Define a function f: R rightarrow R by f(x) = | vedclass

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

Question

(B) Since $f(x)$ is differentiable on $R$,it must be continuous at $x = 0$.For continuity at $x = 0$:$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightar

Vedclass · Accepted Answer

$a = 1, b = 0$

Vedclass · Answer

(B) Since $f(x)$ is differentiable on $R$,it must be continuous at $x = 0$.For continuity at $x = 0$:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^+} f(x) = f(0)$$\lim_{x ightarrow 0^-} \frac{\sin x^2}{x} = \lim_{x ightarrow 0^+} (x^2 + ax + b)$Using the limit $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we have $\lim_{x ightarrow 0^-} (x \cdot \frac{\sin x^2}{x^2}) = 0 \cdot 1 = 0$.Thus,$0 = b$,so $b = 0$.For differentiability at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$:$LHD = \lim_{h ightarrow 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h ightarrow 0^+} \frac{\frac{\sin(-h)^2}{-h} - 0}{-h} = \lim_{h ightarrow 0^+} \frac{\sin h^2}{h^2} = 1$.$RHD = \lim_{h ightarrow 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0^+} \frac{h^2 + ah + b - b}{h} = \lim_{h ightarrow 0^+} (h + a) = a$.Equating $LHD$ and $RHD$,we get $a = 1$.Therefore,$a = 1$ and $b = 0$.

Define a function $f: R \rightarrow R$ by $f(x) = \begin{cases} \frac{\sin x^2}{x}, & \text{for } x < 0 \\ x^2 + ax + b, & \text{for } x \geq 0 \end{cases}$. Suppose $f(x)$ is differentiable on $R$. Then,

Similar Questions

If $f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x > 1 \end{cases}$ is everywhere differentiable, then:

The function $f(x) = a \sin |x| + b e^{|x|}$ is differentiable at $x = 0$ when

Let $f(x) = \begin{cases} 1, & \forall x < 0 \\ 1 + \sin x, & \forall 0 \le x \le \pi/2 \end{cases}$,then what is the value of $f'(x)$ at $x = 0$?

The function $f(x) = \begin{cases} 2x + 1, & x \in \mathbb{Q} \\ x^2 - 2x + 5, & x \notin \mathbb{Q} \end{cases}$ is

Let $f(x) = x|x|$ and $g(x) = \sin x$.
Statement-$1$: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-$2$: $gof$ is twice differentiable at $x=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module