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CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Inverse Trigonometric Functions
Quiz 7
If
tan
−
1
x
=
π
10
for some
x
∈
R
, then the value of
cot
−
1
x
is
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0%
π
5
0%
2
π
5
0%
3
π
5
0%
4
π
5
Explanation
(B) is the correct answer.
We know that
tan
−
1
x
+
cot
−
1
x
=
π
2
.
⇒
cot
−
1
x
=
π
2
−
tan
−
1
x
⇒
cot
−
1
x
=
π
2
−
π
10
=
4
π
10
=
2
π
5
The value of
sin
−
1
{
cos
(
43
π
5
)
}
is
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0%
3
π
5
0%
−
7
π
5
0%
π
10
0%
−
π
10
Explanation
(D) is the correct answer.
We have
sin
−
1
(
cos
40
π
+
3
π
5
)
=
sin
−
1
{
cos
(
8
π
+
3
π
5
}
}
=
sin
−
1
(
cos
3
π
5
)
=
sin
−
1
{
sin
(
π
2
−
3
π
5
)
}
=
sin
−
1
(
sin
(
−
π
10
)
)
=
−
π
10
The value of
cot
(
sin
−
1
x
)
is
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0%
√
1
+
x
2
x
0%
x
√
1
+
x
2
0%
1
x
0%
√
1
−
x
2
x
Explanation
(D) is the correct answer.
Let
sin
−
1
x
=
θ
,
Then
sin
θ
=
x
⇒
csc
θ
=
1
x
⇒
csc
2
θ
=
1
x
2
⇒
1
+
cot
2
θ
=
1
x
2
⇒
cot
θ
=
√
1
−
x
2
x
The domain of
y
=
cos
−
1
(
x
2
−
4
)
is
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0%
[
3
,
5
]
0%
[
0
,
π
]
0%
[
−
√
5
,
−
√
3
]
∩
[
−
√
5
,
√
3
]
0%
[
−
√
5
,
−
√
3
]
∪
[
√
3
,
5
]
Explanation
(D) is the correct answer.
Given
y
=
cos
−
1
(
x
2
−
4
)
⇒
cos
y
=
x
2
−
4
∴
( since
-1 \le \cos y \le 1)
\Rightarrow 3 \le x^2 \le 5
\Rightarrow \sqrt 3 \le |x| \le \sqrt 5
\Rightarrow x \in [ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, \sqrt 5]
Hence the domain of
y
is
[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]
The domain of the function
y=\sin^{-1}(-x^2)
is
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[0, 1]
0%
(0, 1)
0%
[-1, 1]
0%
\phi
Explanation
(C) is the correct answer,
Given
y=\sin^{-1}(-x^2) \Rightarrow \sin y =-x^2
\therefore\ -1 \le -x^2 \le 1
( since
-1 \le \sin y \le 1)
\Rightarrow 1 \ge x^2 \ge -1
\Rightarrow 0 \le x^2 \le 1
\Rightarrow |x| \le 1
\Rightarrow-1 \le x \le 1
Hence the domain is
[-1, 1]
The value of
\sin (2\sin^{-1}(0.6))
is
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0%
0.48
0%
0.96
0%
1.2
0%
\sin 1.2
Explanation
(B) is the correct answer.
Let
\sin^{-1}(0.6) =\theta \Rightarrow\sin \theta =0.6
.
\therefore\ \sin (2\sin^{-1}(0.6))\\=\sin ( 2\theta )\\=2\sin \theta \cos \theta \\=2 \sin\theta\sqrt{1-\sin^2\theta}\\=2(0.6)(\sqrt{1-{0.6}^2})\\=2(0.6)(0.8)=0.96
The value of the expression
\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]
is
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0%
0
0%
1
0%
\dfrac{1}{\sqrt 3}
0%
\sqrt{\dfrac{2}{3}}
Explanation
(D) is the correct answer.
Let
P=\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]
\\=\sin \left[ \cot^{-1}\left( \cos \dfrac {\pi}{4}\right) \right] \qquad\left[\because \tan^{-1}1=\dfrac{\pi}4\right]\\=\sin \left[ \cot^{-1}\dfrac{1}{\sqrt 2}\right] \qquad\left[\because \cos\dfrac{\pi}4=\dfrac1{\sqrt2}\right]\\
Let,
\cot^{-1}\dfrac1{\sqrt2}=y\Rightarrow \cot y=\dfrac1{\sqrt2}
\\ \Rightarrow \sin y=\sqrt{\dfrac23}\Rightarrow y=\sin^{-1}\sqrt{\dfrac23}
\therefore P=\sin \left[ \sin^{-1}\sqrt{\dfrac 23}\right] =\sqrt{\dfrac 23}
The value of
\tan^2 ( \sec^{-1}2)+\cot^2 (\csc^{-1}3)
is
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0%
5
0%
11
0%
13
0%
15
Explanation
(B) is the correct answer.
We have
\tan^2 ( \sec^{-1}2) \cot^2 ( \csc^{-1}3)
=\sec^2 (\sec^{-1}2)-1+\csc^2 (\csc^{-1}3)-1\qquad[\because \tan^2\theta=\sec^2\theta-1,\ \ \ \cot^2\theta=\csc^2\theta-1]
=(2)^2-1+(3)^2-1=11
If
\alpha \le 2\sin^{-1}x+\cos^{-1}x \le \beta
, then
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\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{\pi}{2}
0%
\alpha =0, \beta =\pi
0%
\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{3\pi}{2}
0%
\alpha =0, \beta =2\pi
Explanation
(B) is the correct answer.
We know that
\dfrac {-\pi}{2}\le \sin^{-1}x \le \dfrac {\pi}{2}\\
\Rightarrow \dfrac {-\pi}{2}+\dfrac {\pi}{2} \le \sin^{-1}x + \dfrac {\pi}{2} \le \dfrac {\pi}{2}+\dfrac {\pi}{2}\qquad [
Addidng
\dfrac{\pi}2 ]
\\\Rightarrow 0 \le \sin^{-1}x+ (\sin^{-1}x+\cos^{-1}x) \le \pi\qquad\left[\because(\sin^{-1}x+\cos^{-1}x)= \dfrac{\pi}2 \right]\\
\Rightarrow 0 \le 2\sin^{-1}x+\cos^{-1}x \le \pi
\sin^{-1} (1 -x) - 2 \sin ^{-1} x = \dfrac{\pi }{2} ,
then x
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0 , \dfrac{1 }{2}
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1, \dfrac{1}{2}
0%
0
0%
\dfrac{1 }{2}
Explanation
\sin^{-1} (1 -x) - 2 \sin ^{-1} x = \dfrac{\pi }{2}
-2\sin ^{-1} x = \dfrac{\pi }{2} - \sin^{-1} (1 -x)
-2 \sin ^{-1} x = \cos^{-1} (1 -x)
\cos ( -2 \sin^{-1} x) = 1 -x
\cos ( 2 \sin^{-1} x) = 1 -x
\cos (2 sin ^{-1} x) = 1 -x
1 - 2 \sin^{2} (sin^{-1} x) = 1 - x
1 - 2x^{2} = 1- x\Rightarrow
2x^{2} -x = 0 \\x(2x -1) = 0\\ \Rightarrow x = 0 , x = \dfrac{1}{2}
\dfrac{1}{2}
does not satisfy the equation
\therefore x = 0
is the only solution
Choose the correct answer
\cos ^{-1} ( \cos \dfrac{7\pi }{6})
is equal to
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\dfrac{7\pi }{6}
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\dfrac{5\pi }{6}
0%
\dfrac{\pi }{3}
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\dfrac{\pi }{6}
Explanation
\cos^{-1} \cos\left ( \dfrac{7\pi }{6} \right )
\cos^{-1} \left ( \cos\left ( \pi + \dfrac{\pi }{6} \right ) \right ) = \cos^{-1} \left ( - \cos\dfrac{\pi}{6} \right )
=\pi- \cos^{-1} \left ( \cos\dfrac{\pi }{6} \right )
=\pi-\dfrac{\pi}{6}
= \dfrac{5\pi }{6}
Choose the correct answer :
\sin \left ( \dfrac{\pi }{3} - \sin^{-1}\left ( -\dfrac{1}{2} \right ) \right )
is equal to
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\pi
0%
- \dfrac{\pi }{2}
0%
1
0%
2 \sqrt{3}
Explanation
\sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ]
\left [ \sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ] \right ]
\sin \left ( \dfrac{\pi }{3} + \sin^{-1}\left ( \sin \dfrac{\pi }{6} \right ) \right )
\left ( as \sin^{-1} \left ( \dfrac{-1}{2} \right ) = \sin^{-1}\dfrac{1}{2} and \sin \dfrac{\pi }{6} = \dfrac{1}{2} \right )
= \sin \left ( \dfrac{\pi }{3} + \dfrac{\pi }{6} \right )
= \sin\dfrac{\pi }{2} = 1
If
\sin ^{-1} x = y
, then
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0 \leq y \leq \pi
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- \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2}
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0 < y < \pi
0%
- \dfrac{\pi}{2} < y < \dfrac{\pi }{2}
Explanation
pv of
\sin^{-1} x
is
\left [ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right ]
\therefore (B) - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2}
is correct
\tan^{-1} \sqrt{3} - \sec^{-1} (-2)
is equal to
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\pi
0%
- \dfrac{\pi }{3}
0%
\dfrac{\pi }{3}
0%
- \dfrac{2\pi }{3}
Explanation
\tan^{-1} \sqrt{3} = \dfrac{\pi }{3} \epsilon \left [ \dfrac{-\pi }{2} ,\dfrac{\pi }{2} \right ]
\sec^{-1} (-2) = \pi - \dfrac{\pi }{3} = \dfrac{2 \pi}{3} \epsilon [0, \pi]
\therefore \tan^{-1} \sqrt{3} - sec^{-1} (-2) = \dfrac{\pi}{3} - \dfrac{2 \pi}{3}
= -\dfrac{\pi}{3}
\therefore
Option (B) is correct .
Multiple choice Questions :
2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )
Report Question
0%
\dfrac{24}{25}
0%
- \dfrac{24}{25}
0%
- \dfrac{6}{25}
0%
\dfrac{- 6}{25}
Explanation
2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )
= 2 \sqrt{1-\left ( \dfrac{-4}{5} \right )^{2}} \times\left ( \dfrac{-4}{5} \right )
= -\dfrac{8}{5} \sqrt{\dfrac{9}{25}} = \dfrac{-8}{5} \times\dfrac{3}{5} = \dfrac{-24}{25}
Multiple choice Questions :
\sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )=
Report Question
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\dfrac{\pi}{2}
0%
0
0%
\pi
0%
\dfrac{-\pi}{2}
Explanation
\sin ^{1} x + \cos^{-1} x= \dfrac{\pi}{2}
\sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )= \dfrac{\pi}{2}
Multiple choice Questions :
\cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) =
Report Question
0%
\dfrac{4 \pi}{3}
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\dfrac{2 \pi}{3}
0%
\dfrac{- \pi}{3}
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- \pi
Explanation
\cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) = \cos^{1}\cos \left ( \pi + \dfrac{\pi}{3} \right )
= \cos^{-1} \left [ - cos \dfrac{\pi}{3} \right ]
= \pi - cos^{-1} \left ( \cos\dfrac{\pi}{3} \right ) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}
Multiple choice Questions :
The value of
\tan^{-1} (\tan 5)
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2 \pi
0%
5 - 2 \pi
0%
5
0%
\dfrac{2 \pi}{3}
Explanation
Let
2\pi - 5 = \theta
\Rightarrow 5 = 2 \pi - \theta
\tan 5 = \tan ( 2 \pi - \theta )
\tan^{-1} (\tan 5) = \tan^{-1} \tan ( 2 \pi - \theta)
= \tan^{-1}[-\tan \theta]=-\theta
\tan ^{-1}\left ( \dfrac{x}{y} \right ) - \tan^{-1} \dfrac{x - y}{x + y}
is equal to
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\dfrac{\pi }{2}
0%
\dfrac{\pi }{3}
0%
\dfrac{\pi }{4}
0%
\dfrac{-3\pi }{4}
Explanation
\tan^{-1} \left ( \dfrac{x}{y} \right ) - \left ( \tan^{-1}\dfrac{\frac{x}{y}-1}{1+\dfrac{x}{y}} \right )
=\tan^{-1} \left ( \dfrac{x}{y} \right )-\tan ^{-1} \left ( \dfrac{x}{y} \right )+ \tan^{-1} (1)
\tan^{-1} 1 = \dfrac{\pi }{4}
Multiple choice Questions :
If a > b > c ,
\cot^{-1} \left ( \dfrac{1 + ab}{a - b} \right ) + \cot^{-1}\left ( \dfrac{1 + bc}{b - c} \right ) + \cot^{-1} \left ( \dfrac{1 + ac}{c - a} \right )
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0%
0
0%
\pi
0%
2 \pi
0%
\dfrac{\pi}{2}
Explanation
\tan ^{-1}\left ( \dfrac{a - b}{1 + ab} \right ) + \tan^{-1} \left ( \dfrac{b - c}{1 + bc} \right ) + \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right )
= \tan^{-1} a - \tan^{-1} b + \tan^{-1} b - \tan^{-1} c + \pi - \tan^{-1} a + \tan^{-1} c
= \pi \ \ \ \ \ \ \ \ as \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right ) = \pi - \tan^{-1} \left ( \dfrac{a - c}{1 + ac} \right )
If
\sin^{-1}\left(\dfrac{1}{2}\right)=x
, then general value
x
$ is:
Report Question
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2n\pi \pm \dfrac{\pi}{6}
0%
\dfrac{\pi}{6}
0%
n\pi \pm \dfrac{\pi}{6}
0%
n\pi +(-1)^n\dfrac{\pi}{6}
Explanation
Given,
\sin^{-1}\left(\dfrac{1}{2}\right)=x
\Rightarrow \sin x=\dfrac{1}{2}=\sin \dfrac{\pi}{6}
\Rightarrow x=\dfrac{\pi}{6}
\therefore
General value of
x, \theta=n\pi +(-1)^{n}\dfrac{\pi}{6}
Hence, option
(d)
is correct.
If
\tan^{-1}(1)+\cos^{-1}(\dfrac{1}{\sqrt{2}})=\sin^{-1}x
, then value of
x
is
Report Question
0%
-1
0%
0
0%
1
0%
-\dfrac{1}{2}
Explanation
\tan^{-1}(1)+\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)=\sin^{-1}x
\Rightarrow \sin^{-1}x=\dfrac{\pi}{4}+\dfrac{\pi}{4}
\Rightarrow \sin^{-1}x=\dfrac{\pi}{2}
\Rightarrow x=\sin \dfrac{\pi}{2}
\Rightarrow x=1
Hence, option
(c)
is correct.
2\tan(\tan^{-1}x+\tan^{-1}x^3)
is:
Report Question
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\dfrac{2x}{1-x^2}
0%
1+x^2
0%
2x
0%
none of these
Explanation
2\tan (\tan^{-1}x+\tan^{-1}x^{3})
=2\tan \left\{\tan^{-1}\left(\dfrac{x+x^{3}}{1-x\times x^{3}}\right)\right\}
=2\tan \left\{\tan^{-1}\left(\dfrac{x(1+x^{2})}{1-x^{4}}\right)\right\}
=2\tan \left\{\tan^{-1}\dfrac{x(1+x^{2})}{(1-x^{2})(1+x^{2})}\right\}
=2\tan \dfrac{x}{1-x^{2}}=\dfrac{2x}{1-x^{2}}
Hence, option
(a)
is correct.
Value of
\sin^{-1}(\dfrac{\sqrt{3}}{2})+2\cos^{-1}(\dfrac{\sqrt{3}}{2})
is:
Report Question
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\dfrac{\pi}{2}
0%
\dfrac{\pi}{3}
0%
\dfrac{2\pi}{3}
0%
\pi
Explanation
\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+2\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)
=\left[\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\right]+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)
=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2\pi}{3}\left[\because \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\right]
Hence, option
(c)
is correct.
If
\cot^{-1}x+\tan^{-1}\dfrac{1}{3}=\dfrac{\pi}{2}
then
x
is:
Report Question
0%
1
0%
3
0%
\dfrac{1}{3}
0%
none of these
Explanation
\cot^{-1}(x)+\tan^{-1}\left(\dfrac{1}{3}\right)=\dfrac{\pi}{2}
\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{3}\right)
\Rightarrow \cot^{-1}x=\cot^{-1}\dfrac{1}{3}
On comparing
x=\dfrac{1}{3}
Hence, option
(c)
is correct.
If
\tan^{-1}(3x)+\tan^{-1} 2x=\dfrac{\pi}{4}
, then
x
is:
Report Question
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\dfrac{1}{6}
0%
\dfrac{1}{3}
0%
\dfrac{1}{10}
0%
\dfrac{1}{2}
Explanation
\tan^{-1}(3x)+\tan^{-1}(2x)=\dfrac{\pi}{4}
\Rightarrow \tan^{-1}\left\{\dfrac{3x+2x}{1-3x\times 2x}\right\}=\dfrac{\pi}{4}
\Rightarrow \left(\dfrac{5x}{1-6x^{2}}\right)=\tan \dfrac{\pi}{4}
\Rightarrow \dfrac{5x}{1-6x^{2}}=1
\Rightarrow 1-6x^{2}=5x
\Rightarrow 6x^{2}+5x-1=0
\Rightarrow 6x^{2}+6x-x-1=0
\Rightarrow 6x(x+1)-1(x+1)=0
\Rightarrow (x+1)(6x-1)=0
\Rightarrow x=-1, x=\dfrac{1}{6}
Hence, option
(a)
is correct.
lf the equation
\displaystyle \sin^{-1}(x^{2}+x+1)+\cos^{-1}(\lambda x+1)=\frac{\pi}{2}
has exactly two solutions, then
\lambda
can not have the integral value(s)
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0%
-1
0%
0
0%
1
0%
2
Explanation
\displaystyle \sin ^{-1}(x^{2}+x+1)+\cos ^{-1}(\lambda x+1)=\frac{\pi }{2}
\displaystyle \cos ^{-1}(\lambda x+1)=\frac{\pi }{2} -\sin ^{-1}(x^{2}+x+1)
Taking
\cos
on both sides,
\lambda x+1=x^{2}+x+1
x^{2}+(1-\lambda )x=0
x(x+1-\lambda )=0
x=0
or
\lambda =x+1
\sin ^{-1}(x^{2}+x+1)
is defined for
-1 \leq x^{2}+x+1 \leq 1
x^{2}+x+2 \geq 0
is always true.
x(x+1) \leq 0 \Rightarrow x\in [-1,0]
\Rightarrow x+1 \in [0,1]
\lambda \in [0,1]
\lambda \neq -1,2
Also when
\lambda =1
, there is only one solution to the given equation i.e.,
x=0
So,
\lambda\neq 1
.
Assertion(A):
\cos^{-1}x
and
\tan^{-1}x
are positive for all positive real values of
x
in their domain.
Reason(R): The domain of
f(x)=\cos^{-1}x+\tan^{-1}x
is
[-1, 1].
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0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true
Explanation
Assertion :
\cos ^ {-1} x
range is
\left[0,\pi \right]
and domain is
[-1,1]
\therefore \,\cos^{-1}x>0\,
for all
x>0
\tan ^ {-1} x
range is
\left(\dfrac{ -\pi }2 ,\dfrac \pi 2\right)
and range is
R
\forall x >0 \implies \tan ^ {-1} x >0
Reason :
The domain of
\cos ^ {-1} x
is
[-1,1]
The domain of
\tan ^ {-1} x
is
R
So The domain of
\cos ^ {-1}x +\tan ^ {-1}x
is
R \cap [-1,1]
\implies [-1,1]
\sin^{-1}|\sin x|=\sqrt{\sin^{-1}|\sin x|}
then
x=
Report Question
0%
n\pi-1
0%
n\pi
0%
n\pi+1
0%
n\displaystyle \frac{\pi}{2}+1
Explanation
Case - 1
|sin x|=sin x \ \ x\epsilon [2n\pi , (2n+1)\pi ]
sin ^{-1} sin x =\sqrt{sin ^{-1}sin x}
x=\sqrt{x}
\sqrt{x}=1\ \ or \ \ x=0
x=1
Case - 2
|sin x|=-sin x\ \ x\epsilon [(2n-1)\pi , 2n\pi ]
sin ^{-1}(-sin x)=\sqrt{sin ^{-1}(-sin x)}
-x=\sqrt{-x}
x=-1 , x=0
So,
x=n\pi -1 \ for\ n=0
x=n\pi\ for\ n=0
x=n\pi -1 \ for\ n=0
The number of solutions of:
\displaystyle \sin^{-1}(1+b+b^{2}+\ldots.\infty)+\cos^{-1}(a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+\ldots\infty)=\frac{\pi}{2}
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0%
1
0%
2
0%
3
0%
\infty
Explanation
As
\displaystyle \sin ^{ -1 }{ \theta }
are defined for
\displaystyle \theta \in \left[ -1,1 \right]
\therefore
For
\displaystyle \sin ^{ -1 }{ \left( 1+b+{ b }^{ 3 }+...\infty \right) }
\left| b \right| <1
Hence
\displaystyle 1,b,{ b }^{ 2 },...\infty
is
G.P.
of common ratio
b
\displaystyle \therefore 1+b+{ b }^{ 2 }+...+\infty =\frac { 1 }{ 1-b }
Similarly for
\displaystyle \cos ^{ -1 }{ \left( a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } +...\infty \right) }
\displaystyle\left| \frac { a }{ 3 } \right| <1
Hence
\displaystyle 1,-\frac { { a }^{ 2 } }{ 3 } ,\frac { { a }^{ 3 } }{ 9 } ,...\infty
is
G.P.
of common ration
\displaystyle \left( \frac { -a }{ 3 } \right)
\displaystyle \therefore a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } +...\infty =\frac { a }{ 1+\frac { a }{ 3 } } =\frac { 3a }{ 3+a }
Now, as
\displaystyle \sin ^{ -1 }{ \left( x \right) } +\cos ^{ -1 }{ \left( x \right) } =\frac { \pi }{ 2 }
\displaystyle \therefore \frac { 1 }{ 1-b } =\frac { 3a }{ 3+a }
\displaystyle \Rightarrow 3+a=3a-3ab
\displaystyle \Rightarrow 2a-3=3ab
This has infinite solution.
If
(\tan^{-1}x)^{2}+(\cot^{-1}x)^{2} = \displaystyle \frac{5\pi^{2}}{8}
, then
x=
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0%
-1
0%
1
0%
0
0%
2
Explanation
Let
tan^{-1}x=y
Therefore
y^2+(\frac{\pi}{2}-y)^{2}=\frac{5\pi^2}{8}
2y^2-\frac{2\pi y}{2}+\frac{\pi^2}{4}=\frac{5\pi^2}{8}
2y^2-\pi y-\frac{3\pi^2}{8}=0
y=\frac{-\pi}{4}
and
y=\frac{3\pi}{4}
Hence
tan^{-1}(x)=\frac{-\pi}{4}
x=-1
and
tan^{-1}(x)=\frac{3\pi}{4}
x=-1
However only
x=-1
satisfies the above quadratic equation.
cos^{-1} \left (\sqrt{\dfrac{a-x}{a-b}} \right)
=
sin^{-1} \left (\sqrt{\dfrac{x-b}{a-b}}\right)
is possible if
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a>x>b
or
a< x < b
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a=x=b
0%
a > b
and x takes any value
0%
a < b
and x takes any value
Explanation
In
\cos ^{-1}x , x \in [-1,1]
In
\sin ^{-1}x , x\in [-1,1]
\displaystyle -1 \leq \sqrt{\frac{a-x}{a-b}} \leq 1
\displaystyle 0 \leq \sqrt{\frac{a-x}{a-b}} \leq 1
0 \leq a-x \leq a-b
0\geq x-a \geq b-a
[on multiplying by (-ve) sing direction of equality changes]
a \geq x\geq b
and
\displaystyle 0\leq \sqrt{\frac{x-b}{a-b}}\leq 1
0\leq x-b \leq a-b
b \leq x\leq a
x\epsilon [b,a]
The solution set of the equation
\tan^{-1}x -\cot^{-1}x =\cos^{-1}(2-x)
is
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(0,1)
0%
(-1,1)
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[1,3)
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(1,3)
Explanation
Given,
\tan ^{-1}x-\cot ^{-1}x=\cos ^{-1}(2-x) x\in [1,3]
2\tan ^{-1}x-\dfrac{\pi }{2}=\cos ^{-1}(2-x)
\left[\because \tan ^{-1}x+\cot ^{-1}x=\dfrac{\pi }{2}\right]
2 \tan ^{-1}x=\dfrac{\pi }{2}+\cos ^{-1}(2-x)
Take
\sin
on both sides,
\Rightarrow 2\dfrac{x}{(1+x^{2})}=2-x \quad\quad........2\tan^{-1}x=sin^{-1}\dfrac{2x}{1+x^2}
\Rightarrow x^3-2x^{2}+3x-2=0
\Rightarrow x = 1
So, set containing solution is
[1,3)
.
If
\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma =\displaystyle \frac{3\pi}{2}
, then
\alpha\beta+\alpha\gamma+\beta\gamma
is equal to :
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1
0%
0
0%
3
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-3
Explanation
The above expression is true for
\alpha=1
,
\beta=1
and
\gamma=1
Since
\dfrac{-\pi}{2}\leq sin^{-1}(x) \leq \dfrac{\pi}{2}
Hence
\alpha\beta+\beta\gamma+\gamma\alpha
=(1)+(1)+(1)
=3
The number of positive integral solutions of the equation
tan^{-1}x+cot^{-1}y =tan^{-1} 3
is :
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0
0%
1
0%
2
0%
3
Explanation
cot^{-1}y=tan ^{-1}3-tan ^{-1}x
\displaystyle cot^{-1}y=tan ^{-1}(\dfrac{3-x}{1+3x})
Take tan on both sides
\displaystyle \dfrac{1}{y}=\dfrac{3-x}{1+3x}
\displaystyle y=\dfrac{1+3x}{3-x} \Rightarrow y > 0
Put
x=0 , y=\dfrac{1}{3}
\dfrac{1+3x}{3-x} > 0
x=1 , y=2
\dfrac{3x+1}{x=3} < 0
x=2 , y=7
x\epsilon (-\dfrac{1}{3} , 3)
The integral values of
x=0,1,2
The domain of
\displaystyle \mathrm{f}(\mathrm{x})=\cot^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-[\mathrm{x}^{2}]}}\right)
is
(
[\;.]
denotes the greatest integer function)
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(0,\infty)
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\mathrm{R}-\{0\}
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\mathrm{R}-\{\mathrm x:\mathrm{x}\in \mathrm{Z}\}
0%
(-\infty,0)
Explanation
f(x)=\cot^{-1}\left ( \displaystyle \frac{x}{\sqrt{x^2-[x^2]}} \right )
For
f(x)
to be defined,
x^2-[x^2]>0
\therefore x^2
should not be an integer
Hence,
x
should not be an integer.
Hence, Option C is the answer
If
(tan^{-1} x)^2 +(cot ^{-1}x)^2=\displaystyle \frac{5 \pi^2}{8}
, then
x
=
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-1
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0
0%
1
0%
2
Explanation
We have,
tan^{-1}x + cot^{-1}x =\displaystyle\frac{\pi}{2}
The given equation can be written as :
\displaystyle (tan^{-1}x + cot^{-1}x)^2 - 2tan^{-1}x\left ( \frac{\pi}{2}-tan^{-1}x \right )=\frac{5 \pi^2}{8}
\displaystyle 2(tan^{-1}x)^2 - 2\left ( \frac{\pi}{2} \right )tan^{-1}(x)- \frac{3 \pi ^2}{8}=0
\displaystyle tan^{-1}x=-\frac{\pi}{4} or \frac{3\pi}{4}
Hence,
tan^{-1}x =1
x=-1
The number of integral solutions of
sin^{-1}\sqrt{4x-x^{2}-3}+tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}
is
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zero
0%
infinite
0%
four
0%
None of these
Explanation
Given equation is
\sin^{-1}\sqrt{4x-x^{2}-3}+\tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}
\tan^{ -1 }\sqrt { x^{ 2 }-3x+2 }=\displaystyle\frac { \pi }{ 2 } -\sin^{ -1 }\sqrt { 4x-x^{ 2 }-3 }
\Rightarrow \tan^{ -1 }\sqrt { x^{ 2 }-3x+2 } =\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 }
Since,
\sqrt { x^{ 2 }-3x+2 } \ge 0
\Rightarrow \tan ^{ -1 }{ \sqrt { x^{ 2 }-3x+2 } } <\dfrac { \pi }{ 2 }
Also,
\sqrt { 4x-x^{ 2 }-3 } \ge 0
\Rightarrow 0<\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 } \le \dfrac { \pi }{ 2 }
The value of
sin^{-1}(sin2010^{0})+cos^{-1}(cos2010^{0})+tan^{-1}(tan2010^{0})
is
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\frac{\pi }{6}
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\frac{\pi }{3}
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\frac{2\pi }{3}
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\frac{5\pi }{6}
Explanation
2010^{0}
=11\pi+\frac{\pi}{6}
...(i)
Hence
sin^{-1}(sin(2010^{0}))=sin^{-1}(sin(11\pi+\frac{\pi}{6}))
...(from(i))
cos^{-1}(cos(2010^{0}))=cos^{-1}(cos(11\pi+\frac{\pi}{6}))
tan^{-1}(tan(2010^{0}))=tan^{-1}(tan(11\pi+\frac{\pi}{6}))
Therefore
sin^{-1}(sin(11\pi+\frac{\pi}{6}))+cos^{-1}(cos(11\pi+\frac{\pi}{6}))+tan^{-1}(tan(11\pi+\frac{\pi}{6}))
=-\frac{\pi}{6}+\pi-\frac{\pi}{6}+\frac{\pi}{6}
=\frac{5\pi}{6}
The number of solutions of the equation
1+x^{2}+2x\, sin\: (cos^{-1}y)=0
is
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1
0%
2
0%
3
0%
4
Explanation
1+x^{2}+2x sin cos ^{1-}y=0
\Rightarrow 1+x^{2}+2x\sqrt{1-y^{2}}=0
x^{2}+2x\sqrt{1-y^{2}}+1=0
x^{2}+2x\sqrt{1-y^{2}}+1-y^{2}=-y^{2}
(x+\sqrt{1-y^{2}})=-y^{2} \Rightarrow -y^{2} \geq 0
x+\sqrt{1-y^{2}}=0 \space and \space y=0 (x+\sqrt{1-y^{2}})^{2} \geq 0
\rightarrow y=0 , x=-1
only one solution
The value of
\displaystyle \sec^{-1}\left (\displaystyle \frac{1}{1-2x^{2}}\right)+4{\cos^{-1}}\sqrt{\displaystyle \frac{1+x}{2}}
is equal to
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\pi
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2\pi
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\dfrac{\pi}{2}
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None of these
Explanation
Given
\sec ^{-1} \left(\dfrac 1{1-2x^2}\right)+4\cos ^{-1}\sqrt {\dfrac {1+x}2}
Let
x=\cos \theta
\implies \sec ^{-1} \dfrac 1{1-2\cos ^2 x} +4\cos ^{-1}\left( \sqrt {\dfrac {1+\cos \theta }2}\right)
= \sec ^{-1}\left( \dfrac {-1}{\cos 2\theta }\right) +4\cos ^{-1} (\sqrt {\cos ^2\theta /2})
=\pi -\sec ^{-1} (\sec 2\theta )+4 \cos ^{-1} (\cos \theta/2)
=\pi -2\theta +4 \left(\dfrac \theta 2\right)
=\pi -2\theta +2\theta
=\pi
The largest interval lying in
\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )
for which the function
\left [ f(x)=4^{-x^{2}}+\cos^{-1}\left ( \dfrac{x}{2}-1 \right )+\log (\cos x) \right ]
is defined, is-
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[0,\pi ]
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\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )
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\left [- \dfrac{\pi }{4},\dfrac{\pi }{2} \right )
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\left [0,\dfrac{\pi }{2} \right )
Explanation
For given function to be defined
-1 \leq \cfrac{x}{2}-1 \leq 1
and
\cos x > 0
\Rightarrow 0 \leq \cfrac{x}{2} \leq 2
and
\cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}
\Rightarrow 0 \leq x \leq 4
and
\cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}
Taking common of both we get
x \in \left[0,\cfrac{\pi}{2}\right)
The number of real solutions of
tan^{-1} (\sqrt{x(x+1)}+sin^{-1} \displaystyle \sqrt{(x^{2}+x+1)}=\dfrac{\pi}{2}
is
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0
0%
1
0%
2
0%
infinite
Explanation
From the above expression
-1\leq\sqrt{x^2+x+1}\leq 1
and
x(x+1)\geq 0
\Rightarrow 0\le x^2+x+1\le1 \,and \,x^2+x+1\ge1
\Rightarrow x^2+x+1=1
\Rightarrow x^2+x=0
\Rightarrow x(x+1)=0
Hence there will be two solutions. One at
x=-1
and another at
x=0
.
If
x> 0\,
and
\, cos^{-1}\left ( \dfrac{12}{x} \right )+cos^{-1}\left ( \dfrac{35}{x} \right )=\dfrac{\pi }{2},
then x is
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7
0%
39
0%
37
0%
-37
Explanation
cos^{-1}(a)+cos^{-1}(b)=\dfrac{\pi}{2}
\Rightarrow cos^{-1}(a)=\dfrac{\pi}{2}-cos^{-1}(b)
\Rightarrow cos^{-1}(a)=sin^{-1}(b)
Therefore,
a^2+b^2=1
Substituting the value of a and b we get
\dfrac{12}{x}^{2}+\dfrac{35}{x}^{2}=1
x^2=1369
x=37,-37
However , since
x>0
x=37
The value of
\sin^{-1}
\left \{ \tan\left ( \cos^{-1}\sqrt{\dfrac{2+\sqrt{3}}{4}}+\cos^{-1}\dfrac{\sqrt{12}}{4} -\text{cosec}^{-1}\sqrt{2}\right ) \right \}
, is
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0
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\dfrac{\pi }{2}
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-\dfrac{\pi }{2}
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\pi
Explanation
Given:
E = \sin^{-1} \left(\tan \left(\cos^{-1} \dfrac {\sqrt {2 + \sqrt {3}}}{2} + \cos^{-1} \dfrac {\sqrt 3}{2} - \text{cosec}^{-1} \sqrt 2\right) \right)
We know that,
\cos 15^0 = \sqrt{\dfrac{2+\sqrt3}{4}}
Therefore,
\Rightarrow E= \sin^{-1} (\tan (15^0 + 30^0 - 45^0))
\Rightarrow E= \sin^{-1} (\tan0^0)
\Rightarrow E = \sin^{-1} 0 = 0
If
cosec ^{ -1 }\left(cosec (x) \right)
and
cosec\left(cosec ^{ -1 }(x) \right)
are equal functions, then the maximum range of value of
x
is
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\displaystyle\:\left [ -\frac{\pi }{2},-1\right ]\cup \left [ 1,\frac{\pi }{2} \right ]
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\displaystyle\:\left (-\frac{\pi }{2},-1\right )\cup \left (1,\frac{\pi }{2} \right )
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\displaystyle\:\left ( -\infty ,-1]\cup [1,\infty \right )
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\displaystyle\:\left ( -\infty ,-1)\cup (1,\infty \right )
Explanation
The range of
cosec^{-1}(cosec(x))
=\left[\dfrac{-\pi}{2},0\right)\cup\left(0,\dfrac{\pi}{2}\right]
The range of
cosec(cosec^{-1}(x))
=(-\infty,-1]\cup[1,\infty)
Since it is given that both the functions are equal, then the range of value x common to both will be
=\left[\dfrac{-\pi}{2},-1\right]\cup\left[1,\dfrac{\pi}{2}\right]
If
\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1
, where
[.]
denotes the greatest integer function, the
\theta
lies in the interval
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[\tan { \sin { \cos { 1 } } } ,\sin { \tan { \cos { \sin { 1 } } } } ]
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[\sin { \tan { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]
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[\tan { \sin { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]
0%
None of these
Explanation
We have
\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1
\Rightarrow 1\le \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } }\le \displaystyle\frac { \pi }{ 2 }
\Rightarrow \sin { 1 } \le \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } \le 1
\Rightarrow \cos { \sin { 1 } } \ge \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } \ge \cos { 1 }
\Rightarrow \sin { \cos { \sin { 1 } } } \ge \tan ^{ -1 }{ \theta } \ge \sin { \cos { 1 } }
\Rightarrow \tan { \sin { \cos { \sin { 1 } } } } \ge \theta \ge \tan { \sin { \cos { 1 } } }
\displaystyle \:\cos ^{-1}x+\cos ^{-1}\left ( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} \right )
is equal to
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\displaystyle \:\frac{\pi }{3}
for
x \epsilon \left [ \dfrac{1}{2},1 \right ]
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\displaystyle \:\frac{\pi }{3}
for
x\epsilon \left [ 0,\dfrac{1}{2} \right ]
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\displaystyle \:2\cos ^{-1}x-\cos ^{-1}\dfrac{1}{2}
for
x \epsilon \left [ \dfrac{1}{2},1 \right ]
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\displaystyle \:2\cos ^{-1}x-\cos ^{-1}\dfrac{1}{2}
for
x \epsilon \left [ 0,\dfrac{1}{2}\right ]
Explanation
cos^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+cos^{ -1 }\cfrac { 1 }{ 3 } -cos^{ -1 }x
For
\cfrac { 1 }{ 2 } \le x\le 1
cos^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+\cfrac { \pi }{ 3 } -cos^{ -1 }x=\cfrac { \pi }{ 3 }
And for
0\le x\le \cfrac { 1 }{ 2 }
os^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+cos^{ -1 }x-cos^{ -1 }\cfrac { 1 }{ 2 } =2cos^{ -1 }x-cos^{ -1 }\cfrac { 1 }{ 2 }
The range of values of
p
for which the equation
\sin \cos ^{-1}(\cos (\tan ^{-1}x))= p
has a solution is
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\left ( -\frac{1}{\sqrt{2}},\frac{2}{\sqrt{2}} \right )
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[0,1)
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\left ( \frac{1}{\sqrt{2}1} \right )
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(-1,1)
Explanation
Here The range of
cos(tan^{-1}(x))
will be [-1,1] since the range of
cos\theta
is always
[-1,1]
Hence for
cos(tan^{-1}(x))=1
sin(cos^{-1}(1))
=sin(0)
=0
For
cos(tan^{-1}(x))=0
sin(cos^{-1}(0))
=sin(\frac{\pi}{2})
=1
For
cos(tan^{-1}(x))=-1
sin(cos^{-1}(-1))
=sin(\pi)
=0
Howere p=1 for
x\rightarrow \infty
Hence range of p=[0,1).
The solution set of the equation
\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x
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\displaystyle \left [ -1,\: 1 \right ] - \left \{ 0 \right \}
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\displaystyle \left ( 0, \: 1 \right ] \cup \left \{ -1 \right \}
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\displaystyle \left [ -1,\: 0 \right ) \cup \left \{ 1 \right \}
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\displaystyle \{ 1 \}
Explanation
\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x
\Rightarrow \sin ^{ -1 } \sqrt { 1-x } +\dfrac { \pi }{ 2 } =\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right)
\Rightarrow \dfrac { \pi }{ 2 } -\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x }
\Rightarrow \tan ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x }
Thus is only true when,
x = \{1\}
0:0:1
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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