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JEE Main 2020 · Mathematics · Differentiation and Application of Derivatives · medium

Let f(x)=(sin 3x+asin x)/(x^3 for x≠0. If f can be made continuous at x=0 by defining f(0)=b, then a+b equals

Single correct option

Let f(x)=sin⁡3x+asin⁡xx3f(x)=\dfrac{\sin3x+a\sin x}{x^3} for x≠0x\neq0. If ff can be made continuous at x=0x=0 by defining f(0)=bf(0)=b, then a+ba+b equals
  1. A. 77
  2. B. −1-1
  3. C. 11
  4. D. −7-7
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Answer: (D) −7-7

For a finite limit as x→0x\to0 the numerator must vanish to order x3x^3. Expand: sin⁡3x=3x−27x36+⋯\sin3x=3x-\dfrac{27x^3}{6}+\cdots and sin⁡x=x−x36+⋯\sin x=x-\dfrac{x^3}{6}+\cdots. The coefficient of xx is 3+a3+a, which must be 00, so a=−3a=-3. Then the numerator is −276x3+36x3+⋯=−4x3-\dfrac{27}{6}x^3+\dfrac{3}{6}x^3+\cdots=-4x^3, so lim⁡f(x)=−4=b\lim f(x)=-4=b. Therefore a+b=−3−4=−7a+b=-3-4=-7. Therefore a=−3a=-3 and b=−4b=-4.
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