Practice AP Inter 2nd Year Maths Study Material Chapter 2 Inverse Trigonometric Functions MCQ to identify your strengths and weak areas.
AP Inter 2nd Year Maths Inverse Trigonometric Functions MCQ
I. Select the correct option from the given choices.
Question 1.
If sin-1x = y , then
1) 0 ≤ y ≤ π
2) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
3) 0 < y < π
4) \(-\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)
Solution:
2) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
We know that range of the principle value of \(\sin ^{-1} x=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Given that sin-1 x = y, ∴ \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
Question 2.
tan-1\(\sqrt{3}\) – sec-1(-2) is equal to
1) π
2) \(-\frac{\pi}{3}\)
3) \(\frac{\pi}{3}\)
4) \(\frac{2 \pi}{3}\)
Solution:
2) \(-\frac{\pi}{3}\)
Formula: sec-1(-x) = -sec-1x;
tan-1\(\sqrt{3}\) – sec-1(-2) = tan-1\(\sqrt{3}\) – (π – sec-12) = 60° – 180° = -60° = \(\frac{-\pi}{3}\)
Question 3.
cos-1(cos\(\frac{7 \pi}{6}\)) is equal to
1) \(\frac{7 \pi}{6}\)
2) \(\frac{5 \pi}{6}\)
3) \(\frac{\pi}{6}\)
4) \(\frac{\pi}{6}\)
Solution:
2) \(\frac{5 \pi}{6}\)
\(\frac{7 \pi}{6}=\pi+\frac{\pi}{6}\) and cos-1(-x) = π – cos-1x
∴ cos-1 \(\cos \frac{7 \pi}{6}=\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[-\cos \frac{\pi}{6}\right]=\pi-\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\)
Question 4.
sin\(\left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)\) is equal to
1) \(\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{1}{4}\)
4) 1
Solution:
4) 1
Formula: sin-1(-x) = -sin-1(x)
\(\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=\sin \left[60^{\circ}+\sin \left(\frac{1}{2}\right)\right]\) = sin[60° + 30°] = sin 90° = 1
Question 5.
tan-1\(\sqrt{3}\) – cot-1(\(-\sqrt{3}\)) is equal to
1) π
2) \(\frac{-\pi}{2}\)
3) 0
4) 2\(\sqrt{3}\)
Solution:
2) \(\frac{-\pi}{2}\)
Formula: cot-1(-x) = π – cot-1(x)
tan-1\(\sqrt{3}\) – cot-1(\(-\sqrt{3}\)) = tan-1(\(\sqrt{3}\)) – (π – cot-1(\(\sqrt{3}\)\frac{-\pi}{2})) = 60° – 180° + 30° = -90° = \(\frac{-\pi}{2}\)
Question 6.
sin(tan-1 x), |x| < 1 is equal to
1) \(\frac{x}{\sqrt{1-x^2}}\)
2) \(\frac{1}{\sqrt{1-x^2}}\)
3) \(\frac{1}{\sqrt{1+x^2}}\)
4) \(\frac{x}{\sqrt{1+x^2}}\)
Solution:
4) \(\frac{x}{\sqrt{1+x^2}}\)
tan-1 x = θ ⇒ \(\frac{x}{1}=\tan \theta \Rightarrow \sin \theta=\frac{A B}{B C}=\frac{x}{\sqrt{1+x^2}}\)
Question 7.
If sin-1(1 – x) – 2sin-1 x = \(\frac{\pi}{2}\), then x is equal to
1) 0, \(\frac{1}{2}\)
2) 1, \(\frac{1}{2}\)
3) 0
4) \(\frac{1}{2}\)
Solution:
3) 0
Formula: sin(90° + θ) = cosθ; cos2θ = 1 – 2sin2θ
G.E = \(\sin ^{-1}(1-x)=\frac{\pi}{2}+2 \sin ^{-1} x \Rightarrow(1-x)=\sin \left[\frac{\pi}{2}+2 \sin ^{-1} x\right]\)
⇒ 1 – x = cos(2sin-1 x) = 1 – 2(sin(sin-1 x))2 = 1 – 2x2
∴ 1 – x = 1 – 2x2 ⇒ 2x2 – x = 0 ⇒ x(2x – 1) = 0 x = 1,\(\frac{1}{2}\)
Question 8.
sin\(\left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{-1}{2}\right)\right]\) is equal to:
1) 1
2) \(\frac{1}{2}\)
3) \(\frac{1}{3}\)
4) \(\frac{1}{4}\)
Solution:
2) \(\frac{1}{2}\)
sin-1\(\left(-\frac{1}{2}\right)\) = -30 and cos-1(-x) = π – cos-1x
sin(60° – 30°) = sin30° = \(\frac{1}{2}\)
Question 9.
The principle value of \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) is
1) \(\frac{\pi}{12}\)
2) π
3) \(\frac{\pi}{3}\)
4) \(\frac{\pi}{6}\)
Solution:
1) \(\frac{\pi}{12}\)
cos-1\(\left(\frac{1}{2}\right)\) – sin-1\(\left(\frac{1}{\sqrt{2}}\right)\) = 60° – 45° = 15° = \(\frac{\pi}{12}\)
Question 10.
The principle value of \(\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)\)
1) \(\frac{\pi}{8}\)
2) \(\frac{3\pi}{8}\)
3) \(\frac{-\pi}{8}\)
4) \(\frac{-3\pi}{8}\)
Solution:
1) \(\frac{\pi}{8}\)
\(\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{8}\right)\right)=\tan ^{-1}\left(\tan \left(\frac{\pi}{8}\right)\right)=\frac{\pi}{8}\)
Question 11.
The domain of the function cos-1(2x – 3) is
1) [-1, 1]
2) (1, 2)
3) (-1, 1)
4) [1, 2]
Solution:
4) [1, 2]
Formula: cos-1x is defined for x ∈ (-1, 1)
-1 ≤ (2x – 3) ≤ 1 ⇒ (3 – 1) ≤ 2x ≤ (3 + 1) ⇒ 2 ≤ 2x ≤ 4 ⇒ 1 ≤ x ≤ 2 x ∈ [1, 2]
Question 12.
The value of \(\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)\)
1) \(\frac{\pi}{10}\)
2) \(\frac{3 \pi}{5}\)
3) \(-\frac{\pi}{10}\)
4) \(-\frac{3 \pi}{5}\)
Solution:
3) \(-\frac{\pi}{10}\)
Formula: cosθ = sin(90° – θ)
\(\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]=\sin ^{-1}\left[\sin \left(\frac{-\pi}{10}\right)\right]=-\frac{\pi}{10}\)
Question 13.
The principle value of tan-1(-1) is
1) \(\frac{\pi}{4}\)
2) \(-\frac{\pi}{4}\)
3) \(\frac{\pi}{2}\)
4) \(\frac{\pi}{3}\)
Solution:
2) \(-\frac{\pi}{4}\)
Formula: tan-1(-x) = -tan-1x
tan-1(-1) = -tan-1(1) = \(-\frac{\pi}{4}\)
Question 14.
The principle value of \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\) is
1) \(\frac{13 \pi}{6}\)
2) \(\frac{\pi}{2}\)
3) \(\frac{\pi}{3}\)
4) \(\frac{\pi}{6}\)
Solution:
4) \(\frac{\pi}{6}\)
\(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\)
Question 15.
The simplest form of \(\tan ^{-1}\left[\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}}}\right.\) is
1) \(\frac{\pi}{4}-\frac{\pi}{2}\)
2) \(\frac{\pi}{4}+\frac{\pi}{2}\)
3) \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)
4) \(\frac{\pi}{4}+\frac{\pi}{2} \cos ^{-1} x\)
Solution:
3) \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)
Put x = cos2θ ⇒ 2θ = cos-1x ⇒ θ = \(\frac{1}{2}\)cos-1x
\(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}=\frac{\sqrt{2 \cos ^2 \theta}-\sqrt{2 \sin ^2 \theta}}{\sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta}}\)
= \(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\tan \left(\frac{\pi}{4}-\theta\right)\)
∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)