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Inverse Trigonometric Functions

JEE Mathematics · 73 questions · Page 1 of 8 · Click an option or "Show Solution" to reveal answer

Q1
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x) is equal to :
A 0
B 1
C 12\dfrac{1}{2}
D −12-\dfrac{1}{2}
Correct Answer
Option A
Solution
cos⁡−1x−2sin⁡−1x=cos⁡−12x{\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x

For Domain :

x∈[−12,12]x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]
cos⁡−1x−2(π2−cos⁡−1x)=cos⁡−1(2x){\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)
⇒cos⁡−1x+2cos⁡−1x=π+cos⁡−12x\Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}2x
⇒cos⁡(3cos⁡−1x)=−cos⁡(cos⁡−12x)\Rightarrow \cos (3{\cos ^{ - 1}}x) = - \cos ({\cos ^{ - 1}}2x)
⇒4x3=x\Rightarrow 4{x^3} = x
⇒x=3, ± 12\Rightarrow x = 3,\, \pm \,{1 \over 2}
Q2
If (sin⁡−1x)2−(cos⁡−1x)2=a{({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a; 0 < x < 1, a ≠\ne 0, then the value of 2x2 −- 1 is :
A cos⁡(4aπ)\cos \left( {{{4a} \over \pi }} \right)
B sin⁡(2aπ)\sin \left( {{{2a} \over \pi }} \right)
C cos⁡(2aπ)\cos \left( {{{2a} \over \pi }} \right)
D sin⁡(4aπ)\sin \left( {{{4a} \over \pi }} \right)
Correct Answer
Option B
Solution

Given

a=(sin⁡−1x)2−(cos⁡−1x)2a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}
=(sin⁡−1x+cos⁡−1x)(sin⁡−1x−cos⁡−1x)= ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)
=π2(π2−2cos⁡−1x)= {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)
⇒2cos⁡−1x=π2−2aπ\Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }
⇒cos⁡−1(2x2−1)=π2−2aπ\Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }
⇒2x2−1=cos⁡(π2−2aπ)\Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)
Q3
The domain of the function f(x) = sin⁡−1(∣x∣+5x2+1){\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right) is (– ∞\infty , -a]∪ \cup [a, ∞\infty ). Then a is equal to :
A 17−12{{\sqrt {17} - 1} \over 2}
B 1+172{{1 + \sqrt {17} } \over 2}
C 172+1{{\sqrt {17} } \over 2} + 1
D 172{{\sqrt {17} } \over 2}
Correct Answer
Option B
Solution

f(x) =

sin⁡−1(∣x∣+5x2+1){\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)

∴\therefore

−1≤∣x∣+5x2+1≤1- 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}} \le 1

Since |x| + 5 & x2 + 1 is always positive So

∣x∣+5x2+1≥0{{\left| x \right| + 5} \over {{x^2} + 1}} \ge 0

That means this inequality

−1≤∣x∣+5x2+1- 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}}

always right. So we can ignore it. So for domain :

∣x∣+5x2+1≤1{{\left| x \right| + 5} \over {{x^2} + 1}} \le 1

⇒\Rightarrow

x2−∣x∣−4≥0{{x^2} - \left| x \right| - 4 \ge 0}

⇒\Rightarrow

(∣x∣−1−172)(∣x∣−1+172)≥0\left( {\left| x \right| - {{1 - \sqrt {17} } \over 2}} \right)\left( {\left| x \right| - {{1 + \sqrt {17} } \over 2}} \right) \ge 0

⇒\Rightarrow |x| ≥\ge

1+172{{{1 + \sqrt {17} } \over 2}}

or |x| ≤\le

1−172{{{1 - \sqrt {17} } \over 2}}

As

1−172{{{1 - \sqrt {17} } \over 2}}

is < 0 and |x| always ≥\ge 0. So |x| ≤\le

1−172{{{1 - \sqrt {17} } \over 2}}

not possible. ∴\therefore |x| ≥\ge

1+172{{{1 + \sqrt {17} } \over 2}}

x

∈\in
(−∞,−1+172)∪(1+172,∞)\left( { - \infty , - {{1 + \sqrt {17} } \over 2}} \right) \cup \left( {{{1 + \sqrt {17} } \over 2},\infty } \right)

So, a =

1+172{{{1 + \sqrt {17} } \over 2}}
Q4
A possible value of tan⁡(14sin⁡−1638)\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right) is :
A 7−1\sqrt 7 - 1
B 17{1 \over {\sqrt 7 }}
C 22−12\sqrt 2 - 1
D 122{1 \over {2\sqrt 2 }}
Correct Answer
Option B
Solution
tan⁡(14sin⁡−1638)\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)
sin⁡−1(638)=θ{\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta
sin⁡θ=638\sin \theta = {{\sqrt {63} } \over 8}
cos⁡θ=18\cos \theta = {1 \over 8}
2cos⁡2θ2−1=182{\cos ^2}{\theta \over 2} - 1 = {1 \over 8}
cos⁡2θ2=916{\cos ^2}{\theta \over 2} = {9 \over {16}}
cos⁡θ2=34\cos {\theta \over 2} = {3 \over 4}
1−tan⁡2θ41+tan⁡2θ4=34{{1 - {{\tan }^2}{\theta \over 4}} \over {1 + {{\tan }^2}{\theta \over 4}}} = {3 \over 4}
tan⁡θ4=17\tan {\theta \over 4} = {1 \over {\sqrt 7 }}
Q5
If sin⁡1xa=cos⁡−1xb=tan⁡−1yc{{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}; 0<x<10 < x < 1, then the value of cos⁡(πca+b)\cos \left( {{{\pi c} \over {a + b}}} \right) is :
A 1−y22y{{1 - {y^2}} \over {2y}}
B 1−y2yy{{1 - {y^2}} \over {y\sqrt y }}
C 1−y21 - {y^2}
D 1−y21+y2{{1 - {y^2}} \over {1 + {y^2}}}
Correct Answer
Option D
Solution
sin⁡−1xa=cos⁡−1xb=tan⁡−1yc{{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}
sin⁡−1xa=cos⁡−1xb=sin⁡−1x+cos⁡−1xa+b=π2(a+b){{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}

Now,

tan⁡−1yc=π2(a+b){{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}
2tan⁡−1y=πca+b2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}
⇒cos⁡(πca+b)=cos⁡(2tan⁡−1y)=1−y21+y2\Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}
Q6
If cot−-1(α\alpha) = cot−-1 2 + cot−-1 8 + cot−-1 18 + cot−-1 32 + ...... upto 100 terms, then α\alpha is :
A 1.02
B 1.03
C 1.01
D 1.00
Correct Answer
Option C
Solution
cot⁡−1(α)=cot−12+cot⁡−18+cot⁡−118+cot⁡−132+....100{\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100

terms

=tan⁡−112+tan⁡−118+tan⁡−1118+tan⁡−1132+....100= {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100

term

=∑k=1100tan⁡−112k2= \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \over {2{k^2}}}}
=∑k=1100tan⁡−124k2=∑k=1ntan⁡−1(2k+1)−(2k−1)1+(2k−1)(2k+1)= \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{2 \over {4{k^2}}} = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}{{(2k + 1) - (2k - 1)} \over {1 + (2k - 1)(2k + 1)}}} }
=∑k=1100(tan⁡−1(2k+1)−tan⁡−1(2k−1))= \sum\limits_{k = 1}^{100} {\left( {{{\tan }^{ - 1}}(2k + 1) - {{\tan }^{ - 1}}(2k - 1)} \right)}
=tan⁡−1201−tan⁡−11= {\tan ^{ - 1}}201 - {\tan ^{ - 1}}1
=tan⁡−1200202= {\tan ^{ - 1}}{{200} \over {202}}
=cot⁡−1(1.01)= {\cot ^{ - 1}}(1.01)

Hence

α=1.01\alpha = 1.01
Q7
The domain of the function f(x)=sin⁡−1(3x2+x−1(x−1)2)+cos⁡−1(x−1x+1)f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right) is :
A [0,14]\left[ {0,{1 \over 4}} \right]
B [−2,0]∪[14,12][ - 2,0] \cup \left[ {{1 \over 4},{1 \over 2}} \right]
C [14,12]∪{0}\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\}
D [0,12]\left[ {0,{1 \over 2}} \right]
Correct Answer
Option C
Solution
f(x)=sin⁡−1(3x2+x−1(x−1)2)+cos⁡−1(x−1x+1)f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)
−1≤x−1x+1≤1⇒0≤x<∞- 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty

.... (1)

−1≤3x2+x−1(x−1)2≤1⇒x∈[−14,12]∪{0}- 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left[ {{{ - 1} \over 4},{1 \over 2}} \right] \cup \{ 0\}

.... (2) (1) & (2) ⇒\Rightarrow Domain =

[14,12]∪{0}\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\}
Q8
Let M and m respectively be the maximum and minimum values of the function f(x) = tan−-1 (sin x + cos x) in [0,π2]\left[ {0,{\pi \over 2}} \right], then the value of tan(M −- m) is equal to :
A 2+32 + \sqrt 3
B 2−32 - \sqrt 3
C 3+223 + 2\sqrt 2
D 3−223 - 2\sqrt 2
Correct Answer
Option D
Solution

Let g(x) = sin x + cos x =

2\sqrt 2

sin

(x+π4)\left( {x + {\pi \over 4}} \right)

g(x)

∈\in
[1,2]\left[ {1,\sqrt 2 } \right]

for x

∈\in

[0, π\pi/2] f(x) = tan−-1 (sin x + cos x)

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

tan

(tan⁡−12−π4)=2−11+2×2−12−1=3−22({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2
Q9
If ∑r=150tan⁡−112r2=p\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p} , then the value of tan p is :
A 101102{{101} \over {102}}
B 5051{{50} \over {51}}
C 100
D 5150{{51} \over {50}}
Correct Answer
Option B
Solution
∑r=150tan⁡−1(24r2)=∑r=150tan⁡−1((2r+1)−(2r−1)1+(2r+1)(2r−1))\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} }

=

∑r=150tan⁡−1(2r+1)−tan⁡−1(2r−1)\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)}

=

tan⁡−1(101)−tan⁡−11=tan⁡−15051{\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}
Q10
sin⁡1(sin⁡2π3)+cos⁡−1(cos⁡7π6)+tan⁡−1(tan⁡3π4){\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right) is equal to :
A 11π12{{11\pi } \over {12}}
B 17π12{{17\pi } \over {12}}
C 31π12{{31\pi } \over {12}}
D −-3π4{{3\pi } \over {4}}
Correct Answer
Option A
Solution
sin⁡−1(32)+cos⁡−1(−32)+tan⁡−1(−1){\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)
=π3+5π6−π4= {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}
=4π+10π−3π12=11π12= {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}
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