JEE Main 2026 — Mathematics Calculus

Let the degree of the polynomial $f(x)$f(x) be $n$n.

The degree of $f'(x)$f′(x) is $n-1$n−1 and the degree of $f''(x)$f′′(x) is $n-2$n−2.

Since $f(x) = f'(x) f''(x)$f(x)=f′(x)f′′(x), equating the degrees on both sides gives:

$$n = (n-1) + (n-2) \Rightarrow n = 3$$n=(n−1)+(n−2)⇒n=3

Let $f(x) = ax^3 + bx^2 + cx + d$f(x)=ax3+bx2+cx+d.

Given $f(0) = 0$f(0)=0, we get $d = 0$d=0. Thus, $f(x) = ax^3 + bx^2 + cx$f(x)=ax3+bx2+cx.

Differentiating $f(x)$f(x) with respect to $x$x:

$$f'(x) = 3ax^2 + 2bx + c$$f′(x)=3ax2+2bx+c

$$f''(x) = 6ax + 2b$$f′′(x)=6ax+2b

Substituting these into the given equation $f(x) = f'(x) f''(x)$f(x)=f′(x)f′′(x):

$$ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b)$$ax3+bx2+cx=(3ax2+2bx+c)(6ax+2b)

$$ax^3 + bx^2 + cx = 18a^2x^3 + 18abx^2 + (4b^2 + 6ac)x + 2bc$$ax3+bx2+cx=18a2x3+18abx2+(4b2+6ac)x+2bc

Comparing the coefficients of corresponding powers of $x$x:

For $x^3$x3: $a = 18a^2 \Rightarrow a = \dfrac{1}{18}$a=18a2⇒a=181​ (since $a \neq 0$a=0 for a cubic polynomial)

For the constant term: $0 = 2bc$0=2bc

For $x$x: $c = 4b^2 + 6ac$c=4b2+6ac

Substituting $a = \dfrac{1}{18}$a=181​ into the $x$x coefficient equation:

$$c = 4b^2 + \dfrac{c}{3} \Rightarrow \dfrac{2c}{3} = 4b^2 \Rightarrow c = 6b^2$$c=4b2+3c​⇒32c​=4b2⇒c=6b2

From $0 = 2bc$0=2bc, either $b = 0$b=0 or $c = 0$c=0. In either case, since $c = 6b^2$c=6b2, we get $b = 0$b=0 and $c = 0$c=0.

Thus, the polynomial is $f(x) = \dfrac{1}{18}x^3$f(x)=181​x3.

Now, finding the required values:

$$f'(x) = \dfrac{1}{6}x^2 \Rightarrow f'(2) = \dfrac{4}{6} = \dfrac{2}{3}$$f′(x)=61​x2⇒f′(2)=64​=32​

$$f''(x) = \dfrac{1}{3}x \Rightarrow f''(2) = \dfrac{2}{3}$$f′′(x)=31​x⇒f′′(2)=32​

$$\int_0^2 f(x)\,dx = \int_0^2 \dfrac{1}{18}x^3\,dx = \dfrac{1}{18} \left[ \dfrac{x^4}{4} \right]_0^2 = \dfrac{16}{72} = \dfrac{2}{9}$$∫02​f(x)dx=∫02​181​x3dx=181​[4x4​]02​=7216​=92​

Substituting these into the given expression:

$$36\left(f'(2) + f''(2) + \int_0^2 f(x)\,dx\right) = 36\left(\dfrac{2}{3} + \dfrac{2}{3} + \dfrac{2}{9}\right)$$36(f′(2)+f′′(2)+∫02​f(x)dx)=36(32​+32​+92​)

$$= 36\left(\dfrac{4}{3} + \dfrac{2}{9}\right) = 36\left(\dfrac{14}{9}\right) = 4 \times 14 = 56$$=36(34​+92​)=36(914​)=4×14=56

Answer: $56$56