JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
Let $f(x)$ be a polynomial of degree $5$, and have extrema at $x=1$ and $x=-1$. If $\lim_{x\to 0}\left(\frac{f(x)}{x^{3}}\right)=-5$, then $f(2)-f(-2)$ is equal to :
Differentiation and its Application
Solution
Concept. The limit tells us the lowest power of $x$ in $f$; the two extrema fix the rest. Formulas used. (1) $\lim_{x\to0}\frac{f(x)}{x^3}=-5$ for a polynomial means $f=-5x^3+(\text{higher powers})$(2) An extremum at $a$ needs $f'(a)=0$Step 1 - Form of $f$. By (1), with degree $5$: $$f(x)=-5x^3+ax^4+bx^5,\qquad f'(x)=-15x^2+4ax^3+5bx^4.$$ Step 2 - Use (2). $$f'(1)=-15+4a+5b=0,\qquad f'(-1)=-15-4a+5b=0.$$ So $a=0$, $b=3$, and $f(x)=3x^5-5x^3$. (Indeed $f'=15x^2(x^2-1)$ changes sign at $\pm1$.) Step 3 - Value. $f$ is odd, so $f(2)-f(-2)=2f(2)=2(96-40)=112$. Answer: (D)