If f(x) = begin{cases} frac{sqrt{a^2-ax+x^2}- | vedclass

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

Question

(A) Since $f(x)$ is continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0) = K$.We evaluate the limit: $\lim_{x 	o 0} \frac{\sqrt{a^2-ax+x^2}-\

Vedclass · Accepted Answer

$-\sqrt{a}$

Vedclass · Answer

(A) Since $f(x)$ is continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0) = K$.We evaluate the limit: $\lim_{x 	o 0} \frac{\sqrt{a^2-ax+x^2}-\sqrt{x^2+ax+a^2}}{\sqrt{a+x}-\sqrt{a-x}}$.Rationalizing the numerator and denominator:Multiply by $\frac{\sqrt{a^2-ax+x^2}+\sqrt{x^2+ax+a^2}}{\sqrt{a^2-ax+x^2}+\sqrt{x^2+ax+a^2}} 	imes \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}$.The numerator becomes $(a^2-ax+x^2) - (x^2+ax+a^2) = -2ax$.The denominator becomes $(a+x) - (a-x) = 2x$.So,$K = \lim_{x 	o 0} \frac{-2ax}{2x} 	imes \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a^2-ax+x^2}+\sqrt{x^2+ax+a^2}}$.$K = \lim_{x 	o 0} (-a) 	imes \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a^2-ax+x^2}+\sqrt{x^2+ax+a^2}}$.Substituting $x=0$: $K = (-a) 	imes \frac{\sqrt{a}+\sqrt{a}}{\sqrt{a^2}+\sqrt{a^2}} = (-a) 	imes \frac{2\sqrt{a}}{2a} = -\sqrt{a}$.

If $f(x) = \begin{cases} \frac{\sqrt{a^2-ax+x^2}-\sqrt{x^2+ax+a^2}}{\sqrt{a+x}-\sqrt{a-x}}, & x \neq 0 \\ K, & x=0 \end{cases}$ is continuous at $x=0$,then $K=$

Similar Questions

The function $f(x) = \begin{cases} sgn([x]) & x \notin I \\ [sgn(x)] & x \in I \end{cases}$ is (where $sgn()$ denotes the signum function and $[.]$ denotes the greatest integer function):

Let $f:[-2,2] \rightarrow \mathbb{R}$ be a continuous function such that $f(x)$ assumes only irrational values. If $f(\sqrt{2})=\sqrt{2},$ then

If $\begin{aligned} f(x) &= \frac{4 \sin \pi x}{5 x} \text{ for } x \neq 0 \\ &= 2k \text{ for } x = 0 \end{aligned}$ is continuous at $x = 0$,then the value of $k$ is

Which of the following function$(s)$ not defined at $x = 0$ has/have an irremovable discontinuity at $x = 0$?

If $f(x) = \begin{cases} 1 + x^2, & \text{when } 0 \le x \le 1 \\ 1 - x, & \text{when } x > 1 \end{cases}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module