Practice Start Test

Inverse Trigonometric Functions

JEE Mathematics · 73 questions · Page 7 of 8 · Click an option or "Show Solution" to reveal answer

Q61
If cos⁡−1x−cos⁡−1y2=α,{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha , then 4x2−4xycos⁡α+y24{x^2} - 4xy\cos \alpha + {y^2} is equal to :
A 2sin⁡2α2\sin 2\alpha
B 44
C 4sin⁡2α4{\sin ^2}\alpha
D −4sin⁡2α-4{\sin ^2}\alpha
Correct Answer
Option C
Solution

As we know,

cos⁡−1A−cos⁡−1B{\cos ^{ - 1}}A - {\cos ^{ - 1}}B
=cos⁡−1(AB+1−A2.1−B2)= {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)

Given,

cos⁡−1x−cos⁡−1y2=α{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha
⇒cos⁡−1(x.y2+1−x2.1−y24)=α\Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2}} \over 4}} } \right) = \alpha
⇒xy2+1−x21−y24=cos⁡ x\Rightarrow {{xy} \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{y{}^2} \over 4}} = \cos \,x
⇒(cos⁡x−xy2)2=(1−x2)(1−y24)\Rightarrow {\left( {\cos x - {{xy} \over 2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - {{{y^2}} \over 4}} \right)
⇒cos⁡2+x2y24−2.cos⁡x.xy2\Rightarrow {\cos ^2} + {{{x^2}{y^2}} \over 4} - 2.\cos x.{{xy} \over 2}
                           =1−x2−y24+x2y24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {x^2} - {{{y^2}} \over 4} + {{{x^2}{y^2}} \over 4}
⇒x2+y24−xy cos⁡x=1−cos⁡2x\Rightarrow {x^2} + {{{y^2}} \over 4} - xy\,\cos x = 1 - {\cos ^2}x
⇒x2+y24−xycos⁡x=sin⁡2x\Rightarrow {x^2} + {{{y^2}} \over 4} - xy\cos x = {\sin ^2}x
⇒4x2+y2−4xycos⁡x=4sin⁡2x\Rightarrow 4{x^2} + y{}^2 - 4xy\cos x = 4{\sin ^2}x
Q62
The value of cot⁡−1(1+tan⁡2(2)−1tan⁡(2))−cot⁡−1(1+tan⁡2(12)+1tan⁡(12)) \cot^{-1} \left( \dfrac{\sqrt{1 + \tan^2(2)} - 1}{\tan(2)} \right) - \cot^{-1} \left( \dfrac{\sqrt{1 + \tan^2\left(\dfrac{1}{2}\right)} + 1}{\tan\left(\dfrac{1}{2}\right)} \right) is equal to
A π−32 \pi - \dfrac{3}{2}
B π+52 \pi + \dfrac{5}{2}
C π−54 \pi - \dfrac{5}{4}
D π+32 \pi + \dfrac{3}{2}
Correct Answer
Option C
Solution

cot⁡−1(∣sec⁡2∣−1tan⁡2)−cot⁡−1(∣sec⁡12∣+1tan⁡12)\cot ^{-1}\left(\dfrac{|\sec 2|-1}{\tan 2}\right)-\cot ^{-1}\left(\dfrac{\left|\sec \dfrac{1}{2}\right|+1}{\tan \dfrac{1}{2}}\right)

=cot⁡−1(−1−cos⁡2sin⁡2)−cot⁡−1(1+cos⁡12sin⁡12)=π−cot⁡−1(cot⁡1)−cot⁡−1(cot⁡14)=π−1−14=π−54\begin{aligned} & =\cot ^{-1}\left(\frac{-1-\cos 2}{\sin 2}\right)-\cot ^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right) \\ & =\pi-\cot ^{-1}(\cot 1)-\cot ^{-1}\left(\cot \frac{1}{4}\right) \\ & =\pi-1-\frac{1}{4}=\pi-\frac{5}{4} \end{aligned}
Q63
If x = sin−-1(sin10) and y = cos−-1(cos10), then y −- x is equal to :
A 0
B 10
C 7π\pi
D π\pi
Correct Answer
Option D
Solution

x = sin−-1 sin 10 = 3π\pi −- 10 y = cos−-1cos 10 = 4π\pi −- 10 y −- x = (4π\pi −- 10) −- (3π\pi −- 10) = π\pi

Q64
Given that the inverse trigonometric function assumes principal values only. Let x,yx, y be any two real numbers in [−1,1][-1,1] such that cos⁡−1x−sin⁡−1y=α,−π2≤α≤π\cos ^{-1} x-\sin ^{-1} y=\alpha, \dfrac{-\pi}{2} \leq \alpha \leq \pi. Then, the minimum value of x2+y2+2xysin⁡αx^2+y^2+2 x y \sin \alpha is
A 0
B −-1
C 12\dfrac{1}{2}
D −12\dfrac{-1}{2}
Correct Answer
Option A
Solution
cos⁡−1x−π2+cos⁡−1y=αcos⁡−1x+cos⁡−1y=π2+α∵α∈[−π2,π] then π2+α∈(0,3π2)cos⁡−1(xy−1−x21−y2)=π2+αxy−1−x21−y2=−sin⁡αxy+sin⁡α=1−x21−y2x2y2+sin⁡2α+2xysin⁡α=1−x2−y2+x2y2x2+y2+2xysin⁡α⏟E=cos⁡2α\begin{aligned} & \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\ & \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\ & \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\ & \text{ then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\ & \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\ & x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\ & x y+\sin \alpha=\sqrt{1-x^2} \sqrt{1-y^2} \\ & x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \\ & \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha \end{aligned}

Now, minimum value of

EE

is 0.

Q65
The set of all values of k for which (tan⁡−1x)3+(cot⁡−1x)3=kπ3, x∈R{({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R, is the interval :
A [132,78)\left[ {{1 \over {32}},{7 \over 8}} \right)
B (124,1316)\left( {{1 \over {24}},{{13} \over {16}}} \right)
C [148,1316]\left[ {{1 \over {48}},{{13} \over {16}}} \right]
D [132,98)\left[ {{1 \over {32}},{9 \over 8}} \right)
Correct Answer
Option A
Solution
(tan⁡−1x)3+(cot⁡−1x)3=kπ3{({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}

Let

f(t)=t3+(π2−t)3f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}

Where

t=tan⁡−1xt = {\tan ^{ - 1}}x

;

x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)
=t3+(π2)3−3π2t4+3π2t2−t3= {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3\pi } \over 2}{t^2} - {t^3}
f(t)=3π2t2−3π24 . t+π38f(t) = {{3\pi } \over 2}{t^2} - {{3{\pi ^2}} \over 4}\,.\,t + {{{\pi ^3}} \over 8}

This is a quadratic equation of t. Here, coefficient of t2 term is

3π2{{3\pi } \over 2}

which is > 0. ∴\therefore It is a upward parabola. Now,

f′(t)=3πt−3π24f'(t) = 3\pi t - {{3{\pi ^2}} \over 4}
f′′(t)=3π>0f''(t) = 3\pi > 0

∴\therefore

3πt−3π24=03\pi t - {{3{\pi ^2}} \over 4} = 0
⇒t=π4\Rightarrow t = {\pi \over 4}

(minima) ∴\therefore vertex of graph at

π4{\pi \over 4}

∴\therefore Minimum value at

π4{\pi \over 4}

and maximum value at −-

π2{\pi \over 2}

. ∴\therefore

f(π4)=π364+(π2−π4)3=π332f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}
f(−π2)=−π38+π3f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}
=7π38= {{7{\pi ^3}} \over 8}

∴\therefore

kπ3∈[π332, 7π38)k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)
⇒k∈[132, 78)\Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)
Q66
The value of tan⁡(2tan⁡−1(35)+sin⁡−1(513))\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right) is equal to :
A −18169{{ - 181} \over {69}}
B 22021{{220} \over {21}}
C −29176{{ - 291} \over {76}}
D 15163{{151} \over {63}}
Correct Answer
Option B
Solution
2tan⁡−1(35)=tan⁡−1(6/51−925)=tan⁡−1(651625)=tan⁡−11582{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}

∴\therefore

2tan⁡−1(35)+sin⁡−1(513)=tan⁡−1(158)+tan⁡−1(512)2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) = {\tan ^{ - 1}}\left( {{{15} \over 8}} \right) + {\tan ^{ - 1}}\left( {{5 \over {12}}} \right)
=tan⁡−1(158+5121−158,512)= {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right)
=tan⁡−1(180+4021)=tan⁡−1(22021)= {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)
Q67
Let tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2),{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right), where ∣x∣<13.\left| x \right| < {1 \over {\sqrt 3 }}. Then a value of yy is :
A 3x−x31+3x2{{3x - {x^3}} \over {1 + 3{x^2}}}
B 3x+x31+3x2{{3x + {x^3}} \over {1 + 3{x^2}}}
C 3x−x31−3x2{{3x - {x^3}} \over {1 - 3{x^2}}}
D 3x+x31−3x2{{3x + {x^3}} \over {1 - 3{x^2}}}
Correct Answer
Option C
Solution

Given,

tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2){\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)
⇒tan⁡−1y=tan⁡−1(x+2x1−x21−x(2x1−x2))\Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)
                          =tan⁡−1(x−x3+2x1−x2−2x2)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)

∴\therefore

   \,\,\,
tan⁡−1y=tan⁡−1(3x−x21−3x2){\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)
⇒y=3x−x31−3x2\Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}
Q68
Considering only the principal values of the inverse trigonometric functions, the domain of the function f(x)=cos⁡−1(x2−4x+2x2+3)f(x)=\cos ^{-1}\left(\dfrac{x^{2}-4 x+2}{x^{2}+3}\right) is :
A (−∞,14]\left(-\infty, \dfrac{1}{4}\right]
B [−14,∞)\left[-\dfrac{1}{4}, \infty\right)
C (−1/3,∞)(-1 / 3, \infty)
D (−∞,13]\left(-\infty, \dfrac{1}{3}\right]
Correct Answer
Option B
Solution
−1≤x2−4x+2x2+3≤1- 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1
⇒−x2−3≤x2−4x+2≤x2+3\Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3
⇒2x2−4x+5≥0\Rightarrow 2{x^2} - 4x + 5 \ge 0

&

−4x≤1- 4x \le 1
x∈Rx \in R

&

x≥−14x \ge - {1 \over 4}

So domain is

[−14,∞)\left[ { - {1 \over 4},\infty } \right)
Q69
Considering only the principal values of inverse functions, the set A = { x ≥ \ge 0: tan−-1(2x) + tan−-1(3x) = π4{\pi \over 4}}
A contains two elements
B contains more than two elements
C is an empty set
D is a singleton
Correct Answer
Option D
Solution

tan−-1(2x) + tan−-1(3x) = π\pi/4

⇒  5x1−6x2\Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}

= 1 ⇒\Rightarrow 6x2 + 5x −- 1 = 0 x = −-1 or x =

16{1 \over 6}

x =

16{1 \over 6}

∵\because x > 0

Q70
All x satisfying the inequality (cot–1 x)2– 7(cot–1 x) + 10 > 0, lie in the interval :
A (cot 2, ∞\infty )
B (–∞\infty , cot 5) ∪ \cup (cot 2, ∞\infty )
C (cot 5, cot 4)
D (– ∞\infty , cot 5) ∪ \cup (cot 4, cot 2)
Correct Answer
Option A
Solution

cot−-1 x > 5, cot−-1 x < 2 ⇒\Rightarrow x < cot5, x > cot2

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