Assertion (A): If y = f(x) = (|x| - |x - 1|)^ | vedclass

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

Question

(D) For $x > 1$,$|x| = x$ and $|x - 1| = x - 1$.Then $f(x) = (x - (x - 1))^2 = (1)^2 = 1$.For $0 < x < 1$,$|x| = x$ and $|x - 1| = -(x - 1) = 1 - x$.T

Vedclass · Accepted Answer

$A$ is false,$R$ is true.

Vedclass · Answer

(D) For $x > 1$,$|x| = x$ and $|x - 1| = x - 1$.Then $f(x) = (x - (x - 1))^2 = (1)^2 = 1$.For $0 Then $f(x) = (x - (1 - x))^2 = (2x - 1)^2 = 4x^2 - 4x + 1$.The left-hand derivative at $x = 1$ is $LHD = \lim_{h ightarrow 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h ightarrow 0^-} \frac{(2(1+h) - 1)^2 - 1}{h} = \lim_{h ightarrow 0^-} \frac{(2h + 1)^2 - 1}{h} = \lim_{h ightarrow 0^-} \frac{4h^2 + 4h}{h} = 4$.The right-hand derivative at $x = 1$ is $RHD = \lim_{h ightarrow 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h ightarrow 0^+} \frac{1 - 1}{h} = 0$.Since $LHD 
eq RHD$,the derivative at $x = 1$ does not exist.Thus,Assertion $(A)$ is false.The Reason $(R)$ is the standard definition of the derivative,which is true.Therefore,$(A)$ is false and $(R)$ is true.

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then:

Similar Questions

If $f(x) = \begin{cases} 2x+3, & x \leq 1 \\ ax^{2}+bx, & x > 1 \end{cases}$ is differentiable $\forall x \in R$,then $f(2) = $ . . . . . . .

Let $f(x) = |x - \alpha| + |x - \beta|$, where $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 2 = 0$. Then the number of points in $[\alpha, \beta]$ at which $f$ is not differentiable is

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,

The number of points,where the function $f: R \rightarrow R, f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$ is $NOT$ differentiable,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.Then: