JEE Main 2026 — Mathematics Calculus

Let the polynomial of degree $5$5 be $f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + k$f(x)=ax5+bx4+cx3+dx2+ex+k.

Given $\displaystyle \displaystyle\lim_{x \to 0} \left(\dfrac{f(x)}{x^3}\right) = -5$x→0lim​(x3f(x)​)=−5, the terms of degree less than $3$3 must be zero, and the coefficient of $x^3$x3 must be $-5$−5.

Thus, $k = 0$k=0, $e = 0$e=0, $d = 0$d=0, and $c = -5$c=−5.

The polynomial becomes $f(x) = ax^5 + bx^4 - 5x^3$f(x)=ax5+bx4−5x3.

Differentiating with respect to $x$x, we get:

$$f'(x) = 5ax^4 + 4bx^3 - 15x^2$$f′(x)=5ax4+4bx3−15x2

Since $f(x)$f(x) has extrema at $x = 1$x=1 and $x = -1$x=−1, we have $f'(1) = 0$f′(1)=0 and $f'(-1) = 0$f′(−1)=0.

$$f'(1) = 5a + 4b - 15 = 0$$f′(1)=5a+4b−15=0

$$f'(-1) = 5a - 4b - 15 = 0$$f′(−1)=5a−4b−15=0

Adding both equations, we get $10a - 30 = 0 \Rightarrow a = 3$10a−30=0⇒a=3.

Subtracting the equations, we get $8b = 0 \Rightarrow b = 0$8b=0⇒b=0.

So, the polynomial is $f(x) = 3x^5 - 5x^3$f(x)=3x5−5x3.

Now, we find $f(2)$f(2) and $f(-2)$f(−2):

$$f(2) = 3(2)^5 - 5(2)^3 = 3(32) - 5(8) = 96 - 40 = 56$$f(2)=3(2)5−5(2)3=3(32)−5(8)=96−40=56

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

Therefore, $f(2) - f(-2) = 56 - (-56) = 112$f(2)−f(−2)=56−(−56)=112.

Answer: $112$112