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JEE Main 2017 · Mathematics · Differentiation and Application of Derivatives · medium

Let f(x)=(sin 2x+asin x)/(x^3 for x≠0 and f(0)=b. If f is continuous at x=0, then a+b equals

Single correct option

Let f(x)=sin⁡2x+asin⁡xx3f(x)=\dfrac{\sin2x+a\sin x}{x^3} for x≠0x\ne0 and f(0)=bf(0)=b. If ff is continuous at x=0x=0, then a+ba+b equals
  1. A. −3-3
  2. B. −1-1
  3. C. 11
  4. D. 33
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Answer: (A) −3-3

For the limit to exist, the numerator must vanish faster than xx. Expand: sin⁡2x+asin⁡x=(2+a)x−8+a6x3+…\sin2x+a\sin x=(2+a)x-\dfrac{8+a}{6}x^3+\dots. The coefficient of xx must vanish, so a=−2a=-2. Then the numerator is −66x3=−x3-\dfrac{6}{6}x^3=-x^3 to leading order, so the limit is −1-1. Continuity requires b=−1b=-1. Thus a+b=−2−1=−3a+b=-2-1=-3.
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