TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type

Question 1.
Is the function f, defined by f(x) = \(\left\{\begin{array}{l}
x^2 \text { if } x \leq 1 \\
x \text { if } x>1
\end{array}\right.\), continuous on R. [May ’15 (AP), ’11]
Solution:
We find the limit at a = 1

∴ f is continuous at x = 1
Hence f is continuous on R.

Question 2.
Is f defined by f(x) = \(\begin{cases}\frac{\sin 2 x}{x} & \text { if } x \neq 0 \\ 1, & \text { if } x=0\end{cases}\), continuous on ‘0’? [May ’12, ’10, ’04; Mar. ’05]
Solution:
Given, f(x) = \(\begin{cases}\frac{\sin 2 x}{x} & \text { if } x \neq 0 \\ 1, & \text { if } x=0\end{cases}\)
Take a = 0

∴ f is discontinuous at x = 0.

Question 3.
Check the continuity of the following function at 2.
f(x) = \(\begin{cases}\frac{1}{2}\left(x^2-4\right) & \text { if } 0<x<2 \\ 0, & \text { if } x=2 \\ 2-8 x^{-3}, & \text { if } x>2\end{cases}\). [Mar. ’19 (TS): Mar. ’17 (AP): May ’15 (TS), ’08]
Solution:

Question 4.
Check the continuity of f given by f(x) = \(f(x)= \begin{cases}\frac{x^2-9}{x^2-2 x-3} & \text { if } 0<x<5 \text { and } x \neq 3 \\ 1.5 & \text { if } x=3\end{cases}\) at the point 3. [Mar. ’15 (AP), ’14, ’13, ’02; May ’04]
Solution:

Question 5.
Prove that the functions sin x and cos x are continuous on R. [May ’08]
Solution:
(i) Let f(x) = sin x and a ∈ R
\(\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} \sin x\) = sin a = f(a)
∴ \(\lim _{x \rightarrow a} f(x)\) = f(a)
∴ f is continuous at x = a
∴ Since a is arbitrary, f is continuous on R.
(ii) Let g(x) = cos x and a ∈ R
\(\lim _{x \rightarrow a} g(x)=\lim _{x \rightarrow a} \cos x\) = cos a = g(a)
∴ \(\lim _{x \rightarrow a} g(x)\) = g(a)
∴ g is continuous at x = a
∴ since a is arbitrary, g is continuous on R.

Question 6.
Find real constants a, b so that the function f is given by f(x) = \(\begin{cases}\sin x & \text { if } x \leq 0 \\ x^2+\mathbf{a} & \text { if } 0<x<1 \\ \mathbf{b x}+3 & \text { if } 1 \leq x \leq 3 \\ -3 & \text { if } x>3\end{cases}\) is continuous on R. [Mar. ’18 (AP & TS); May ’13]
Solution:
Given, f(x) = \(\begin{cases}\sin x & \text { if } x \leq 0 \\ x^2+\mathbf{a} & \text { if } 0<x<1 \\ \mathbf{b x}+3 & \text { if } 1 \leq x \leq 3 \\ -3 & \text { if } x>3\end{cases}\)
Since f is continuous on R.
f is continuous at 0, 3.

Since f is continuous at x = 3 then
LHL = RHL
3b + 3 = -3
3b = -3 – 3
b = -2
∴ a = 0, b = -2

Question 7.
Show that f(x) = \(\left\{\begin{array}{cl}
\frac{\cos a x-\cos b x}{x^2} & \text { if } x \neq 0 \\
\frac{1}{2}\left(b^2-a^2\right) & \text { if } x=0
\end{array}\right.\) where a and b are real constants, is continuous at ‘0’. [Mar. ’17 (TS), ’13(old), ’09; May ’14; B.P.]
Solution:

Some More Maths 1B Limits and Continuity Important Questions Short Answer Type

Question 8.
Find \(\lim _{x \rightarrow 3} \frac{x^3-6 x^2+x}{x^2-9}\)
Solution:

Question 9.
Find \(\lim _{x \rightarrow 3} \frac{x^3-3 x^2}{x^2-5 x+6}\)
Solution:

Question 10.
Compute \(\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3}\)
Solution:
Given, \(\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3}\)

Question 11.
Compute \(\lim _{x \rightarrow 3} \frac{x^2-8 x+15}{x^2-9}\). [Mar. ’16 (AP & TS)]
Solution:

Question 12.
If f(x) = \(-\sqrt{25-x^2}\) then find \(\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}\)
Solution:

Question 13.
Compute \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\), b ≠ 0, a ≠ b. [Mar. ’18 (TS)]
Solution:
Given, \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\)
Now dividing the numerator and denominator by x, we get

Question 14.
Compute \(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\). [Mar. ’18 (AP); May ’15 (TS)]
Solution:

Question 15.
Evaluate \(\lim _{x \rightarrow 1} \frac{\log _e x}{x-1}\).
Solution:

Question 16.
Compute \(\lim _{x \rightarrow 3} \frac{e^x-e^3}{x-3}\)
Solution:

Question 17.
Compute \(\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}\)
Solution:

Question 18.
Compute \(\lim _{x \rightarrow 1} \frac{(2 x-1)(\sqrt{x}-1)}{2 x^2+x-3}\)
Solution:

Question 19.
Compute \(\lim _{x \rightarrow 0} \frac{\log _e(1+5 x)}{x}\)
Solution:

Question 20.
Compute \(\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{8}}-(1-x)^{\frac{1}{8}}}{x}\)
Solution:

Question 21.
Compute \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x}\)
Solution:

Question 22.
Compute \(\lim _{x \rightarrow 0} \frac{\sec x-1}{x^2}\)
Solution:

Question 23.
Compute \(\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}\), n ≠ 0.
Solution:

Question 24.
Compute \(\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}\). [May ’14]
Solution:
Given, \(\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}\)

Question 25.
Compute \(\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{\sin ^3 x}\)
Solution:

Question 26.
Compute \(\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}\)
Solution:
Given, \(\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}\)

Question 27.
Compute \(\lim _{x \rightarrow \infty} \frac{x^2-\sin x}{x^2-2}\). [May ’16 (AP)]
Solution:

Question 28.
Compute \(\lim _{x \rightarrow 3} \frac{x^2+3 x+2}{x^2-6 x+9}\). [Mar. ’19 (AP)]
Solution:

Question 29.
Compute \(\lim _{x \rightarrow \infty} \frac{3 x^2+4 x+5}{2 x^3+3 x-7}\). [May ’15 (AP)]
Solution:

Question 30.
Compute \(\lim _{x \rightarrow \infty} \frac{6 x^2-x+7}{x+3}\).
Solution:

Question 31.
Compute \(\lim _{x \rightarrow \infty} \frac{x^2+5 x+2}{2 x^2-5 x+1}\). [May ’14; Mar. ’17 (AP)]
Solution:

Question 32.
Compute \(\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{4}{x^2-4}\right]\)
Solution:

Question 33.
Compute \(\lim _{x \rightarrow-\infty} \frac{5 x^3+4}{\sqrt{2 x^4+1}}\)
Solution:

Question 34.
Compute \(\lim _{x \rightarrow \infty} \frac{2+\cos ^2 x}{x+2007}\)
Solution:
Given, \(\lim _{x \rightarrow \infty} \frac{2+\cos ^2 x}{x+2007}\)
We know that
-1 ≤ cos x ≤ 1
0 ≤ cos²x ≤ 1
2 + 0 ≤ 2 + cos²x ≤ 2 + 1
2 ≤ 2 + cos²x ≤ 3
\(\frac{2}{x+2007} \leq \frac{2+\cos ^2 x}{x+2007} \leq \frac{3}{x+2007}\)

Question 35.
Compute \(\lim _{x \rightarrow-\infty} \frac{6 x^2-\cos 3 x}{x^2+5}\)
Solution:
We know that
-1 ≤ cos x ≤ 1
-1 ≥ cos 3x ≥ 1
1 ≥ -cos 3x ≥ -1
-1 ≤ -cos 3x ≤ 1

Question 36.
Show that f, given by f(x) = \(\frac{\mathbf{x}-|\mathbf{x}|}{\mathbf{x}}\) (x ≠ 0), is continuous on R – {0}.
Solution:

∴ f is discontinuous at x = 0.
∴ Hence f is continuous on R – {0}.

Question 37.
If f is a function defined by f(x) = \(\begin{cases}\frac{x-1}{\sqrt{x}-1} & \text { if } x>1 \\ 5-3 x & \text { if }-2 \leq x \leq 1 \\ \frac{6}{x-10} & \text { if } x<-2\end{cases}\) then discuss the continuity of ‘f’.
Solution:

Question 38.
If f, given by f(x) = \(\begin{cases}\mathbf{k}^2 x-k & \text { if } \mathbf{k} \geq 1 \\ 2 & \text { if } x<1\end{cases}\) is a continuous function on R, then find the values of k. [Mar. ’15 (TS)]
Solution:
Given, \(\begin{cases}\mathbf{k}^2 x-k & \text { if } \mathbf{k} \geq 1 \\ 2 & \text { if } x<1\end{cases}\)
∴ f is continuous on R
∴ f is continuous at x = 1
at x = 1, LHL = RHL = f(1) ………(1)

From (1), LHL = RHL
⇒ 2 = k² – k
⇒ k² – k – 2 = 0
⇒ k² – 2k + k – 2 = 0
⇒ k(k – 2) + 1(k – 2) = 0
⇒ (k – 2)(k + 1) = 0
⇒ k – 2 = 0 (or) k + 1 = 0
⇒ k = 2 (or) k = -1

Question 39.
Check the continuity of ‘f’ given by f(x) = \(\left\{\begin{array}{rlr}
4-x^2, & \text { if } & x \leq 0 \\
\mathbf{x}-5, & \text { if } & 0 4 x^2-9, & \text { if } & 1<x<2 \\
3 x+4, & \text { if } & x \geq 2
\end{array}\right.\) at points x = 0, 1, 2. [Mar. ’16 (TS)]
Solution: