If f(x)=≤ft\{{array}{ccl}a/2(x-|x|) & & \\ 0, & { for } & x 0{array}. is continuous at x=0, then

Correct Answer is : a is any real value and b is any real value

a is only rational value and b is any real value

a is only irrational value and b is any real value

a is only rational value and b is only rational value

Correct Answer is : a is any real value and b is any real value

Test Index

MHTCET 11 May 2024 Shift 1 Paper

Question : 108 of 150

Marks: +1, -0

f(x)=\left\{\begin{array}{ccl}\frac{a}{2}(x-|x|) & & \\ 0, & \text { for } & x < 0 \\ 0, & \text { for } & x=0 \\ b x^{2} \sin \left(\frac{1}{x}\right) & \text { for } & x > 0\end{array}\right.

x=0

, then

Solution:

Concept:

For continuity at

x=0

, the left-hand limit, right-hand limit, and

f(0)

must all be equal.

Explanation:

Given

f(0) = 0

For

x < 0

, we have

|x| = -x

f(x) = \frac{a}{2}(x - |x|) = \frac{a}{2}(x + x) = ax

Therefore,

\lim_{x \to 0^-} f(x) = \lim_{x \to 0} ax = 0

for any real value of

a

For

x > 0

f(x) = b x^{2} \sin\left(\frac{1}{x}\right)

Since

-1 \le \sin\left(\frac{1}{x}\right) \le 1

, we have

\left|b x^{2} \sin\left(\frac{1}{x}\right)\right| \le |b| x^{2}

x \to 0

|b| x^{2} \to 0

, so

\lim_{x \to 0^+} f(x) = 0

for any real value of

b

Thus, all three values are equal to

0

for any real

a

and

b

Answer:

Option A:

a

is any real value and

b

is any real value.

Go to Question:

Prev Question

Next Question