Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 51

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Question 1
1 / -0

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 $$[-1, 1].$$

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

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 $$ R \cap [-1,1]$$

$$\implies [-1,1]$$
Question 2
1 / -0

The number of solutions of:
$$\displaystyle \sin^{-1}(1+b+b^{2}+\ldots.\infty)+\cos^{-1}(a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+\ldots\infty)=\frac{\pi}{2}$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\infty$$

Solution

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.
Question 3
1 / -0

If $$(\tan^{-1}x)^{2}+(\cot^{-1}x)^{2} = \displaystyle \frac{5\pi^{2}}{8}$$, then $$x=$$

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$2$$

Solution

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.
Question 4
1 / -0

$$cos^{-1} \left (\sqrt{\dfrac{a-x}{a-b}} \right)$$ =$$sin^{-1} \left (\sqrt{\dfrac{x-b}{a-b}}\right)$$ is possible if

A
$$a>x>b$$ or $$ a< x < b $$

B
$$a=x=b$$

C
$$a > b$$ and x takes any value

D
$$a < b$$ and x takes any value

Solution

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
$$\displaystyle 0\leq \sqrt{\frac{x-b}{a-b}}\leq 1$$

$$0\leq x-b \leq a-b$$

$$b \leq x\leq a$$

$$x\epsilon [b,a]$$
Question 5
1 / -0

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 :

A
$$1$$

B
$$0$$

C
$$3$$

D
$$-3$$

Solution

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
$$\alpha\beta+\beta\gamma+\gamma\alpha$$
$$=(1)+(1)+(1)$$
$$=3$$
Question 6
1 / -0

The solution set of the equation $$\tan^{-1}x -\cot^{-1}x =\cos^{-1}(2-x)$$ is

A
$$(0,1)$$

B
$$(-1,1)$$

C
$$[1,3)$$

D
$$(1,3)$$

Solution

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 $$[1,3)$$.
Question 7
1 / -0

The number of real solutions of $$tan^{-1} (\sqrt{x(x+1)}+sin^{-1} \displaystyle \sqrt{(x^{2}+x+1)}=\dfrac{\pi}{2}$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
infinite

Solution

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$$ x=0$$.
Question 8
1 / -0

The number of positive integral solutions of the equation $$tan^{-1}x+cot^{-1}y =tan^{-1} 3$$ is :

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$cot^{-1}y=tan ^{-1}3-tan ^{-1}x$$
$$\displaystyle cot^{-1}y=tan ^{-1}(\dfrac{3-x}{1+3x})$$
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 $$x=0,1,2$$
Question 9
1 / -0

If $$(tan^{-1} x)^2 +(cot ^{-1}x)^2=\displaystyle \frac{5 \pi^2}{8}$$, then $$x$$ =

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

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, $$tan^{-1}x =1 $$
$$x=-1$$
Question 10
1 / -0

The number of integral solutions of $$sin^{-1}\sqrt{4x-x^{2}-3}+tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}$$ is

A
zero

B
infinite

C
four

D
None of these

Solution

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, $$\sqrt { 4x-x^{ 2 }-3 } \ge 0$$
$$\Rightarrow 0<\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 } \le \dfrac { \pi }{ 2 } $$

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Inverse Trigonometric Functions Test - 51

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Inverse Trigonometric Functions Test - 51

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SHARING IS CARING

Question 1

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$\infty$$

Solution

Question 3

$$-1$$

$$1$$

$$0$$

$$2$$

Solution

Question 4

$$a>x>b$$ or $$ a< x < b $$

$$a=x=b$$

$$a > b$$ and x takes any value

$$a < b$$ and x takes any value

Solution

Question 5

$$1$$

$$0$$

$$3$$

$$-3$$

Solution

Question 6

$$(0,1)$$

$$(-1,1)$$

$$[1,3)$$

$$(1,3)$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

infinite

Solution

Question 8

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 9

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 10

zero

infinite

four

None of these

Solution

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