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MHTCET 15 May 2024 Shift 2 Paper

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Question : 138 of 150

Marks: +1, -0

If

0 < x < 1

\sqrt{1+x^{2}}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^{2}-1\right]^{\frac{1}{2}}

is equal to

$x^{2} \sqrt{1+x^{2}}$
$x$
$x \sqrt{1+x^{2}}$
$\sqrt{1+x^{2}}$

Solution:

Concept:

Use the substitution

\cot^{-1} x = \theta

and simplify using trigonometric identities.

Explanation:

Let

\cot^{-1} x = \theta

. Then

x = \cot \theta

.

Since

0 < x < 1

, we get

\frac{\pi}{4} < \theta < \frac{\pi}{2}

, so

\cot \theta > 0

.

Also,

\sqrt{1+x^2} = \sqrt{1+\cot^2 \theta} = \csc \theta

.

Now,

x\cos(\cot^{-1}x) + \sin(\cot^{-1}x) = \cot\theta \cos\theta + \sin\theta

.

This equals

\frac{\cos^2\theta}{\sin\theta} + \sin\theta = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta} = \csc\theta

.

So the given expression becomes

\sqrt{1+x^2}\left[(\csc\theta)^2 - 1\right]^{1/2}

.

Since

\csc^2\theta - 1 = \cot^2\theta

, we get

\sqrt{1+x^2} \cdot |\cot\theta|

.

As

\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)

,

\cot\theta > 0

, hence

|\cot\theta| = \cot\theta = x

.

Therefore, the expression simplifies to

x\sqrt{1+x^2}

.

Answer:

Option C:

x\sqrt{1+x^2}

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