Q103 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

If $f(x)=3 x^{15}-5 x^{10}+7 x^5+50 \cos (x-1)$, then $\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^3+3 h}=$

  • A
    -25
  • B
    25
  • Correct answer
    C
    -10
  • D
    10

Solution

$\begin{aligned} & f(x)=3 x^{15}-5 x^{10}+7 x^5+50 \cos (x-1) \\ & \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^2+3 h}=\lim _{h \rightarrow 0}\left(\frac{f(1-h)-f(1)}{-h}\right) \\ & \lim _{h \rightarrow 0}\left(\frac{-1}{h^2+3}\right)=f^{\prime}(1)\left(\frac{-1}{3}\right) \\ & f^{\prime}(x)=45 x^{14}-50 x^9+35 x^4+50 \sin (x-1) \\ & f^{\prime}(1)=45-50+35=30 \\ & \therefore \lim _{\rightarrow \rightarrow} \frac{f(1-h)-f(1)}{h}=-10\end{aligned}$

Hence, the correct answer is option: 'C'.

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