If y=≤ft(x²+1)^{ x} for x > 0 such that d y/d x=y≤ft[2 x x/g(x)+ x · [g(x)]], then the function 1/g(x) is...

Correct Answer is : strictly decreasing

Correct Answer is : strictly decreasing

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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

y=\left(x^{2}+1\right)^{\sin x}

x > 0

\frac{d y}{d x}=y\left[\frac{2 x \sin x}{g(x)}+\cos x \cdot \log [g(x)]\right]

\frac{1}{g(x)}

is...

Solution:

Concept:

Use logarithmic differentiation because the variable appears in both base and exponent.

Explanation:

Given

y=\left(x^{2}+1\right)^{\sin x}

Take logarithm on both sides.

\log y=\sin x \cdot \log\left(x^{2}+1\right)

Differentiate with respect to

x

\frac{1}{y}\frac{dy}{dx}=\cos x \cdot \log\left(x^{2}+1\right)+\sin x \cdot \frac{2x}{x^{2}+1}

Thus,

\frac{dy}{dx}=y\left[\frac{2x\sin x}{x^{2}+1}+\cos x \cdot \log\left(x^{2}+1\right)\right]

Compare with the given expression.

\frac{dy}{dx}=y\left[\frac{2x\sin x}{g(x)}+\cos x \cdot \log[g(x)]\right]

So,

g(x)=x^{2}+1

Therefore,

\frac{1}{g(x)}=\frac{1}{x^{2}+1}

Let

f(x)=\frac{1}{x^{2}+1}

Then,

f'(x)=\frac{-2x}{\left(x^{2}+1\right)^{2}}

For

x>0

f'(x)<0

Hence,

\frac{1}{g(x)}

is strictly decreasing.

Answer:

Option D: strictly decreasing.

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