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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 132 of 150

Marks: +1, -0

f^{\prime}(x)=\frac{(\sqrt{x}+1) e^{\sqrt{x}}}{\sqrt{x}}

f(0)=e

f(1)=\ldots \ldots \ldots

. .

Solution:

Concept:

Use the fundamental theorem of calculus and substitution to evaluate the definite integral of

f'(x)

Explanation:

Given

f'(x)=\frac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}}

and

f(0)=e

Using

f(1)-f(0)=\int_0^1 f'(x)\,dx

, we write:

f(1)-e=\int_0^1 \frac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}}\,dx

Let

\sqrt{x}=t

, so

x=t^2

and

dx=2t\,dt

Limits: when

x=0

t=0

; when

x=1

t=1

The integral becomes:

\int_0^1 \frac{(t+1)e^t}{t}\cdot 2t\,dt = 2\int_0^1 (t+1)e^t\,dt

Since

\frac{d}{dt}(t e^t)=(t+1)e^t

, we get:

2\int_0^1 (t+1)e^t\,dt = 2\left[t e^t\right]_0^1 = 2e

Thus,

f(1)-e=2e

, so

f(1)=3e

Answer:

f(1)=3e

Therefore, the correct option is C.

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