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QJEEM18Ma9a3f: JEE Main 2018 Mathematics - Class 12 JEEM18Ma9a3f Solution: If f ( x ) =

If f(x)=|cos⁡xx12sin⁡xx22xtan⁡xx1|, then lim⁡x→0f′(x)x\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}

Multiple Choice Options

  • A does not exist.
  • B exists and is equal to 2.
  • C existsand is equal to 0.
  • D exists and is equal to −- 2.

Step-by-Step Verified Solution

Given,

f(x)=|cos⁡xx12sin⁡xx22xtan⁡xx1|

= cosx(x2 - 2x2) - x(2 sinx - 2x tanx) + (2x sinx - x2 tanx)
= x2 (tanx - cosx)
∴\therefore f′(x){f^{'}}(x) = 2x (tanx - cosx) + x2(sec2x + sinx)

∴\therefore lim⁡x→0f′(x)x\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}

= lim⁡x→o2x(tan⁡x−cos⁡x)+x2(sec⁡2x+sin⁡x)x\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}

= lim⁡x→o  2(tan⁡x−cos⁡x)+x(sec⁡2x+sin⁡x)\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)

= 2 (0-1) + 0

= -2

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