Question

Let $f\left(x\right)=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$. Then $\frac{f(1-h)-f\left(1\right)}{h^3+3h}$:

• A. does not exist
• B. is $\frac{50}{3}$
• C. is $\frac{53}{3}$
• D. is $\frac{22}{3}$

Answer 1

Answer: (C)

is $\frac{53}{3}$

we know $f^{\prime }\left(x\right)=lim\left(\frac{f(x+h)-f\left(x\right)}{h}\right)$ , h $\to$ 0 by this

$\Rightarrow$ $lim\left(\frac{f(1-h)-f\left(1\right)}{-h\times (h^2+3)}\right)=-\frac{f^{\prime }\left(1\right)}{h^2+3}$..........equation (1)

now $f^{\prime }\left(x\right)=30x^9-56x^7+30x^5-63x^2+6x$

$\Rightarrow$ $f^{\prime }\left(1\right)=30-56+30-63+6$

$\Rightarrow$ $f^{\prime }\left(1\right)=-53$ put the value in equation 1

$\Rightarrow$ $lim\left(\frac{f(1-h)-f\left(1\right)}{-h\times (h^2+3)}\right)=-\frac{f^{\prime }\left(1\right)}{h^2+3}$ = $\frac{53}{h^2+3}$ , h $\to$0

$\Rightarrow$ $\frac{53}{0+3}=\frac{53}{3}$

ANS [C]