KCET 2025 | Limits, Continuity and Differentiability Question 1 | Mathematics

Column - I		Column - II
(a)	$x \| x \|$	(i)	continuous in (-1, 1)
(b)	$\sqrt{\| x \|}$	(ii)	differentiable in (-1, 1)
(c)	$x + [x]$	(iii)	strictly increasing in (-1, 1)
(d)	$\| x - 1 \| + \| x + 1 \|$	(iv)	not differentiable at, at least one point in (-1, 1)

$\mathrm{a}-\mathrm{i}, \mathrm{b}-\mathrm{ii}, \mathrm{c}-\mathrm{iv}, \mathrm{d}-\mathrm{iii}$

$a-i v, b-i i i, c-i, d-i i$

a - ii, b - iv, c - iii, d - i

a - iii, b - ii, c - iv, d - i

KCET 2025

MCQ (Single Correct Answer)

-0

The function $f(x)=\left\{\begin{array}{ll}e^x+a x & , x<0 \\ b(x-1)^2 & , x \geq 0\end{array}\right.$ is differentiable at $x=0$. Then

$a=1, b=1$

$a=3, b=1$

$a=-3, b=1$

$a=3, b=-1$

KCET 2025

MCQ (Single Correct Answer)

-0

$$ \text { A function } f(x)=\left\{\begin{array}{cl} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. $$

continuous at $\mathrm{x}=0$

not continuous at $x=0$

differentiable at $x=0$

differentiable at $\mathrm{x}=0$, but not continuous at $\mathrm{x}=0$

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