SATHEE: Application of Derivatives 4 Question 37

Application of Derivatives 4 Question 37

####39. Which of the following is true?

(a) $(2 + a)^{2} f^{''} (1) + (2 - a)^{2} f^{''} (- 1) = 0$

(b) $(2 - a)^{2} f^{''} (1) - (2 + a)^{2} f^{''} (- 1) = 0$

(d) $f^{'} (1) f^{'} (- 1) = - (2 + a)^{2}$

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Answer:

Correct Answer: 39. 6

Solution:

$f (x) = \frac{(x^{2} + a x + 1) - 2 a x}{x^{2} + a x + 1} = 1 - \frac{2 a x}{x^{2} + a x + 1}$

$\begin{aligned} f^{'} (x) & = - \frac{(x^{2} + a x + 1) \cdot 2 a - 2 a x (2 x + a)}{{(x^{2} + a x + a)}^{2}} \\ = \frac{- 2 a x^{2} + 2 a}{{(x^{2} + a x + a)}^{2}} = 2 a \frac{(x^{2} - 1)}{{(x^{2} + a x + 1)}^{2}} \end{aligned}$

$\begin{aligned} {(x^{2} + a x + 1)}^{2} (2 x) - 2 (x^{2} - 1) \\ f^{''} (x) = 2 a \frac{(x^{2} + a x + 1) (2 x + a)}{{(x^{2} + a x + 1)}^{4}} \\ = 2 a \frac{2 x (x^{2} + a x + 1) - 2 (x^{2} - 1) (2 x + a)}{{(x^{2} + a x + 1)}^{3}} \end{aligned}$

Now, $f^{''} (1) = \frac{4 a (a + 2)}{(a + 2)^{3}} = \frac{4 a}{(a + 2)^{2}}$

and $f^{''} (- 1) = \frac{4 a (a - 2)}{(2 - a)^{3}} = \frac{- 4 a}{(a - 2)^{2}}$

$∴ (2 + a)^{2} f^{''} (1) + (2 - a)^{2} f^{''} (- 1) = 4 a - 4 a = 0$

Application of Derivatives 4 Question 37

Answer:

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Application of Derivatives 4 Question 37

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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