SATHEE: Limit Continuity and Differentiability 7 Question 24

Limit Continuity and Differentiability 7 Question 24

24. There exists a function $f (x)$ satisfying $f (0) = 1$ , $f^{'} (0) = - 1, f (x) > 0, \forall x$ and

$(1982, 2 M)$

(a) $f^{''} (x) < 0, \forall x$

(b) $- 1 < f^{''} (x) < 0, \forall x$

(d) $f^{''} (x) < - 2, \forall x$

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Answer:

Correct Answer: 24. $(b, c)$

Solution:

Let $u = \sec^{- 1} - \frac{1}{2 x^{2} - 1}$ and $v = \sqrt{1 - x^{2}}$

Put $x = \cos θ$

$∴ u = \sec^{- 1} (- \sec 2 θ)$ and $v = \sin θ$

$\Rightarrow u = π - 2 θ [∵ \sec^{- 1} (- x) = π - \sec^{- 1} x]$

and $v = \sin θ$

$\Rightarrow \frac{d u}{d θ} = - 2$

an $d \frac{d v}{d θ} = \cos θ$

$\Rightarrow \frac{d u}{d v} = - \frac{2}{\cos θ}, \frac{d u}{d v}_{θ = π / 3} = - 4$

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Limit Continuity and Differentiability 7 Question 24

24. There exists a function f(x) satisfying f(0)=1, f′(0)=−1,f(x)>0,∀x and

Answer:

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24. There exists a function $f (x)$ satisfying $f (0) = 1$ , $f^{'} (0) = - 1, f (x) > 0, \forall x$ and