Let f(x) = begin{cases} a sin(x + b) & x ge 0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For $f(x)$ to be differentiable at $x = 0$,it must be continuous at $x = 0$. Thus,$\lim_{x 	o 0^-} f(x) = f(0) \implies 6(0)^7 - 0 + 1 = a \sin(b

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) For $f(x)$ to be differentiable at $x = 0$,it must be continuous at $x = 0$. Thus,$\lim_{x 	o 0^-} f(x) = f(0) \implies 6(0)^7 - 0 + 1 = a \sin(b) \implies a \sin b = 1$. Also,the left-hand derivative must equal the right-hand derivative at $x = 0$. $f'(x) = \begin{cases} a \cos(x + b) & x > 0 \ 42x^6 - 1 & x Equating the derivatives at $x = 0$: $a \cos b = -1$. Squaring and adding the two equations: $(a \sin b)^2 + (a \cos b)^2 = 1^2 + (-1)^2 \implies a^2(\sin^2 b + \cos^2 b) = 2 \implies a^2 = 2 \implies a = \pm \sqrt{2}$. Case $I$: If $a = \sqrt{2}$,then $\sin b = \frac{1}{\sqrt{2}}$ and $\cos b = -\frac{1}{\sqrt{2}}$. This corresponds to $b = \frac{3\pi}{4}$. Case $II$: If $a = -\sqrt{2}$,then $\sin b = -\frac{1}{\sqrt{2}}$ and $\cos b = \frac{1}{\sqrt{2}}$. This corresponds to $b = \frac{7\pi}{4}$. Thus,the ordered pairs are $(\sqrt{2}, \frac{3\pi}{4})$ and $(-\sqrt{2}, \frac{7\pi}{4})$. There are $2$ such ordered pairs.

Let $f(x) = \begin{cases} a \sin(x + b) & x \ge 0 \\ 6x^7 - x + 1 & x < 0 \end{cases}$ be differentiable for all real $x$. If $a \in \mathbb{R}$ and $b \in [0, 2\pi]$,then the number of ordered pairs $(a, b)$ is:

Similar Questions

If $f(x) = \begin{cases} ax^2 + b; & x \le 0 \\ x^2; & x > 0 \end{cases}$ possesses a derivative at $x = 0$,then:

If $f(x) = \begin{cases} ax^2 + bx - \frac{13}{8}, & x \leq 1 \\ 3x - 3, & 1 < x \leq 2 \\ bx^3 + 1, & x > 2 \end{cases}$ is differentiable $\forall x \in R$,then $a - b =$

If $\alpha \in R - \{-1\}$ and $f(x) = |(|x| + \alpha)(|x| - 1)|$,then the number of points at which $f(x)$ is not differentiable is:

Number of points where the function $f(x) = (x^2 - 1) | x^2 - x - 2 | + \sin(|x|)$ is not differentiable,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module