = sin 0 – cos 0 = 0 – 1 = -1
R.H.L= \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) (sin x – cos x) = sin 0 – cos 0 = 0 – 1 = -1
Here L.H.L = R.H.L
\(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) f(x)= \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) f(x) = f(0)
∴ f is continuous at x = 0
Hence f is continuous at every point of the real line. Thus, f is a continuous function on R.

Question 20.
Find the values of k. so that the function f is continuous at the indicated point f(x) = \(\begin{cases}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2}\end{cases}\) at x = \(\frac{\pi}{2}\)
Solution:
The given function is f(x) = \(\begin{cases}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2}\end{cases}\)
It is clear that f is defined at x = \(\frac{\pi}{2}\) and f(\(\frac{\pi}{2}\)) = 3
From the given f(x) to be continuous at x = \(\frac{\pi}{2}\), we have

Question 21.
Find the values of k so that the function f is continuous at the indicated point f(x) = \(\begin{cases}k x^2, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{cases}\) at x = 2
Solution:
The given function is f(x) = \(\begin{cases}k x^2, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{cases}\)
f is defined at x = and f(2) = k(2)² = 4k
For the given function f(x) to be continuous at x = 2 we have
\(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) f(x) = f(2)
⇒ \(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) (kx²) = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) (3) = 4k ⇒ 4k = 3 ⇒ k = \(\frac{3}{4}\)
∴ the value of k = \(\frac{3}{4}\)

Question 22.
Find the values of k so that the function f is continuous at the indicated point f(x) = \(\begin{cases}k x+1, & \text { if } x \leq \pi \\ \cos x, & \text { if } x>\pi\end{cases}\) at x = π
Solution:
The given function is f(x) = \(\begin{cases}k x+1, & \text { if } x \leq \pi \\ \cos x, & \text { if } x>\pi\end{cases}\)
It is clear that f is defined at x = π and f(π) = kπ + 1
For the given function f(x) to be continuous at x = π we have
\(\underset{\mathrm{x} \rightarrow \pi-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow \pi+}{\mathrm{Lt}}\) f(x) = f(π)
⇒ \(\underset{\mathrm{x} \rightarrow \pi-}{\mathrm{Lt}}\) (kx + 1) = \(\underset{\mathrm{x} \rightarrow \pi+}{\mathrm{Lt}}\) (cosx)kπ + 1
⇒ kπ + 1 = cos π = kπ + 1
⇒ kπ + 1 = -1 = kπ + 1
⇒ k = –\(\frac{2}{\pi}\)
∴ the value of k = –\(\frac{2}{\pi}\)

Question 23.
Find the values of k so that the function f is continuous at the indicated point f(x) = \(\begin{cases}k x+1, & \text { if } x \leq 5 \\ 3 x-5, & \text { if } x>5\end{cases}\) at x =
Solution:
The given function is f(x) = \(\begin{cases}k x+1, & \text { if } x \leq 5 \\ 3 x-5, & \text { if } x>5\end{cases}\)
It is clear that f is defined at x = 5 and f(5) = kx + 1 = 5k + 1
For the given function f(x) to be continuous at x = 5, we have
\(\underset{\mathrm{x} \rightarrow 5-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 5+}{\mathrm{Lt}}\) f(x) = f(5)
⇒ \(\underset{\mathrm{x} \rightarrow 5-}{\mathrm{Lt}}\) (kx + 1) = \(\underset{\mathrm{x} \rightarrow 5-}{\mathrm{Lt}}\) (3x – 5) = 5k + 1
⇒ 5k + 1 = 3(5) – 5 = 5k + 1
⇒ 5k + 1 = 15 – 5 = 5k + 1
⇒ 5k + 1 = 10 = 5k + 1
⇒ 5k + 1 = 10 = 5k + 1
⇒ 5k + 1 = 10 ⇒ 5k = 9 ⇒ k = \(\frac{9}{5}\)
∴ the value of k = \(\frac{9}{5}\)

Question 24.
Find the values of a and b such that the function defined by f(x) = \(\begin{cases}5, & \text { if } x \leq 2 \\ a x+b, & \text { if } 2<x<10 \\ 21, & \text { if } x \geq 10\end{cases}\) is a continuous function.
Solution:
The given function is f(x) = \(\begin{cases}5, & \text { if } x \leq 2 \\ a x+b, & \text { if } 2<x<10 \\ 21, & \text { if } x \geq 10\end{cases}\)
It is clear that f is defined at all points of the real line.
If f is a continuous function, then f is continuous at all real numbers.
In particular, f is continuous at x = 2 and x = 10
(i) When f is continuous at x = 2 , we obtain
\(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) f(x) = f(2)
⇒ \(\underset{\mathrm{x} \rightarrow 2-}{\mathrm{Lt}}\) (5) = \(\underset{\mathrm{x} \rightarrow 2+}{\mathrm{Lt}}\) (ax + b) = 5
⇒ 5 = 2a + b = 5
⇒ 2a + b = 5 …………. (1)

(ii) When f is continuous at x = 10, we obtain
\(\underset{\mathrm{x} \rightarrow 10-}{\mathrm{Lt}}\) f(x) = \(\underset{\mathrm{x} \rightarrow 10+}{\mathrm{Lt}}\) f(x) = f(10)
⇒ \(\underset{\mathrm{x} \rightarrow 10-}{\mathrm{Lt}}\) (ax + b) = \(\underset{\mathrm{x} \rightarrow 10+}{\mathrm{Lt}}\) (21) = 21
⇒ 10a + b = 21 ………… (2)
On subtracting equation (1) from equation (2), we obtain 8a = 16 ⇒ a = 2
By putting a = 2 in equation (1), we get 2(2) + b = 5 ⇒ 4 + b = 5 ⇒ b = 1
∴ the values of a and b for which is a continuous function are 2 and 1 respectively.

Question 25.
Show that the function defined by f(x) = cos (x²) is a continuous function.
Solution:
The given function is f(x) = cos (x²).
This function f is defined for every real number and f can be written as the composition of two functions as,f = goh,where g(x) = cos x and h(x) = x²
[∵ (goh)(x) = g(h(x)) = g(x²) = cos(x²) = f(x)]
It has to be proved first that g(x) = cosx and h(x) = x² are continuous functions.
It is clear that g is defined for every real number.
Let c be a real number.
(i) Let g(c) = cosc. Put x = c + h
If x → c, then, h → 0
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) g(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) cos x
= \(\underset{\mathrm{x} \rightarrow \mathrm{h}}{\mathrm{Lt}}\) cos(c + h) = cos(c + 0) = cos c = g(c)
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) g(x) = g(c)
∴ g(x) = cosx is a continuous function.

(ii) Let h(x) = x²
It is evident that h is defined for every real number.
Let k be a real number, then h(k) = k²
\(\underset{\mathrm{x} \rightarrow \mathrm{k}}{\mathrm{Lt}}\) h(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{k}}{\mathrm{Lt}}\) x² = k²
∴ h is a continuous function.
∴ f(x) = (goh)(x) = cos(x²) is a continuous function.

Question 26.
Show that the function defined by f(x) = |cos x| is a continuous function.
Solution:
We may rewrite f as f(x) = \(\begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}\)
By Example 3, we know that f is continuous at x = 0.
LH.L = \(\lim _{x \rightarrow c-}\) f(x) = \(\lim _{x \rightarrow c-}\) (x) = c
R.H.L = \(\lim _{x \rightarrow c+}\) f(x) = \(\lim _{x \rightarrow c+}\) x = c
Here L.H.L = R.H.L = f(c) for all real values of c.
Hence, f is continuous at all points.
In Ex 21 Part 2 we proved that cosx is a continuous function.
Thus, their composite functions |cosx| is also a continuous function.

Question 27.
Examine that sin is a continuous function.
Solution:
Let f(x) = sin x and g(x) = x
We know that sin x and |x| are continuous functions. .. f and g are continuous.
Now (fog) (x) = f(g(x)) = f(|x|) = sin |x|
We know that composite function of two continuous functions ¡s continuous.
∴ fog is continuous.
Hence, sin |x| is continuous.

Question 28.
Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|
Solution:
We may rewrite f as f(x) = \(\begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}\)
By Example 3, we know that f is continuous at x = 0.
LH.L = \(\lim _{x \rightarrow c-}\) f(x) = \(\lim _{x \rightarrow c-}\) (x) = c
R.H.L = \(\lim _{x \rightarrow c+}\) f(x) = \(\lim _{x \rightarrow c+}\) x = c
Here L.H.L = R.H.L = f(c) for all real values of c.
Hence, f is continuous at all points.

Let h(x) = |x + 1|can be written as h(x) = \(\left\{\begin{array}{l}
-(x+1), \text { if } x<-1 \\
x+1, \text { if } x \geq-1
\end{array}\right.\)
It is clear that h is defined for every real number.
Let c be a real number.

Case I:
If c < -1, then h(c) = -(c + 1)
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) h(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) [-(x + 1)] = -(c + 1)
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) h(x) = h(c)
∴ h is continuous at all points x, such that x < -1.

Case II:
If c > -1, then h(c) = c + 1
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) h(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) (x + 1) = c + 1
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) h(x) = h(c)
∴ h is continuous at all points x, such that x > -1.

Case III:
If c = -1, then h(c) = h(-1) = -1 + 1 = 0
L.H.L = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) h(x) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) [-(x + 1)] = -(-1 + 1) = 0
R.H.L = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) h(x) = \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) (x + 1) = (-1 + 1) = 0
Here L.H.L = R.H.L
∴ \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) h(x)= \(\underset{\mathrm{x} \rightarrow 1+}{\mathrm{Lt}}\) h(x) = h(-1)
∴ h is continuous at x = -1
Thus h is continuous at all points. It concludes that g and h are continuous functions.
∴ f = g – h is also a continuous function.

Question 29.
Discuss the continuity of the following functions:
(a) f(x) = sin x – cos x
(b) f(x) = sin x . cos x
Solution:
We know that if g and h are two continuous functions, then g – h, gh are also continuous.
Let g(x) = sinx and h(x) = cosx.
(a) Let h(x) = cosx
It is clear that h(x) = cosx is defined for every real number.
Let c be a real number.
Put x = c + h. If x → c, then h → 0
h(c) = cosc
\(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) h(x) = \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) cos x = \(\underset{\mathrm{x} \rightarrow 0}{\mathrm{Lt}}\) cos(c + h) = cos(c + 0) = cos c = h(c)
∴ \(\underset{\mathrm{x} \rightarrow \mathrm{c}}{\mathrm{Lt}}\) h(x) = h(c)
∴ h(x) = cos x is a continuous function.
Here, we conclude that,
f(x) = g(x) – h(x) = sinx – cosx is a continuous function.
f(x) = g(x) × h(x) = sinx × cosx is a continuous function.

III.

Question 1.
Discuss the continuity of the cosine function and cosecant function.
Solution:
We know that if h(x), g(x) and are two continuous functions, then we have
(i) \(\frac{\mathrm{h}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\), g(x) ≠ 0
(ii) \(\frac{1}{\mathrm{~g}(\mathrm{x})}\), g(x) ≠ 0
(iii) \(\frac{1}{\mathrm{~h}(\mathrm{x})}\), h(x) ≠ 0 is continuous.
In the previous problem we proved that sinx and cosx are continous functions.
\(\frac{1}{\sin x}\) = csc x is continuous except at x = nπ (n ∈ Z)

Question 2.
Discuss the continuity of the secant function and cotangent function.
Solution:
We know that if h(x), g(x) and are two continous functions, then we have
(i) \(\frac{\mathrm{h}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\), g(x) ≠ 0
(ii) \(\frac{1}{\mathrm{~g}(\mathrm{x})}\), g(x) ≠ 0
(iii) \(\frac{1}{\mathrm{~h}(\mathrm{x})}\), h(x) ≠ 0 is continuous.
In the previous problem we proved that sinx and cosx are continous functions.
\(\frac{1}{\sin x}\) = csc x is continuous except at x = nπ (n ∈ Z)
secx = \(\frac{1}{\cos x}\) is continuous when cos x ≠ 0
⇒ sec x is continuous when x ≠ (2n + 1)\(\frac{\pi}{2}\) (n ∈ Z)
secant is continuous except at x = (2n + 1)\(\frac{\pi}{2}\) (n ∈ Z)
cot x = \(\frac{\cos x}{\sin x}\). when sin x ≠ 0 cot x is continuous When x ≠ nπ (n ∈ Z)
cotangent is continuous except at x = nπ (n ∈ Z)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7d

I.

Question 1.
Find the integral of \(\frac{3 x^2}{x^6+1}\)
Solution:
Put x³ = t ⇒ 3x² dx = dt

Question 2.
Find the integral of \(\frac{3 x^2}{x^6+1}\)
Solution:
Put 2x = t ⇒ 2dx = dt

Question 3.
Find the integral of \(\frac{1}{\sqrt{(2-x)^2+1}}\)
Solution:
Put 2 – x = t ⇒ -dx = dt

Question 4.
Find the integral of \(\frac{1}{\sqrt{9-25 x^2}}\)
Solution:
Put 5x = t ⇒ 5dx = dt \(\left[ \int \frac{1}{\sqrt{a^2-x^2}} d x={Sin}^{-1}\left(\frac{x}{a}\right)+C\right]\)

Question 5.
Find the integral of \(\frac{3 x}{1+2 x^4}\)
Solution:
Put \(\sqrt{2}\)x² = t ⇒ 2\(\sqrt{2}\)xdx = dt \(\left[\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\right]\)

Question 6.
Find the integral of \(\frac{x^2}{1-x^6}\)
Solution:
Put x³ = t ⇒ 3x² dx = dt
∴ \(\int \frac{x^2}{1-x^6} d x=\frac{1}{3} \int \frac{d t}{1-t^2}=\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]+C=\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|+C\)

Question 7.
Find the integral of \(\frac{x-1}{\sqrt{x^2-1}}\)
Solution:

Question 8.
Find the integral of \(\frac{x^2}{\sqrt{x^6+a^6}}\)
Solution:
Put x³ = t ⇒ 3x² dx = dt \(\left[\int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|\right]\)

Question 9.
Find the integral of \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
Solution:
Put tan x = t ⇒ sec² dx = dt
∴ \(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x=\int \frac{d t}{\sqrt{t^2+2^2}}=\log \left|t+\sqrt{t^2+4}\right|+C=\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|+C\)

Question 10.
Find the integral of \(\frac{\cos x}{\sqrt{4-\cos ^4 x}}\)
Solution:
Put sin x = t ⇒ cos xdx = dt
∴ \(\int \frac{\cos x}{\sqrt{4-\cos ^4 x}} d x=\int \frac{d t}{\sqrt{2^2-(t)^2}}=\sin ^{-1}\left(\frac{t}{2}\right)+C=\sin ^{-1}\left(\frac{\sin x}{2}\right)+C\)

II.

Question 1.
Find the integral of \(\frac{1}{\sqrt{x^2+2 x+2}}\)
Solution:
We have x² + 2x + 2 = x² + 2x + 1 – 1 + 2 = (x + 1)² + 1²

Question 2.
Find the integral of \(\frac{1}{9 x^2+6 x+5}\)
Solution:
We have 9x² + 6x + 5 = 9x² + 6x + 1 + 4 = (3x + 1)² + 2²

Question 3.
Find the integral of \(\frac{1}{\sqrt{7-6 x-x^2}}\)
Solution:
7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9) [∵ \(\int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+C\)]
Thus 7 – (x² + 6x + 9 – 9) = 16 – (x² + 6x + 9) = 16 – (x + 3)² = 4² – (x + 3)²
∴ \(\int \frac{1}{\sqrt{7-6 x-x^2}} d x=\int \frac{1}{\sqrt{4^2-(x+3)^2}} d x\) Put x + 3 = t ⇒ dx = dt
∴ \(\int \frac{1}{\sqrt{4^2-(x+3)^2}} d x=\int \frac{1}{\sqrt{4^2-t^2}} d t=\sin ^{-1}\left(\frac{t}{4}\right)+C=\sin ^{-1}\left(\frac{x+3}{4}\right)+C\)

Question 4.
Find the integral of \(\frac{1}{\sqrt{(x-1)(x-2)}}\)
Solution:
We have (x – 1)(x – 1) = x² – 3x + 2

Question 5.
Find the integral of \(\frac{1}{\sqrt{8+3 x-x^2}}\)
Solution:
We have 8 + 3x – x² = -(x² – 3x – 8) = \(-\left(x^2-3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-8\right)\)

Question 6.
Find the integral of \(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Solution:
We have (x – a)(x – b) = x² – (a + b)x + ab
⇒ x² – (a + b)x + ab = x² – (a + b)x + \(\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}+a b=\left[x-\left(\frac{a+b}{2}\right)\right]^2-\frac{(a-b)^2}{4}\)

Question 7.
Find the integral of \(\frac{x+2}{\sqrt{x^2-1}}\)
Solution:
Let x + 2 = A\(\frac{d}{d x}\)(x² – 1) + B …………..(1) ⇒ x + 2 = A(2x) + B
Equating the coefficientsof x and constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\); B = 2
From (1) we get \(\int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}=\int \frac{\frac{1}{2}(2 \mathrm{x})+2}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}=\frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}+\int \frac{2}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}\) ……………….(2)
To find \(\frac{1}{2} \int \frac{2 x}{\sqrt{x^2-1}} d x\) we take x² – 1 = t ⇒ 2xdx = dt

Question 8.
Find the integral of \(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
Solution:
Put 2x² + x – 3 = t ⇒ (4x + 1)dx = dt
∴ \(\int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x=\int \frac{1}{\sqrt{t}} d t=2 \sqrt{t}+C=2 \sqrt{2 x^2+x-3}+C\)
\(\int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x=2 \sqrt{f(x)}+c\)

III.

Question 1.
Find the integral of \(\frac{5 x-2}{1+2 x+3 x^2}\)
Solution:
Let 5x – 2 = A\(\frac{d}{d x}\)(1 + 2x + 3x²) + B ⇒ 5x – 2 = A(2 + 6x) + B
Equating the coefficients of x and constant term on both sides, we get
5 = 6A ⇒ A = \(\frac{5}{6}\)
2A + B = -2 ⇒ B = \(-\frac{11}{3}\)
∴ 5x – 2 = \(\frac{5}{6}\)(2 + 6x) + (\(-\frac{11}{3}\))
∴ \(\int \frac{5 x-2}{1+2 x+3 x^2} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^2} d x=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^2} d x\)
Let I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx and I₂ = \(\int \frac{1}{1+2 x+3 x^2}\)dx
∴ \(\int \frac{5 x-2}{1+2 x+3 x^2} d x=\frac{5}{6} I_1-\frac{11}{3} I_2\) …..(1)
First we find I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx
Put 1 + 2x + 3x² = t ⇒ (2+ 6x)dx = dt
∴ I₁ = \(\int \frac{d t}{t}\) ⇒ I₁ = log |t| ⇒ I₁ = log |1 + 2x + 3x² | ……(2)
Now I₂ = \(\int \frac{1}{1+2 x+3 x^2} d x\)
Here, 1 + 2x + 3x² = 1 + 3(x² + \(\frac{2}{3}\)x) [\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]

Question 2.
Find the integral of \(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
Solution:
We have \(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^2-9 x+20}}\)
Let 6x + 7 = A\(\frac{d}{d x}\)(x – 9x + 20) + B ⇒ 6x + 7 = A(2x – 9) + B
Equating the coefficients of x and constant term, we get
2A = 6 ⇒ A = 3; -9A + B = 7 ⇒ B = 34
∴ 6x+ 7 = 3(2x – 9) + 34

Let x² – 9x + 20 = t ⇒ (2x – 9)dx = dt
∴ I₁ = \(\int \frac{d t}{\sqrt{t}} \Rightarrow I_1=2 \sqrt{t} \Rightarrow I_1=2 \sqrt{x^2-9 x+20}\) ……………..(2)
Now I₂ = \(\int \frac{1}{\sqrt{x^2-9 x+20}}\)
x² – 9x + 20 = x² – 9x + 20 + \(\frac{81}{4}-\frac{81}{4}\)
x² – 9x + 20 + \(\frac{81}{4}-\frac{81}{4}=\left(x-\frac{9}{2}\right)^2-\frac{1}{4}=\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
⇒ I₂ = \(\int \frac{1}{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2} d x=\log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right|\) ………….(3)
Substituting equations (2) and (3) in (1), we get
\(\int \frac{6 x+7}{\sqrt{x^2-9 x+20}} d x=3\left[2 \sqrt{x^2-9 x+20}\right]+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]+C\)
= \(6 \sqrt{x^2-9 x+20}+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]+C\)

Question 3.
Find the integral of \(\frac{x+2}{\sqrt{4 x-x^2}}\)
Solution:
Let x + 2 = A\(\frac{d}{d x}\)(4x – x²) + B
Equating the coefficients of x and constant term on both sides, we get
-2A = 1 ⇒ A = \(-\frac{1}{2}\); 4A + B = 2 ⇒ B = 4 ⇒ (x + 2) = \(-\frac{1}{2}\)(4 – 2x) + 4
∴ \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{\left(4 x-x^2\right)}} d x=-\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{\left(4 x-x^2\right)}} d x+4 \int \frac{1}{\sqrt{\left(4 x-x^2\right)}} d x\)
let I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) and I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
∴ \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2} I_1+4 I_2\) …….(1)
First we find I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) [∵ \(\int \frac{1}{\sqrt{x}} d x=2 \sqrt{x}+C\)]
Let 4x – x² = t ⇒ (4 – 2x)dx = dt ⇒ I₁ = \(\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=2 \sqrt{\mathrm{t}}=2 \sqrt{4 \mathrm{x}-\mathrm{x}^2}\) …….(2)
Now I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
⇒ 4x – x² = -(-4x + x²) = (-4x + x² + 4 – 4) = 4 – (x – 2)² = (2)² – (x – 2)²
∴ I₂ = \(\int \frac{1}{\sqrt{(2)^2-(x-2)^2}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right)\) …..(3)
Substituting (2) and (3) in (1), we get
\(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2}\left(2 \sqrt{4 x-x^2}\right)+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C\)
= \(-\sqrt{4 x-x^2}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C\)

Question 4.
Find the integral of \(\frac{x+2}{\sqrt{x^2+2 x+3}}\)
Solution:

First we find I₁ = \(\int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x\)
Put x² + 2x + 3 = t ⇒ (2x + 2)dx = dt
∴ I₁ = \(\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{x^2+2 x+3}\) ………..(2)
Now I₂ = \(\int \frac{1}{\sqrt{x^2+2 x+3}} d x\),
Consider x² + 2x + 3 = x² + 2x + 1 + 2 = (x + 1)² + \((\sqrt{2})^2\)
Now I₂ = \(\int \frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}} d x=\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\) ….(3)
Substituting (2) and (3) in (1), we get
\(\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x=\frac{1}{2}\left[2 \sqrt{x^2+2 x+3}\right]+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|+C\)
= \(\sqrt{x^2+2 x+3}+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|+C\)

Question 5.
Find the integral of \(\frac{x+3}{x^2-2 x-5}\)
Solution:
Let (x + 3) = A\(\frac{d}{d x}\)(x² – 2x – 5) + B
⇒ (x + 3) = A(2x – 2) + B
Equating the coefficients of x and constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\)
-2A + B = 3 ⇒ B = 4 ⇒ (x = 3) = \(\frac{1}{2}\)(2x – 2) + 4

Substituting (2) and (3) in (1), we get
\(\begin{aligned}
\int \frac{x+3}{x^2-2 x-5} d x & =\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{A}{\not 2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C \\
& =\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C
\end{aligned}\)

Question 6.
Find the integral of \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\)
Solution:
Let 5x + 3 = A\(\frac{d}{d x}\)(x² + 4x + 10) + B ⇒ 5x + 3 = A(2x + 4) + B
Equating the coefficients of x and constant term on both sides, we get 2A = 5 ⇒ A = \(\frac{5}{2}\)
4A + B = 3 ⇒ B = -7 ⇒ 5x + 3 = \(\frac{5}{2}\)(2x + 4) – 7

Question 7.
Find the integral of \(\frac{1}{x \sqrt{a x-x^2}}\) [Hint : Put x = \(\frac{a}{t}\)]
Solution:
Put x = \(\frac{a}{t}\) ⇒ dx = \(-\frac{a}{t^2}\)dt
∴ \(\int \frac{1}{x \sqrt{a x-x^2}} d x=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-\left(\frac{a}{t}\right)^2}}\left(-\frac{a}{t^2} d t\right)=-\int \frac{1}{a t} \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{t^2}}} d t=-\frac{1}{at} \int \frac{d t}{\frac{\sqrt{t-1}}{t}}\)
= \(-\frac{1}{a} \int \frac{1}{\sqrt{t-1}} d t=-\frac{1}{a}[2 \sqrt{t-1}]+C=-\frac{1}{a}\left[2 \sqrt{\frac{a}{x}-1}\right]+C=-\frac{2}{a}\left(\sqrt{\frac{a-x}{x}}\right)+C\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4f Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4f

I.

Question 1.
Prove that the determinant \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\) is independent of θ.
Solution:
∆ = \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\)
= x(-x² – 1) – sin θ(-x sin θ – cos θ) + cos θ(-sin θ + x cos θ)
= -x³ – x + x sin²θ – sin θ cos θ sin θ cos θ + x cos²θ
= -x³ – x + x(sin²θ + cos²θ) = -x³ – x + x(1) = -x³
Thus, ∆ is independent of θ.

Question 2.
Evaluate \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\)
Expanding along C₃, we get
∆ = -sin α(-sin αsin²β – cos²β sin α) + cos α(cosα(cos²β + cosαsin²β)
= sin²α(sin²β + cos²β) + cos²α(cos²β + sin²β)
= sin²α(1) + cos²α(1) = 1

Question 3.
Evaluate \(\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|\)
Solution:
∆ = \(\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|\) [R₂ → R₂ – R₁ and R₃ → R₃ – R₁]
= 1 (xy – 0) [Expanding along C₁]
= xy

II.

Question 1.
If A^-1 = \(\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\), find (AB)^-1
Solution:
We know that (AB)^-1 = B^-1A^-1.
Given that B = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)
∴ |B| = 1(3) – 2(-1) – 2(2) = 3 + 2 – 4 = 5 – 4 = 1
Now, B₁₁ = 3; B₁₂ = 1; B₁₃ = 2
B₂₁ = 2; B₂₂ = 1; B₂₃ = 2
B₃₁ = 6; B₃₂ = 2; B₃₃ = 5

Question 2.
Evaluate \(\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|\)
Solution:
∆ = \(\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|\)

Question 3.
Show that \(\left|\begin{array}{ccc}
y+z & x & x \\
y & z+x & y \\
z & z & x+y
\end{array}\right|\) = 4xyz
Solution:
L.H.S = \(\left|\begin{array}{ccc}
y+z & x & x \\
y & z+x & y \\
z & z & x+y
\end{array}\right|\) = \(\left|\begin{array}{ccc}
0 & -2 z & -2 y \\
y & z+x & y \\
z & z & x+y
\end{array}\right|\) (∵ R₁ → (R₁ – (R₂ + R₃))
= -2\(\left|\begin{array}{ccc}
0 & \mathrm{z} & \mathrm{y} \\
\mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{y} \\
\mathrm{z} & \mathrm{z} & \mathrm{x}+\mathrm{y}
\end{array}\right|\) = -2\(\left|\begin{array}{lll}
0 & z & y \\
y & x & 0 \\
z & 0 & x
\end{array}\right|\) (∵ R₂ → R₂ – R₁ ; R₃ → R₃ – R₁)
= -2[0(x² – 0) – z(xy – 0) + y(0 – xz)]
= -2[0 – xyz – xyz] = -2(-2 xyz) = 4xyz
= R.H.S

Question 4.
Show that \(\left|\begin{array}{ccc}
\mathrm{a} & \mathrm{~b} & \mathrm{c} \\
\mathrm{~b} & \mathrm{c} & \mathrm{a} \\
\mathrm{c} & \mathrm{a} & \mathrm{~b}
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)²
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
\mathrm{a} & \mathrm{~b} & \mathrm{c} \\
\mathrm{~b} & \mathrm{c} & \mathrm{a} \\
\mathrm{c} & \mathrm{a} & \mathrm{~b}
\end{array}\right|\) = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ + b³ + abc + abc – c³
= -(a³+ b³ + c³ – 3abc)
⇒ ∆² = (a³+ b³ + c³ – 3abc)² ………. (1)

III.

Question 1.
Let A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]\). Verify that (i) |adj A|^-1 = adj (A^-1)
(ii) (A^-1)^-1 = A
Solution:
Given that A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]\)
⇒|A| = 1(15 – 1) – 2(10 – 1) + 1(2 – 3) = 14 – 18 – 1 = -5
Cofactor matrix of A:
Now, A₁₁ = 14; A₁₂ = -9; A₁₃ = -1
A₂₁ = -9; A₂₂ = 4; A₂₃ = 1
A₃₁ = -1; A₃₂ = 1; A₃₃ = -1
Hence, adj A = \(\left[\begin{array}{ccc}
14 & -9 & -1 \\
-9 & 4 & 1 \\
-1 & 1 & -1
\end{array}\right]\) …………. (1)
∴ A^-1 = \(\frac{1}{\mathrm{~A}}\) (adj A) = –\(\frac{1}{5}\left[\begin{array}{ccc}
14 & -9 & -1 \\
-9 & 4 & 1 \\
-1 & 1 & -1
\end{array}\right]\) = \(\frac{1}{5}\left[\begin{array}{ccc}
-14 & 9 & 1 \\
9 & -4 & -1 \\
1 & -1 & 1
\end{array}\right]\) …………… (2)
(i) Using (1) & (2) we prove the result [adj A]^-1 = adj (A^-1)
(i) |adjA| = 14(-4 – 1) + 9(9 + 1) – 1(-9 + 4)
= 14(-5) + 9(10) + (-1)(-5)= -70 + 90 + 5 = 25
From (1) Cofactor matrix of AdjA
A₁₁ = -5; A₁₂ = -10; A₁₃ = -5
A₂₁ = -10; A₂₂ = -15; A₂₃ = -5
A₃₁ = -5; A₃₂ = -5; A₃₃ = -25

Hence, (A^-1)^-1 = A is proved.

Question 2.
Solve the system of equations \(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}\) = 4, \(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}\) = 1, \(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}\) = 2
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, and \(\frac{1}{z}\) = r
∴ The given system of equations is
2p + 3q + 10r = 4
4p – 6q + 5r = 1
6p + 9q – 20r = 2
This system can be written in the form of AX = B, where
A = \(\left[\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\), X = \(\left[\begin{array}{l}
\mathrm{p} \\
\mathrm{q} \\
\mathrm{r}
\end{array}\right]\) B = \(\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\)
∴ |A = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36)
= 150 + 330 + 720 = 1200
Thus, A is non-singular
∴ A^-1 exists.
Here, A₁₁ = 75; A₁₂ = 110; A₁₃ = 72
A₂₁ = 150; A₂₂ = -100; A₂₃ = 0
A₃₁ = 75; A₃₂ = 30; A₃₃ = -24

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7c

I.

Question 1.
Find the integral of sin²(2x + 5)
Solution:
sin²(2x + 5) = \(\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2}\) [∵ sin² A = \(\frac{1-\cos 2 \mathrm{~A}}{2}\)]
∴ ∫sin²(2x + 5) = ∫ \(\frac{1-\cos (4 x+10)}{2}\)dx
= \(\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x=\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+C=\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+C\)

Question 2.
Find the integral of sin 3xcos 4x
Solution:
I = ∫sin 3x cos 4xdx = \(\frac{1}{2}\)∫2sin3x cos4x dx
= \(\frac{1}{2}\)∫(sin(3x + 4x) + sin(3x – 4x)dx [∵ 2sin A cos B = sin(A + B) + sin(A – B)]
= \(\frac{1}{2}\)∫(sin 7x + sin(-x)) dx = \(\frac{1}{2}\)∫(sin 7x – sin x) dx [∵ sin(θ) = -sinθ]
= \(\frac{1}{2}\)[∫sin 7xdx – ∫sinxdx = \(\frac{1}{2}\) \(\left[\frac{-\cos 7 x}{7}-(-\cos x)\right]\) + c
= \(\frac{-1}{14}\)cos 7x + \(\frac{1}{2}\)cos x + C

Question 3.
Find the integral of sin4x sin8x
Solution:
We know that SinA SinB = \(\frac{1}{2}\)[cos(A – B) – cos(A + B)]
∴ ∫sin4x sin 8x dx = ∫[\(\frac{1}{2}\)cos(4x – 8x) – \(\frac{1}{2}\)cos(4x + 8x)] dx
= \(\frac{1}{2}\)∫(cos(-4x) – cos12x) dx = \(\frac{1}{2}\)∫(cos 4x – cos 12x) dx = \(\frac{1}{2}\left[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right]\)

Question 4.
Find the integral of \(\frac{1-\cos x}{1+\cos x}\)
Solution:

Question 5.
Find the integral of \(\frac{\cos x}{1+\cos x}\)
Solution:

Question 6.
Find the integral of \(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\)
Solution:
\(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}=\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x}=\frac{\sin x}{\cos ^2 x}+\frac{\cos x}{\sin ^2 x}\) = tan xsec x + cot xcosec x
∴ \(\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x\) = ∫(tan x sec x + cot x cosec x) dx = sec x – cosecx + C

Question 7.
Find the integral of \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\)
Solution:

Question 8.
Find the integral of \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\)
Solution:
Let I = \(\int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x\)

Put cos x + sin x = t ⇒ (-sinx + cosx)dx = dt
∴ From (i), I = \(\int \frac{d t}{t}\) = log|t| + C = log|cos x + sinx| + C

Question 9.
Find the integral of sin^-1(cos x)
Solution:
\(\int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} \sin \left(\frac{\pi}{2}-x\right) d x\) [∵ sin^-1 sin θ = θ]
\(=\int\left(\frac{\pi}{2}-x\right) d x=\int \frac{\pi}{2} d x-\int x d x=\frac{\pi}{2} \int 1 d x-\int x^1 d x=\frac{\pi}{2} x-\frac{x^2}{2}+c\)

Question 10.
Find the integral of \(\frac{\sin ^2 x}{1+\cos x}\)
Solution:

II.

Question 1.
Find the integral of cos 2x cos 4x cos 6x
Solution:
We know that Cos A cos B = \(\frac{1}{2}\)[cos(A + B) + cos(A – B)]
∫cos 2x(cos 4x cos 6x) dx = ∫cos 2x[\(\frac{1}{2}\)[cos(4x + 6x) + cos(4x – 6x)]] dx
= ∫\(\frac{1}{2}\)[cos 2x cos 10x + cos 2x cos(-2x) dx = \(\frac{1}{2}\)∫\(\frac{1}{2}\)[cos 2x cos 10x + cos² 2x] dx
= \(\frac{1}{2}\)∫[(\(\frac{1}{2}\)cos(2x + 10x) + \(\frac{1}{2}\)cos(2x – 10x)) + (\(\frac{1+\cos 4 x}{2}\))] dx
= \(\frac{1}{4}\)∫(cos 12x + cos8x + 1 + cos4x) dx = \(\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]\) + C

Question 2.
Find the integral of sin x sin 2x sin 3x
Solution:
We know that Sin A sin B = \(\frac{1}{2}\)[cos(A – B) – cos(A + B)]

Question 3.
Find the integral of sin³(2x + 1)
Solution:
Let I = ∫sin³(2x + 1) = ∫sin²(2x + 1) sin(2x + 1) dx
= ∫(1 – cos²(2x + 1)) sin(2x + 1) dx
Put cos(2x + 1) = t ⇒ -2 sin(2x + 1) dx = dt ⇒ sin(2x + 1) dx = \(\frac{-\mathrm{dt}}{2}\)
∴ \(I=\frac{-1}{2} \int\left(1-t^2\right) d t=\frac{-1}{2}\left[t-\frac{t^3}{3}\right]=\frac{-1}{2}\left[\cos (2 x+1)-\frac{\cos ^3(2 x+1)}{3}\right]\)
\(=\frac{-\cos (2 x+1)}{2}+\frac{\cos ^3(2 x+1)}{6}+C=\frac{\cos ^3(2 x+1)}{6}-\frac{\cos (2 x+1)}{2}++C\)

Question 4.
Find the integral of sin³x cos³ x
Solution:
Let I = ∫sin³x cos³x dx = ∫cos³ xsin²x sin x dx = ∫cos³x(1 – cos² x) sin x dx
Put cos x = t ⇒ -sin x dx = dt
∴ I = \(-\int t^3\left(1-t^2\right) d t=-\int\left(t^3-t^5\right) d t=-\left[\frac{t^4}{4}-\frac{t^6}{6}\right]+C\)
\(=-\left[\frac{\cos ^4 x}{4}-\frac{\cos ^6 x}{6}\right]+C=\frac{\cos ^6 x}{6}-\frac{\cos ^4 x}{4}+C\)

Question 5.
Find the integral of sin⁴x.
Solution:
We have sin⁴ x = sin² x sin² x = \(\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)=\frac{1}{4}(1-\cos 2 x)^2\)

Question 6.
Find the integral of cos⁴ x .
Solution:
We have cos ⁴2x = (cos ² 2x)² = \(\left(\frac{1+\cos 4 x}{2}\right)^2\)

Question 7.
Find the integral of \(\frac{\cos 2 x-\cos 2 u}{\cos x-\cos \alpha}\)
Solution:

Question 8.
Find the integral of \(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution:
We have \(\frac{\cos x-\sin x}{1+\sin 2 x}=\frac{\cos x-\sin x}{\left(\sin ^2 x+\cos ^2 x\right)+2 \sin x \cos x}\) [∵ sin² x + cos² x = 1; sin 2x = 2sin x cosx ]
= \(=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}\)
Put sin x + cos x = t ⇒ (cos x – sin x) dx = dt
∴ \(\int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^2} d x\)
\(=\int \frac{d t}{t^2}=\int t^{-2} d t=-t^{-1}+C=-\frac{1}{t}+C=\frac{-1}{\sin x+\cos x}+C\)

Question 9.
Find the integral of tan³2x sec 2x
Solution:
Let I = ∫tan³2xsec 2x dx = ∫tan² 2x tan 2x sec 2x dx
= ∫(sec² 2x – 1) sec 2x tan ex dx [∵ tan² θ = sec² θ – 1]
= \(\frac{1}{2}\)∫(sec² 2x – 1)(2 sec 2x tan 2x) dx ……(i)
Put sec 2x = t ⇒ sec 2x tan 2x \(\frac{d}{dx}\)(2x) = \(\frac{dt}{dx}\) ⇒ 2 sec 2x tan 2x dx = dt
∴ From (i), I = \(\frac{1}{2} \int\left(\mathrm{t}^2-1\right) \mathrm{dt}=\frac{1}{2}\left(\int \mathrm{t}^2 \mathrm{dt}-\int 1 \mathrm{dt}\right)\)
= \(\frac{1}{2}\left(\frac{t^3}{3}-t\right)+c=\frac{1}{6} t^3-\frac{1}{2} t+c=\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c\)   [∵ t = sec 2x]

Question 10.
Find the integral of tan⁴ x.
Solution:
∫tan⁴ x dx = ∫tan² x tan² x dx = ∫tan² x(sec² x – 1) dx
= ∫(tan² x sec² x – tan² x) dx = ∫tan² x sec² dx – ∫tan² dx
= ∫tan² x sec² x dx – ∫(sec²x – 1)dx
= ∫tan² x sec²xdx – ∫sec²dx + 1dx
= ∫tan² x sec² x dx – tan x + x + C₁ …….(i)
For this integral, put tan x = t, ⇒ sec²x dx = dt
∴ \(\int \tan ^2 x \sec ^2 x d x=\int t^2 d t=\frac{t^3}{3}+C_2=\frac{\tan ^3 x}{3}+C_2\)
From (i) I = \(\frac{\tan ^3 x}{3}\) – tan x + x + C, where C = C₁ + C₂

Question 11.
Find the integral of \(\frac{1}{\sin x \cos ^3 x}\)
Solution:

Question 12.
Find the integral of \(\frac{1}{\cos (x-a) \cos (x-b)}\)
Solution:

Question 13.
Find the integral of \(\frac{\sin x}{\sin (x-a)}\)
Solution:
Put x – a = t ⇒ x = t + a ⇒ dx = dt
∴ \(\int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin (t+a)}{\sin t} d t=\int \frac{\sin t \cos a+\cos t \sin a}{\sin t} d t=\int(\cos a+\cot t \sin a) d t\)
= tcos a + sin a log|sin t| + C₁ = (x – 1)cos a + sin alog|sin(x – a)| + C₁
= x cos a + sin a log|sin(x – a)| – a cos a + C₁ = sin a log|sin(x – a)| + x cos a + C

Question 14.
Find the integral of \(\frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}\)
Solution:

Question 15.
Find the integral of \(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution:
Given integral is \(\frac{1}{\cos (x+a) \cos (x+b)}\) Multiplying and dividing by sin(a – b), we get