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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6
Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
The value of sin-1(cos x), 0 ≤ x ≤ π is …………
(a) π – x
(b) x – \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{2}\) – x
(d) π – x
Solution:
(c) \(\frac{\pi}{2}\) – x
Hint:
Question 2.
If sin-1 + sin-1 y = \(\frac{2 \pi}{3}\); then cos-1x + cos-1y is equal to ………….
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) π
Solution:
(b) \(\frac{\pi}{3}\)
Hint:
Question 3.
……………
(a) 2π
(b) π
(c) 0
(d) tan-1\(\frac{12}{65}\)
Solution:
(c) 0
Hint:
Question 4.
If sin-1x = 2 sin-1 α has a solution, then ……………
Solution:
(a) |α| ≤ \(\frac{1}{\sqrt{2}}\)
Hint:
Question 5.
sin-1 (cos x) = \(\frac{\pi}{2}\) – x is valid for ……………..
(a) -π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
(d) \(-\frac{\pi}{4}\) ≤ x ≤ \(\frac{3 \pi}{4}\)
Solution:
(b) 0 ≤ x ≤ π
Question 6.
If sin-1 x + sin-1 y + sin-1 z = \(\frac{3 \pi}{2}\), the value of x2017 + y2018 + z2019 – \(\frac{9}{x^{101}+y^{101}+z^{101}}\) is ……………….
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
The maximum value of sin-1 x is \(\frac{\pi}{2}\) and sin-1 1 = \(\frac{\pi}{2}\)
Here it is given that
sin-1 x + sin-1 y + sin-1 z = \(\frac{3 \pi}{2}\)
⇒ x = y = z = 1
and so 1 + 1 + 1 – \(\frac{9}{1+1+1}\) = 3 – 3 = 0
Question 7.
If cot-1 x = \(\frac{2 \pi}{5}\) for some x ∈ R, the value of tan-1 x is …………
(a) \(-\frac{\pi}{10}\)
(b) \(\frac{\pi}{5}\)
(c) \(\frac{\pi}{10}\)
(d) \(-\frac{\pi}{5}\)
Solution:
(c) \(\frac{\pi}{10}\)
Hint:
Question 8.
The domain of the function defined by f(x) = sin-1 \(\sqrt{x-1}\) is …………….
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Solution:
(a) [1, 2]
Hint:
The domain for sin-1 x is [0, 1]
So \(\sqrt{x-1}\) = 0 ⇒ x – 1 = 0 ⇒ x = 1
\(\sqrt{x-1}\) = 1 ⇒ x – 1 = 0 ⇒ x = 2
∴ The domain is [1, 2]
Question 9.
If x = \(\frac{1}{5}\), the value of cos(cos-1 x + 2 sin-1 x) is …………….
Solution:
(d) \(-\frac{1}{5}\)
Hint:
Question 10.
\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\) is equal to …………
Solution:
(d) tan-1\(\frac{1}{2}\)
Hint:
Question 11.
If the function f(x) = sin-1(x2 – 3), then x belongs to …………..
(a) [1, -1]
(b) [\(\sqrt{2}\), 2]
(c) \([-2,-\sqrt{2}] \cup[\sqrt{2}, 2]\)
(d) \([-2,-\sqrt{2}] \cap[\sqrt{2}, 2]\)
Solution:
(c) \([-2,-\sqrt{2}] \cup[\sqrt{2}, 2]\)
Hint:
f(x) = sin-1(x2 – 3)
Domain of sin-1 (x) is [-1, 1]
⇒ -1 ≤ x2 – 3 ≤ 1 ⇒ 2 ≤ x2 ≤ 4
⇒ \(\sqrt{2}\) ≤ x ≤ 2 ⇒ \(\sqrt{2}\) ≤ |x| ≤ 2
x ∈ \([-2,-\sqrt{2}] \cup[\sqrt{2}, 2]\)
Question 12.
If cot-1 2 and cot-1 3 are two angles of a triangle, then the third angle is …………..
Solution:
(b) \(\frac{3 \pi}{4}\)
Hint:
Question 13.
\(\sin ^{-1}\left(\tan \frac{\pi}{4}\right)-\sin ^{-1}(\sqrt{\frac{3}{x}})=\frac{\pi}{6}\). Then x is a root of the equation …………..
(a) x2 – x – 6 = 0
(b) x2 – x – 12 = 0
(c) x2 + x – 12 = 0
(d) x2 + x – 6 = 0
Solution:
(b) x2 – x – 12 = 0
Hint:
Question 14.
sin-1(2 cos2 x – 1) + cos-1(1 – 2 sin2 x) = ……………
Solution:
(a) \(\frac{\pi}{2}\)
Hint:
2 cos2x – 1 = cos 2x
1 – 2 sin2 x = cos 2x
∴ sin-1 x(cos 2x) + cos-1(cos 2x) = \(\frac{\pi}{2}\) (∵ sin-1 x + cos-1 x = \(\frac{\pi}{2}\))
Question 15.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})\) = u, then cos 2u is equal to …………..
(a) tan2 α
(b) 0
(c) -1
(d) tan 2α
Solution:
(c) -1
Hint:
cot-1 x + tan-1 x = \(\frac{\pi}{2}\) ⇒ u = \(\frac{\pi}{2}\) so 2u = π
∴ cos 2u = cos π = -1
Question 16.
If |x| ≤ 1, then 2tan-1 x – sin-1 \(\frac{2 x}{1+x^{2}}\) is equal to ………….
(a) tan-1 x
(b) sin-1 x
(c) 0
(d) π
Solution:
(c) 0
Hint:
Let x = tan θ so \(\frac{2 x}{1+x^{2}}\) = sin 2θ.
Now 2 tan-1(tanθ) – sin-1(sin 2θ) = 2θ – 2θ = 0
Question 17.
The equation tan-1 x – cot-1 x = tan-1 \(\left(\frac{1}{\sqrt{3}}\right)\) has …………..
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Solution:
(b) unique solution
Hint:
Question 18.
If sin-1 x + cot-1 \(\left(\frac{1}{2}\right)=\frac{\pi}{2}\), then x is equal to …………
Solution:
(b) \(\frac{1}{\sqrt{5}}\)
Hint:
sin-1 x + cot-1 \(\left(\frac{1}{2}\right)=\frac{\pi}{2}\)
Question 19.
If sin-1\(\frac{x}{5}\) + cosec-1\(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x is ………
(a) 4
(b) 5
(c) 2
(d) 3
Solution:
(d) 3
Hint:
Question 20.
sin(tan-1), |x| < 1 is equal to ………….
Solution:
(d) \(\frac{x}{\sqrt{1+x^{2}}}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6 Additional Questions
Question 1.
Find the principal value of
Solution:
Question 2.
Find the principal value of
Solution:
Question 3.
Solution:
Question 4.
Evaluate
Solution:
Question 5.
Evaluate
Solution:
Question 6.
Question 7.
Solution:
Question 8.
Solution:
x = \(\frac{1}{6}\)
Question 9.
Find the values of each of the following:
Solution:
Question 10.
Solve for x:
Solution:
Question 11.
Prove:
Question 12.
Evaluate: sin(tan-1 x + cot-1 x)
Question 13.
The value of sin-1(1) + sin-1(0) is …….
Solution:
(a) \(\frac{\pi}{2}\)
Hint:
Question 14.
Solution:
(d) \(\frac{\pi}{2}\)
Hint:
Question 15.
tan-1x + cot-1x = ……..
(a) 1
(b) – π
(c) \(\frac{\pi}{2}\)
(d) π
Solution:
(c) \(\frac{\pi}{2}\)
Hint:
Question 16.
Solution:
(a) \(\frac{-\pi}{2}\)
Hint:
Question 17.
Solution:
(b) \(\frac{\pi}{2}\)
Hint:
Question 18.
Solution:
(a) \(\sin ^{-1} \frac{1}{\sqrt{2}}\)
Hint:
Question 19.
Solution:
(d) 2π
Hint:
Question 20.
Solution:
(c) 2π
Hint: