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Cubic with given turning points and ∫f=14/3 | JEE

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

Question
A cubic f(x)f(x) vanishes at x=−2x=-2 and has local minimum/maximum at x=−1x=-1 and x=13x=\dfrac13. If ∫−11f(x) dx=143\displaystyle\int_{-1}^{1}f(x)\,dx=\dfrac{14}{3}, then f(1)=f(1)=
A11
B22
C33correct
D44
Solution
Step 1: Use the turning points to fix f′f'. Since ff is cubic with stationary points at x=−1x=-1 and x=13x=\dfrac13,
f′(x)=A(x+1) ⁣(x−13)=A3(3x2+2x−1).f'(x)=A(x+1)\!\left(x-\dfrac13\right)=\dfrac{A}{3}\big(3x^{2}+2x-1\big).
 Step 2: Integrate to get ff:
f(x)=A3(x3+x2−x)+B.f(x)=\dfrac{A}{3}\big(x^{3}+x^{2}-x\big)+B.
 Apply f(−2)=0f(-2)=0:
A3(−8+4+2)+B=0 ⇒ A3(−2)+B=0 ⇒ B=2A3.\dfrac{A}{3}(-8+4+2)+B=0\ \Rightarrow\ \dfrac{A}{3}(-2)+B=0\ \Rightarrow\ B=\dfrac{2A}{3}.
 Step 3: Apply the integral condition. Only the even part A3x2+B\dfrac{A}{3}x^{2}+B remain over [−1,1][-1,1]:
∫−11f(x) dx=2∫01 ⁣(A3x2+B)dx=2(A9+B)=143.\int_{-1}^{1}f(x)\,dx=2\int_{0}^{1}\!\left(\dfrac{A}{3}x^{2}+B\right)dx=2\left(\dfrac{A}{9}+B\right)=\dfrac{14}{3}.
 With B=2A3B=\dfrac{2A}{3}: A9+2A3=A+6A9=7A9\dfrac{A}{9}+\dfrac{2A}{3}=\dfrac{A+6A}{9}=\dfrac{7A}{9}, so 2⋅7A9=143⇒A=3.2\cdot\dfrac{7A}{9}=\dfrac{14}{3}\Rightarrow A=3. Step 4: Then B=2B=2 and f(x)=x3+x2−x+2f(x)=x^{3}+x^{2}-x+2, giving
f(1)=1+1−1+2=3.f(1)=1+1-1+2=3.
 Correct answer: (3)
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