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Maths Question 18 – JEE-MAIN 2026

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Let f(x) be a polynomial of degree 5, and have extrema at x=1 and x=−1. If limx→0(f(x)x3)=−5, then f(2)−f(−2) is equal to :

The limit condition limx→0(f(x)x3)=−5 provides crucial information about the lowest degree terms of the polynomial f(x).

Step 1: Determine the form of f(x) using the limit condition✦ Active

Given that f(x) is a polynomial of degree 5 and limx→0(f(x)x3)=−5, for the limit to be finite and non-zero, the terms x0,x1,x2 in f(x) must be zero. Thus, f(x) can be written as f(x)=Ax5+Bx4+Cx3. Applying the limit:

limx→0(Ax5+Bx4+Cx3x3)=limx→0(Ax2+Bx+C)=C

Since the limit is given as −5, we have C=−5. Therefore, f(x)=Ax5+Bx4−5x3.

Step 2: Use the extrema conditions to find coefficients A and B○ Expand

The polynomial has extrema at x=1 and x=−1, which means f′(1)=0 and f′(−1)=0. First, find the derivative of f(x):

f′(x)=5Ax4+4Bx3−15x2

Setting f′(1)=0:

5A(1)4+4B(1)3−15(1)2=0⇒5A+4B−15=0⇒5A+4B=15(1)

Setting f′(−1)=0:

5A(−1)4+4B(−1)3−15(−1)2=0⇒5A−4B−15=0⇒5A−4B=15(2)

Adding equations (1) and (2):

(5A+4B)+(5A−4B)=15+15⇒10A=30⇒A=3

Substitute A=3 into equation (1):

5(3)+4B=15⇒15+4B=15⇒4B=0⇒B=0

Thus, the polynomial is f(x)=3x5−5x3.

Step 3: Calculate f(2)−f(−2)○ Expand

First, evaluate f(2):

f(2)=3(2)5−5(2)3=3(32)−5(8)=96−40=56

Next, evaluate f(−2):

f(−2)=3(−2)5−5(−2)3=3(−32)−5(−8)=−96+40=−56

Finally, calculate f(2)−f(−2):

f(2)−f(−2)=56−(−56)=56+56=112

Alternatively, observe that f(x)=3x5−5x3 is an odd function, meaning f(−x)=−f(x). Therefore, f(2)−f(−2)=f(2)−(−f(2))=2f(2)=2(56)=112.

💡 Teacher's Secret Hint

Recognizing that f(x) is an odd function can simplify the final calculation.

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