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Degree-5 Polynomial with Limit Condition: f(2)-f(-2) = 112 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let f(x)f(x) be a polynomial of degree 5, and have extrema at x=1x=1 and x=−1x=-1. If lim⁡x→0(f(x)x3)=−5\displaystyle\lim_{x\to0}\left(\dfrac{f(x)}{x^3}\right)=-5, then f(2)−f(−2)f(2)-f(-2) is equal to
A00
B5050
C9292
D112112correct
Solution
Step 1: Extrema at x=±1x=\pm1 mean f′(1)=0f'(1)=0 and f′(−1)=0f'(-1)=0. Step 2: Since lim⁡x→0f(x)x3=−5\displaystyle\lim_{x\to0}\dfrac{f(x)}{x^3}=-5 is finite, the numerator must vanish to order 3 at 00:
f(0)=0,f′(0)=0,f′′(0)=0,f′′′(0)3!=−5⇒f′′′(0)=−30.f(0)=0,\quad f'(0)=0,\quad f''(0)=0,\quad \frac{f'''(0)}{3!}=-5\Rightarrow f'''(0)=-30.
 Step 3: Because f′f' (degree 4) vanishes at x=0x=0 (double, from f′(0)=f′′(0)=0f'(0)=f''(0)=0) and at x=±1x=\pm1, write
f′(x)=(ax+b) x (x−1)(x+1)=ax4+bx3−ax2−bx.f'(x)=(ax+b)\,x\,(x-1)(x+1)=ax^4+bx^3-ax^2-bx.
 Step 4: Then f′′(x)=4ax3+3bx2−2ax−bf''(x)=4ax^3+3bx^2-2ax-b. From f′′(0)=0f''(0)=0: −b=0⇒b=0-b=0\Rightarrow b=0. Step 5: f′′′(x)=12ax2+6bx−2af'''(x)=12ax^2+6bx-2a; from f′′′(0)=−2a=−30⇒a=15f'''(0)=-2a=-30\Rightarrow a=15. Hence
f′(x)=15x2(x2−1)=15x4−15x2.f'(x)=15x^2(x^2-1)=15x^4-15x^2.
 Step 6: Integrate: f(x)=3x5−5x3+Cf(x)=3x^5-5x^3+C; with f(0)=0f(0)=0, C=0C=0, so f(x)=3x5−5x3f(x)=3x^5-5x^3. Step 7:
f(2)−f(−2)=2f(2)=2(3⋅32−5⋅8)=2(96−40)=2(56)=112.f(2)-f(-2)=2f(2)=2(3\cdot32-5\cdot8)=2(96-40)=2(56)=112.
 Correct answer: (4)
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