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Maths Question 17 – JEE-MAIN 2026

Let f be a real polynomial of degree n such that f(x)=f(x)f(x), for all xR. If f(0)=0, then 36(f(2)+f(2)+02f(x)dx) is equal to:

The problem involves a differential equation where the function is a polynomial. The first step is to determine the degree of this polynomial by comparing the degrees of both sides of the equation.

Step 1: Determine the Polynomial f(x)✦ Active

Let f(x) be a polynomial of degree n. By comparing the degrees of both sides of the equation f(x)=f(x)f(x), we get n=(n1)+(n2), which simplifies to n=3.

Let f(x)=ax3+bx2+cx+d. Comparing the leading coefficients in the equation gives a=(3a)(6a)=18a2, which implies a=1/18 (since a0).

The condition f(0)=0 implies d=0. By substituting the general form of f(x) into the differential equation and comparing the remaining coefficients, we find that b=0 and c=0.

f(x)=118x3
Step 2: Calculate the Required Values○ Expand

Now we find the values of the derivatives at x=2 and the definite integral.

f(x)=ddx(118x3)=16x2f(2)=16(22)=23
f(x)=ddx(16x2)=13xf(2)=13(2)=23
02f(x)dx=02118x3dx=118[x44]02=118(164)=418=29
Step 3: Compute the Final Expression○ Expand

Substitute the calculated values into the given expression.

36(f(2)+f(2)+02f(x)dx)=36(23+23+29)
=36(43+29)=36(12+29)=36(149)=4×14=56
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