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Functions — JEE Maths practice question

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

Question
If aa and bb are constants, f(x)=asin⁡x+bxcos⁡x+2x2f(x) = a\sin x + bx\cos x + 2x^2. If f(2)=15f(2) = 15, then f(−2)f(-2) is.
Solution
Answer: 1
Step 1: Use odd-even symmetry Let h(x)=asin⁡x+bxcos⁡xh(x) = a\sin x + bx\cos x (odd part) and p(x)=2x2p(x) = 2x^2 (even part).
f(−x)=−asin⁡x−bxcos⁡x+2x2  ⟹  f(x)+f(−x)=4x2f(-x) = -a\sin x - bx\cos x + 2x^2 \implies f(x)+f(-x) = 4x^2
 Step 2: Evaluate at x=2x = 2
f(2)+f(−2)=4(4)=16  ⟹  f(−2)=16−15=1f(2)+f(-2) = 4(4) = 16 \implies f(-2) = 16-15 = 1
 Answer: 1
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