Application of Derivatives 4 Question 37
####39. Which of the following is true?
(a) $(2+a)^{2} f^{\prime \prime}(1)+(2-a)^{2} f^{\prime \prime}(-1)=0$
(b) $(2-a)^{2} f^{\prime \prime}(1)-(2+a)^{2} f^{\prime \prime}(-1)=0$
(c) $f^{\prime}(1) f^{\prime}(-1)=(2-a)^{2}$
(d) $f^{\prime}(1) f^{\prime}(-1)=-(2+a)^{2}$
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Answer:
Correct Answer: 39. 6
Solution:
- $f(x)=\frac{\left(x^{2}+a x+1\right)-2 a x}{x^{2}+a x+1}=1-\frac{2 a x}{x^{2}+a x+1}$
$ \begin{aligned} f^{\prime}(x) & =-\frac{\left(x^{2}+a x+1\right) \cdot 2 a-2 a x(2 x+a)}{\left(x^{2}+a x+a\right)^{2}} \\ & =\frac{-2 a x^{2}+2 a}{\left(x^{2}+a x+a\right)^{2}}=2 a \frac{\left(x^{2}-1\right)}{\left(x^{2}+a x+1\right)^{2}} \end{aligned} $
$ \begin{aligned} & \left(x^{2}+a x+1\right)^{2}(2 x)-2\left(x^{2}-1\right) \\ & f^{\prime \prime}(x)=2 a \frac{\left(x^{2}+a x+1\right)(2 x+a)}{\left(x^{2}+a x+1\right)^{4}} \\ & =2 a \frac{2 x\left(x^{2}+a x+1\right)-2\left(x^{2}-1\right)(2 x+a)}{\left(x^{2}+a x+1\right)^{3}} \end{aligned} $
Now, $f^{\prime \prime}(1)=\frac{4 a(a+2)}{(a+2)^{3}}=\frac{4 a}{(a+2)^{2}}$
and $f^{\prime \prime}(-1)=\frac{4 a(a-2)}{(2-a)^{3}}=\frac{-4 a}{(a-2)^{2}}$
$\therefore \quad(2+a)^{2} f^{\prime \prime}(1)+(2-a)^{2} f^{\prime \prime}(-1)=4 a-4 a=0$
