SATHEE: Inverse Circular Functions 2 Question 5

Inverse Circular Functions 2 Question 5

5. If $0 < x < 1$ , then $Missing \left or extra \right$ is equal to

(2008, 3M)

(a) $\frac{x}{\sqrt{1 + x^{2}}}$

(b) $x$

(d) $\sqrt{1 + x^{2}}$

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Answer:

Correct Answer: 5. (c)

Solution:

We have, $0 < x < 1$

Let $\cot^{- 1} x = θ$

$\begin{array}{ll} \Rightarrow & \cot θ = x \\ \Rightarrow & \sin θ = \frac{1}{\sqrt{1 + x^{2}}} = \sin (\cot^{- 1} x) \\ and & \cos θ = \frac{x}{\sqrt{1 + x^{2}}} = \cos (\cot^{- 1} x) \end{array}$

Now, $\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{1 / 2}$

$\begin{aligned} = \sqrt{1 + x^{2}} x \frac{x}{\sqrt{1 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}} - 1 \\ = \sqrt{1 + x^{2}} \frac{1 + x^{2}}{\sqrt{1 + x^{2}}} - 1 \\ = \sqrt{1 + x^{2}} {[1 + x^{2} - 1]}^{1 / 2} = x \sqrt{1 + x^{2}} \end{aligned}$

Inverse Circular Functions 2 Question 5

5. If $0 < x < 1$ , then $Missing \left or extra \right$ is equal to

Answer:

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Inverse Circular Functions 2 Question 5

5. If 0<x<1, then Missing \left or extra \rightMissing \left or extra \right is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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5. If $0 < x < 1$ , then $Missing \left or extra \right$ is equal to