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MHTCET 19 Apr 2026 Shift 2 Paper

Question : 118 of 150

Marks: +1, -0

Let

f(x)=1-\frac{1}{x}, g_{2}(x)=f(f(x)), g_{3}(x)=f(f(f(x)))

\int x \cdot g_{2026}(x) d x=\int g_{2025}(x) d x+h(x)+c_{1}

h(x)=\ldots

Solution:

Concept:

The repeated composition of

f(x)

has a periodic pattern of length 3, so we can simplify

g_{2025}(x)

and

g_{2026}(x)

easily.

Explanation:

Given

f(x)=1-\frac{1}{x}

First,

g_2(x)=f(f(x))=\frac{1}{1-x}

Next,

g_3(x)=f(g_2(x))=x

So the function repeats after every 3 compositions:

f^3(x)=x

Since

2025

is divisible by

3

, we get

g_{2025}(x)=x

Therefore,

g_{2026}(x)=f(g_{2025}(x))=f(x)=1-\frac{1}{x}

Now substitute in the given integral:

\int x \cdot g_{2026}(x)\,dx=\int x\left(1-\frac{1}{x}\right)dx=\int (x-1)\,dx

Also,

\int g_{2025}(x)\,dx=\int x\,dx

Thus,

\int (x-1)\,dx=\int x\,dx-x+C

Comparing with

\int x\,dx+h(x)+c_1

, we get

h(x)=-x

Answer:

h(x)=-x

, so the correct option is B.

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