SATHEE: Limit Continuity and Differentiability 7 Question 29

Limit Continuity and Differentiability 7 Question 29

29. Let $a, b \in R$ and $f : R \to R$ be defined by $f (x) = a \cos (| x^{3} - x |) + b | x | \sin (| x^{3} + x |)$ . Then, $f$ is

(2016 Adv.)

(a) differentiable at $x = 0$ , if $a = 0$ and $b = 1$

(b) differentiable at $x = 1$ , if $a = 1$ and $b = 0$

(d) not differentiable at $x = 1$ , if $a = 1$ and $b = 1$

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Answer:

Correct Answer: 29. (A) $\to$ p, q, r, s; (B) $\to p, s$ ; (C) $\to r, s$ ; (D) $\to p, s$

Solution:

Here, $(\sin y)^{\sin \frac{π}{2} x} + \frac{\sqrt{3}}{2} \sec^{- 1} (2 x) + 2^{x} \tan \log (x + 2) = 0$

On differentiating both sides, we get

$\begin{aligned} (\sin y)^{\sin \frac{π}{2} x} \cdot \log (\sin y) \cdot \cos \frac{π}{2} x \cdot \frac{π}{2} \\ + \sin \frac{π}{2} x (\sin y)^{\sin \frac{π}{2} x - 1} \cdot \cos y \cdot \frac{d y}{d x} \\ + \frac{\sqrt{3}}{2} \cdot \frac{2}{(2 | x |) \sqrt{4 x^{2} - 1}} + \frac{2^{x} \cdot \sec^{2} \log (x + 2)}{(x + 2)} \\ + 2^{x} \log 2 \cdot \tan \log (x + 2) = 0 \end{aligned}$

Putting $x = - 1, y = - \frac{\sqrt{3}}{π}$ , we get

Limit Continuity and Differentiability 7 Question 29

29. Let $a, b \in R$ and $f : R \to R$ be defined by $f (x) = a \cos (| x^{3} - x |) + b | x | \sin (| x^{3} + x |)$ . Then, $f$ is

Answer:

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Limit Continuity and Differentiability 7 Question 29

29. Let a,b∈R and f:R→R be defined by f(x)=acos⁡(|x3−x|)+b|x|sin⁡(|x3+x|). Then, f is

Answer:

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29. Let $a, b \in R$ and $f : R \to R$ be defined by $f (x) = a \cos (| x^{3} - x |) + b | x | \sin (| x^{3} + x |)$ . Then, $f$ is