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Method of Differentiation: Values

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

Question
If f(x)=(ax+b)sin⁡x+(cx+d)cos⁡xf(x) = (ax+b)\sin x+(cx+d)\cos x, then the values of a,b,c,da, b, c, d such that f′(x)=xcos⁡xf'(x) = x\cos x for all xx, then (a+b+c+d)(a+b+c+d) is.
Solution
Answer: 2
Step 1: Compute f′(x)f'(x)
f′(x)=(ax+b)cos⁡x+asin⁡x+ccos⁡x−(cx+d)sin⁡xf'(x) = (ax+b)\cos x+a\sin x+c\cos x-(cx+d)\sin x
=(ax+b+c)cos⁡x+(a−d−cx)sin⁡x= (ax+b+c)\cos x+(a-d-cx)\sin x
 Step 2: Match coefficients with xcos⁡xx\cos x Coefficient of xcos⁡xx\cos x: a=1a = 1. Constant in cos⁡x\cos x: b+c=0b+c = 0. Coefficient of xsin⁡xx\sin x: −c=0  ⟹  c=0-c = 0 \implies c = 0. Constant in sin⁡x\sin x: a−d=0  ⟹  d=1a-d = 0 \implies d = 1. From b+c=0b+c = 0: b=0b = 0. Step 3: Sum
a+b+c+d=1+0+0+1=2a+b+c+d = 1+0+0+1 = 2
 Answer: 2
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