Differentiability Proofs PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2023
00:00
Show that $f(x) = |x - 3|$ is continuous but not differentiable at $x = 3$. 2 Marks
Continuity: $LHL = \lim_{h \to 0} |(3-h)-3| = 0$. $RHL = \lim_{h \to 0} |(3+h)-3| = 0$. $f(3) = 0$. So continuous.
Differentiability: $LHD = \lim_{h \to 0} \frac{|(3-h)-3| - |3-3|}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1$.
$RHD = \lim_{h \to 0} \frac{|(3+h)-3| - |3-3|}{h} = \lim_{h \to 0} \frac{h}{h} = 1$.
Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x=3$.
Proved
Q2
2022 (Term 2)
00:00
Check differentiability of $f(x) = |x - 1|$ at $x = 1$. 2 Marks
$LHD = \lim_{h \to 0} \frac{|(1-h)-1| - 0}{-h} = -1$.
$RHD = \lim_{h \to 0} \frac{|(1+h)-1| - 0}{h} = 1$.
Since $LHD \neq RHD$, not differentiable at $x=1$.
Not differentiable at x=1
Q3
2020
00:00
Check whether $f(x) = x|x|$ is differentiable at $x = 0$. 2 Marks
$f(x) = \begin{cases} x^2 & x \ge 0 \\ -x^2 & x < 0 \end{cases}$.
$LHD = \lim_{x \to 0^-} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^-} \frac{-x^2}{x} = 0$.
$RHD = \lim_{x \to 0^+} \frac{x^2}{x} = 0$.
Since $LHD = RHD = 0$, it is differentiable at $x=0$.
Differentiable at x=0
Q4
2018 (All India)
00:00
Find $a$ and $b$ if $f(x) = \begin{cases} x^2 + 3x + a & x \le 1 \\ bx + 2 & x > 1 \end{cases}$ is differentiable at $x = 1$. 4 Marks
1) For continuity at $x=1$: $1+3+a = b+2 \Rightarrow a - b = -2$.
2) For differentiability: $LHD = \frac{d}{dx}(x^2+3x)|_{x=1} = 2x+3|_{x=1} = 5$.
$RHD = \frac{d}{dx}(bx+2)|_{x=1} = b$.
Differentiability $\Rightarrow b = 5$.
Substitute $b=5$ in (1): $a - 5 = -2 \Rightarrow a = 3$.
a = 3, b = 5
Q5
2025 Board
00:00
If $f(x) = |x| + |x - 1|$, which of the following is correct? 1 Mark
The sum of two continuous functions ($|x|$ and $|x-1|$) is always continuous.
However, $|x|$ is not differentiable at $x=0$, and $|x-1|$ is not differentiable at $x=1$.
Thus, their sum is continuous but not differentiable at these critical points.
(c) f(x) is continuous but not differentiable at x = 0 and x = 1
Q6
2019
00:00
Show that $f(x) = x^{2/3}$ is not differentiable at $x = 0$. 2 Marks
Check $f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{h^{2/3}}{h} = \lim_{h \to 0} \frac{1}{h^{1/3}}$.
As $h \to 0^+$, limit is $\infty$. As $h \to 0^-$, limit is $-\infty$.
Since the limit is not finite, $f(x)$ is not differentiable at $x=0$.
Proved
Q7
2016 (Foreign)
00:00
Find $a$ and $b$ so that $f(x) = \begin{cases} x^2 + ax + b & 0 \le x < 2 \\ 3x + 2 & 2 \le x \le 4 \end{cases}$ is differentiable at $x = 2$. 4 Marks
1) Continuity at $x=2$: $4 + 2a + b = 3(2) + 2 = 8 \Rightarrow 2a + b = 4$.
2) Differentiability at $x=2$: $LHD = 2x + a |_{x=2} = 4 + a$. $RHD = 3$.
$4 + a = 3 \Rightarrow a = -1$.
Substitute $a=-1$ in (1): $2(-1) + b = 4 \Rightarrow b = 6$.
a = -1, b = 6
Q8
2015 (Foreign)
00:00
Check the differentiability of the function $f(x) = \begin{cases} 2x + 3 & -3 \le x < -2 \\ x^2 + 1 & -2 \le x < 0 \\ x+1 & 0 \le x \le 3 \end{cases}$ at $x = -2$ and $x = 0$. 4 Marks
At $x=-2$: $LHD = \frac{d}{dx}(2x+3) = 2$. $RHD = \frac{d}{dx}(x^2+1) = 2x|_{x=-2} = -4$.
Since $2 \neq -4$, not differentiable at $x=-2$.
At $x=0$: $LHD = \frac{d}{dx}(x^2+1) = 2x|_{x=0} = 0$. $RHD = \frac{d}{dx}(x+1) = 1$.
Since $0 \neq 1$, not differentiable at $x=0$.
Not differentiable at x=-2 and x=0