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Inverse Trigonometric Functions

JEE Mathematics · 73 questions · Page 4 of 8 · Click an option or "Show Solution" to reveal answer

Q31
Let x∗y=x2+y3x * y = {x^2} + {y^3} and (x∗1)∗1=x∗(1∗1)(x * 1) * 1 = x * (1 * 1). Then a value of 2sin⁡−1(x4+x2−2x4+x2+2)2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right) is :
A π4{\pi \over 4}
B π3{\pi \over 3}
C π2{\pi \over 2}
D π6{\pi \over 6}
Correct Answer
Option B
Solution

The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷).

It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem.

In this case, the operation "*" is defined by the equation x∗y=x2+y3x * y = x^2 + y^3, which means if you have two numbers xx and yy, then the result of applying the "*" operation to them is x2+y3x^2 + y^3.

The problem also specifies an additional rule for this operation: (x∗1)∗1=x∗(1∗1)(x * 1) * 1 = x * (1 * 1), which needs to be taken into account when solving the problem.

This is a type of "associativity" condition.

Given,

x ∗ y=x2+y3x\, * \,y = {x^2} + {y^3}

∴\therefore

x ∗ 1=x2+13=x2+1x\, * \,1 = {x^2} + {1^3} = {x^2} + 1

Now,

(x ∗ 1) ∗ 1=(x2+1) ∗ 1(x\, * \,1)\, * \,1 = ({x^2} + 1)\, * \,1
⇒(x ∗ 1) ∗ 1=(x2+1)2+13\Rightarrow (x\, * \,1)\, * \,1 = {({x^2} + 1)^2} + {1^3}
⇒(x ∗ 1) ∗ 1=x4+1+2x2+1\Rightarrow (x\, * \,1)\, * \,1 = {x^4} + 1 + 2{x^2} + 1

Also,

x ∗ (1 ∗ 1)x\, * \,(1\, * \,1)
=x ∗ (12+13)= x\, * \,({1^2} + {1^3})
=x ∗ 2= x\, * \,2
=x2+23= {x^2} + {2^3}
=x2+8= {x^2} + 8

Given that,

(x ∗ 1) ∗ 1=x ∗ (1 ∗ 1)(x\, * \,1)\, * \,1 = x\, * \,(1\, * \,1)

∴\therefore

x4+1+2x2+1=x2+8{x^4} + 1 + 2{x^2} + 1 = {x^2} + 8
⇒x4+x2−6=0\Rightarrow {x^4} + {x^2} - 6 = 0
⇒x4+3x2−2x2−6=0\Rightarrow {x^4} + 3{x^2} - 2{x^2} - 6 = 0
⇒x2(x2+3)−2(x3+3)=0\Rightarrow {x^2}({x^2} + 3) - 2({x^3} + 3) = 0
⇒(x2+3)(x2−2)=0\Rightarrow ({x^2} + 3)({x^2} - 2) = 0
⇒x2=2, −3\Rightarrow {x^2} = 2,\, - 3

[

x2=−3{x^2} = -3

not possible as square of anything should be always positive] ∴\therefore

x2=2{x^2} = 2

∴\therefore Now,

2sin⁡−1(x4+x2−2x4+x2+2)2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)
=2sin⁡−1(22+2−222+2+2)= 2{\sin ^{ - 1}}\left( {{{{2^2} + 2 - 2} \over {{2^2} + 2 + 2}}} \right)
=2sin⁡−1(48)= 2{\sin ^{ - 1}}\left( {{4 \over 8}} \right)
=2sin⁡−1(12)= 2{\sin ^{ - 1}}\left( {{1 \over 2}} \right)
=2×π6= 2 \times {\pi \over 6}
=π3= {\pi \over 3}
Q32
The domain of the function cos⁡−1(2sin⁡−1(14x2−1)π){\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right) is :
A R−{−12,12}R - \left\{ { - {1 \over 2},{1 \over 2}} \right\}
B (−∞,−1]∪[1,∞)∪{0}( - \infty , - 1] \cup [1,\infty ) \cup \{ 0\}
C (−∞,−12)∪(12,∞)∪{0}\left( { - \infty ,{{ - 1} \over 2}} \right) \cup \left( {{1 \over 2},\infty } \right) \cup \{ 0\}
D (−∞,−12]∪[12,∞)∪{0}\left( { - \infty ,{{ - 1} \over {\sqrt 2 }}} \right] \cup \left[ {{1 \over {\sqrt 2 }},\infty } \right) \cup \{ 0\}
Correct Answer
Option D
Solution
−1≤2sin⁡−1(14x2−1)π≤1- 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi } \le 1
⇒−π2≤sin⁡−1(14x2−1)≤π2\Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right) \le {\pi \over 2}
⇒−1≤14x2−1≤1\Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1

∴\therefore

14x2−1+1≥0{1 \over {4{x^2} - 1}} + 1 \ge 0
⇒1+4x2−14x2−1≥0\Rightarrow {{1 + 4{x^2} - 1} \over {4{x^2} - 1}} \ge 0
⇒4x24x2−1≥0\Rightarrow {{4{x^2}} \over {4{x^2} - 1}} \ge 0

⇒\Rightarrow

4x2(2x+1)(2x−1)≥0{{4{x^2}} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0

...... (1) ∴\therefore

x∈(−α,−12)∪{0}∪(12,α)x \in \left( { - \alpha , - {1 \over 2}} \right) \cup \{ 0\} \cup \left( {{1 \over 2},\alpha } \right)

.....(2) And

14x2−1−1≤0{1 \over {4{x^2} - 1}} - 1 \le 0
⇒1−4x2+14x2−1≤0\Rightarrow {{1 - 4{x^2} + 1} \over {4{x^2} - 1}} \le 0
⇒2−4x24x2−1≤0\Rightarrow {{2 - 4{x^2}} \over {4{x^2} - 1}} \le 0
⇒2x2−14x2−1≥0\Rightarrow {{2{x^2} - 1} \over {4{x^2} - 1}} \ge 0

⇒\Rightarrow

(2x+1)(2x−1)(2x+1)(2x−1)≥0{{\left( {\sqrt 2 x + 1} \right)\left( {\sqrt 2 x - 1} \right)} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0

...... (3)

x∈(−α,−12)∪(−12,12)∪(12,α)x \in \left( { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left( { - {1 \over 2},{1 \over 2}} \right) \cup \left( {{1 \over {\sqrt 2 }},\alpha } \right)

.....(4) From (3) and (4), we get ∴\therefore

x∈[−α,−12)∪[12,α)∪{0}x \in \left[ { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left[ {{1 \over {\sqrt 2 }},\alpha } \right) \cup \{ 0\}
Q33
The value of cot(cos⁡ec−153+tan⁡−123)cot\left( {\cos e{c^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right) is :
A 617{{6 \over 17}}
B 317{{3 \over 17}}
C 417{{4 \over 17}}
D 517{{5 \over 17}}
Correct Answer
Option A
Solution

Given,

Cot(sosec⁡−153+tan⁡−123)Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)
=cot⁡(tan⁡−134+tan⁡−123)= \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)
=cot(tan⁡−134+231−34−23)= cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)
=cot⁡(tan⁡−1(176))= \cot \left( {{{\tan }^{ - 1}}\left( {{{17} \over 6}} \right)} \right)
=cot⁡(cot⁡−1617)= \cot \left( {{{\cot }^{ - 1}}{6 \over {17}}} \right)
=617= {6 \over {17}}
Q34
The sum of the absolute maximum and absolute minimum values of the function f(x)=tan⁡−1(sin⁡x−cos⁡x)f(x)=\tan ^{-1}(\sin x-\cos x) in the interval [0,π][0, \pi] is :
A 0
B tan⁡−1(12)−π4\tan ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)-\dfrac{\pi}{4}
C cos⁡−1(13)−π4\cos ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)-\dfrac{\pi}{4}
D −π12\dfrac{-\pi}{12}
Correct Answer
Option C
Solution
f(x)=tan⁡−1(sin⁡x−cos⁡x),     [0,π]f(x) = {\tan ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ]

Let

g(x)=sin⁡x−cos⁡xg(x) = \sin x - \cos x
=2sin⁡(x−π4)= \sqrt 2 \sin \left( {x - {\pi \over 4}} \right)

and

x−π4∈[−π4, 3π4]x - {\pi \over 4} \in \left[ {{{ - \pi } \over 4},\,{{3\pi } \over 4}} \right]

∴\therefore

g(x)∈[−1, 2]g(x) \in \left[ { - 1,\,\sqrt 2 } \right]

and

tan⁡−1x{\tan ^{ - 1}}x

is an increasing function ∴\therefore

f(x)∈[tan⁡−1(−1), tan⁡−12]f(x) \in \left[ {{{\tan }^{ - 1}}( - 1),\,{{\tan }^{ - 1}}\sqrt 2 } \right]
∈[−π4, tan⁡−12]\in \left[ { - {\pi \over 4},\,{{\tan }^{ - 1}}\sqrt 2 } \right]

∴\therefore Sum of

fmax⁡{f_{\max }}

and

fmin⁡=tan⁡−12−π4{f_{\min }} = {\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}
=cos⁡−1(13)−π4= {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) - {\pi \over 4}
Q35
If α=cos⁡−1(35)\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right), β=tan⁡−1(13)\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right) where 0<α,β<π20 < \alpha ,\beta < {\pi \over 2} , then α\alpha - β\beta is equal to :
A tan⁡−1(914){\tan ^{ - 1}}\left( {{9 \over {14 }}} \right)
B sin⁡−1(9510){\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)
C cos⁡−1(9510){\cos ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)
D tan⁡−1(9510){\tan ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)
Correct Answer
Option B
Solution

Here

cos⁡α=35\cos \alpha = {3 \over 5}

∴\therefore

tan⁡α=43\tan \alpha = {4 \over 3}

and

tan⁡β=13\tan \beta = {1 \over 3}

We know,

tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡α.tan⁡β\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}

=

43−131+43.13{{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}

=

913{9 \over {13}}

∴\therefore

(α−β)\left( {\alpha - \beta } \right)

=

tan⁡−1(913){\tan ^{ - 1}}\left( {{9 \over {13}}} \right)

=

sin⁡−1(9510){\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)
Q36
For α,β,γ≠0\alpha, \beta, \gamma \neq 0, if sin⁡−1α+sin⁡−1β+sin⁡−1γ=π\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi and (α+β+γ)(α−γ+β)=3αβ(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta, then γ\gamma equals
A 3\sqrt{3}
B 32\dfrac{\sqrt{3}}{2}
C 12\dfrac{1}{\sqrt{2}}
D 3−122\dfrac{\sqrt{3}-1}{2 \sqrt{2}}
Correct Answer
Option B
Solution

Let

sin⁡−1α=A,sin⁡−1β=B,sin⁡−1γ=C\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=C
A+B+C=π(α+β)2−γ2=3αβα2+β2−γ2=αβα2+β2−γ22αβ=12⇒cos⁡C=12sin⁡C=γcos⁡C=1−γ2=12γ=32\begin{aligned} & \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\ & (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\ & \alpha^2+\beta^2-\gamma^2=\alpha \beta \\ & \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\ & \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\ & \sin \mathrm{C}=\gamma \\ & \cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2} \\ & \gamma=\frac{\sqrt{3}}{2} \end{aligned}
Q37
If the inverse trigonometric functions take principal values then cos⁡−1(310cos⁡(tan⁡−1(43))+25sin⁡(tan⁡−1(43))){\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right) is equal to :
A 0
B π4{\pi \over 4}
C π3{\pi \over 3}
D π6{\pi \over 6}
Correct Answer
Option C
Solution
cos⁡−1(310cos⁡(tan⁡−1(43))+25sin⁡(tan⁡−1(43))){\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)
=cos⁡−1(310 . 35+25 . 45)= {\cos ^{ - 1}}\left( {{3 \over {10}}\,.\,{3 \over 5} + {2 \over 5}\,.\,{4 \over 5}} \right)
=cos⁡−1(12)=π3= {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}
Q38
Considering only the principal values of inverse trigonometric functions, the number of positive real values of xx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan ^{-1}(x)+\tan ^{-1}(2 x)=\dfrac{\pi}{4} is :
A more than 2
B 2
C 0
D 1
Correct Answer
Option D
Solution
tan⁡−1x+tan⁡−12x=π4;x>0⇒tan⁡−12x=π4−tan⁡−1x\begin{aligned} & \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} ; x>0 \\ & \Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x \end{aligned}

Taking tan both sides

⇒2x=1−x1+x⇒2x2+3x−1=0x=−3±9+88=−3±178\begin{aligned} & \Rightarrow 2 \mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}} \\ & \Rightarrow 2 \mathrm{x}^2+3 \mathrm{x}-1=0 \\ & \mathrm{x}=\frac{-3 \pm \sqrt{9+8}}{8}=\frac{-3 \pm \sqrt{17}}{8} \end{aligned}

Only possible

x=−3+178x=\frac{-3+\sqrt{17}}{8}
Q39
If the domain of the function sin⁡−1(3x−222x−19)+log⁡e(3x2−8x+5x2−3x−10)\sin ^{-1}\left(\dfrac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\dfrac{3 x^2-8 x+5}{x^2-3 x-10}\right) is (α,β](\alpha, \beta], then 3α+10β3 \alpha+10 \beta is equal to:
A 95
B 100
C 97
D 98
Correct Answer
Option C
Solution
sin⁡−1(3x−222x−19)+log⁡e(3x2−8x+5x2−3x−10)−1≤3x−222x−19≤13x−222x−19+1≥0 and 3x−222x−19−1≤03x−22+2x−192x−19≥0 and 3x−22−2x+192x−19≤0⇒5x−412x−19≥0 and x−32x−19≤0x∈(−∞,415]∪(192,∞) and x∈[3,192)\begin{aligned} & \sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _e\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right) \\ & -1 \leq \frac{3 x-22}{2 x-19} \leq 1 \\ & \frac{3 x-22}{2 x-19}+1 \geq 0 \text{ and } \frac{3 x-22}{2 x-19}-1 \leq 0 \\ & \frac{3 x-22+2 x-19}{2 x-19} \geq 0 \text{ and } \frac{3 x-22-2 x+19}{2 x-19} \leq 0 \\ & \Rightarrow \frac{5 x-41}{2 x-19} \geq 0 \text{ and } \frac{x-3}{2 x-19} \leq 0 \\ & x \in\left(-\infty, \frac{41}{5}\right] \cup\left(\frac{19}{2}, \infty\right) \text{ and } x \in\left[3, \frac{19}{2}\right) \end{aligned}
⇒x∈[3,415].... (1) and, 3x2−8x+5x2−3x−10>0(3x−5)(x−1)(x−5)(x−2)>0⇒x∈(−∞,−2)∪[1,53]∪(5,∞)…\begin{aligned} & \Rightarrow \quad x \in\left[3, \frac{41}{5}\right] \quad \text{.... (1)}\\ & \text{ and, } \frac{3 x^2-8 x+5}{x^2-3 x-10}>0 \\ & \frac{(3 x-5)(x-1)}{(x-5)(x-2)}>0 \\ & \Rightarrow \quad x \in(-\infty,-2) \cup\left[1, \frac{5}{3}\right] \cup(5, \infty) \ldots \end{aligned}

Taking intersection of individual domains

x∈(5,415]x \in\left(5, \frac{41}{5}\right]
⇒α=5 and β=415⇒3α+10β=15+82=97\begin{aligned} & \Rightarrow \quad \alpha=5 \text{ and } \beta=\frac{41}{5} \\ & \Rightarrow 3 \alpha+10 \beta=15+82 \\ & =97 \end{aligned}
∴\therefore \quad

Option (4) is correct

Q40
Let SS be the set of all solutions of the equation cos⁡−1(2x)−2cos⁡−1(1−x2)=π,x∈[−12,12]\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]. Then ∑x∈S2sin⁡−1(x2−1)\sum\limits_{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right) is equal to :
A π−2sin⁡−1(34)\pi-2 \sin ^{-1}\left(\dfrac{\sqrt{3}}{4}\right)
B π−sin⁡−1(34)\pi-\sin ^{-1}\left(\dfrac{\sqrt{3}}{4}\right)
C −2π3\dfrac{-2 \pi}{3}
D None
Correct Answer
Option D
Solution
cos⁡−1(2x)=π+2cos⁡−11−x2 Since cos⁡−1(2x)∈[0,π] R.H.S. ≥ππ+2cos⁡−11−x2=π⇒cos⁡−11−x2=0⇒1−x2=1⇒x=0 but at x=0cos⁡−1(2x)=cos⁡−1(0)=π2\begin{aligned} & \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\ & \text{ Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\ & \text{ R.H.S. } \geq \pi \\\\ & \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\ & \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\ & \Rightarrow \sqrt{1-\mathrm{x}^2}=1 \\\\ & \Rightarrow \mathrm{x}=0 \\\\ & \text{ but at } \mathrm{x}=0 \\\\ & \cos ^{-1}(2 \mathrm{x})=\cos ^{-1}(0)=\frac{\pi}{2} \end{aligned}

∴\therefore No solution possible for given equation.

x∈ϕ\mathrm{x} \in \phi
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