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

The function $f(x) = \max \{(1 - x), (1 + x), 2\},$ $x \in ( - \infty , \infty ),$ is

Consider the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ are respectively the number of points at which $f$ is not continuous and $f$ is not differentiable,then $m+n$ is

Number of points where the function $f(x) = \text{maximum}(\sqrt{2x - x^2}, 2 - x)$ is non-differentiable is:

In the interval $[0, 3]$,the function $f(x) = |x - 1| + |x - 2|$ is

If $f(x) = \begin{cases} x + 2, & -1 < x < 3 \\ 5, & x = 3 \\ 8 - x, & x > 3 \end{cases}$,then at $x = 3$,$f'(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module