JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

Let

f(x)= \begin{cases}1 /|x| & \text { for }|x| \geq 1 \\ a x^2+b & \text { for }|x|<1\end{cases}

The coefficients

a

and

b

so that

f

is continuous and differentiable at any point, are equal to

View answer and explanation

Correct option: A

a=-1 / 2, b=3 / 2

Option A (

a=-1 / 2, b=3 / 2

) is correct because these values ensure that the function is both continuous and differentiable at

x = 1

and

x = -1

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