JEE Main 2026 — Application of Derivatives Question with Solution

Let the polynomial of degree $5$ be $f(x)=a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex+k$.

Given $\underset{x\to 0}{\lim }\left(\frac{f(x)}{x^3}\right)=-5$, the terms of degree less than $3$ must be zero, and the coefficient of $x^3$ must be $-5$.

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

The polynomial becomes $f(x)=a{x}^5+b{x}^4-5x^3$.

Differentiating with respect to $x$, we get:

$f^{\prime }(x)=5a{x}^4+4b{x}^3-15x^2$

Since $f(x)$ has extrema at $x=1$ and $x=-1$, we have $f^{\prime }(1)=0$ and $f^{\prime }(-1)=0$.

$f^{\prime }(1)=5a+4b-15=0$

$f^{\prime }(-1)=5a-4b-15=0$

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

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

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

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

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

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

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

Answer: $112$