Logo
Jeeify
MCQMediumJEE 2026Functions

JEE Mathematics 2026 Question with Solution

Let ff be a polynomial function such that

f(x2+1)=x4+5x2+2,for all x∈R.f(x^2+1)=x^4+5x^2+2,\quad \text{for all } x\in\mathbb{R}.

Then

∫03f(x) dx\int_0^3 f(x)\,dx

is equal to:

  • A

    53\dfrac{5}{3}

  • B

    272\dfrac{27}{2}

  • C

    332\dfrac{33}{2}

  • D

    413\dfrac{41}{3}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

f(x2+1)=x4+5x2+2f(x^2+1)=x^4+5x^2+2

for all x∈Rx\in\mathbb{R}.

Find:

∫03f(x) dx\int_0^3 f(x)\,dx

Use the substitution

t=x2+1t=x^2+1

so that

x2=t−1x^2=t-1

Then

f(t)=x4+5x2+2f(t)=x^4+5x^2+2

and since

x4=(t−1)2=t2−2t+1x^4=(t-1)^2=t^2-2t+1

we get

f(t)=(t2−2t+1)+5(t−1)+2f(t)=(t^2-2t+1)+5(t-1)+2

Therefore,

f(t)=t2+3t−2f(t)=t^2+3t-2

so

f(x)=x2+3x−2f(x)=x^2+3x-2

Now evaluate the integral:

∫03f(x) dx=∫03(x2+3x−2) dx\int_0^3 f(x)\,dx=\int_0^3 (x^2+3x-2)\,dx

Hence,

∫03(x2+3x−2) dx=[x33+3x22−2x]03\int_0^3 (x^2+3x-2)\,dx=\left[\frac{x^3}{3}+\frac{3x^2}{2}-2x\right]_0^3

Substituting the limits,

=(273+272−6)−0=\left(\frac{27}{3}+\frac{27}{2}-6\right)-0 =9+272−6=272=9+\frac{27}{2}-6=\frac{27}{2}

Therefore, the value of the integral is 272\dfrac{27}{2}, so the correct option is B.

Polynomial Identity Approach

Given: The identity

f(x2+1)=x4+5x2+2f(x^2+1)=x^4+5x^2+2

holds for all real xx.

Find: The value of

∫03f(x) dx\int_0^3 f(x)\,dx

Because the expression on the right depends only on x2x^2, set

y=x2+1y=x^2+1

Then

x2=y−1x^2=y-1

and

x4=(y−1)2x^4=(y-1)^2

So,

f(y)=(y−1)2+5(y−1)+2f(y)=(y-1)^2+5(y-1)+2

Expanding,

f(y)=y2−2y+1+5y−5+2f(y)=y^2-2y+1+5y-5+2 f(y)=y2+3y−2f(y)=y^2+3y-2

Hence the polynomial is

f(x)=x2+3x−2f(x)=x^2+3x-2

Integrate term by term:

∫03f(x) dx=∫03(x2+3x−2) dx\int_0^3 f(x)\,dx=\int_0^3 (x^2+3x-2)\,dx =∫03x2 dx+3∫03x dx−2∫031 dx=\int_0^3 x^2\,dx+3\int_0^3 x\,dx-2\int_0^3 1\,dx =[x33]03+3[x22]03−2[x]03=\left[\frac{x^3}{3}\right]_0^3+3\left[\frac{x^2}{2}\right]_0^3-2[x]_0^3 =273+3⋅92−2⋅3=\frac{27}{3}+3\cdot\frac{9}{2}-2\cdot 3 =9+272−6=272=9+\frac{27}{2}-6=\frac{27}{2}

Therefore, the correct option is B.

Common mistakes

  • A common mistake is to replace f(x2+1)f(x^2+1) by f(x)2+1f(x)^2+1. This is incorrect because ff is evaluated at the input x2+1x^2+1. First set a new variable such as t=x2+1t=x^2+1, then rewrite the entire identity in terms of tt.

  • Students often compute x4x^4 incorrectly after substitution. If x2=t−1x^2=t-1, then x4=(x2)2=(t−1)2x^4=(x^2)^2=(t-1)^2, not t2−1t^2-1. Expand carefully before combining like terms.

  • Another mistake is to find f(x)=x2+3x−2f(x)=x^2+3x-2 correctly but then integrate incorrectly. The antiderivative of 3x3x is 3x22\frac{3x^2}{2}, not 3x23x^2. Integrate each term separately and apply the limits at the end.

Practice more Functions questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Functions

Related questions