If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\int_{0}^{\mathrm{x}}(5+|1-\mathrm{t}|) \mathrm{dt}, & x>2 \\ 5 \mathrm{x}+1, & x \leq 2\end{array}\right.$, then

(C) $f(x)$ is continuous but not differentiable at $x=2$

(A) $f(x)$ is not continuous at $x=2$

(B) $f(x)$ is not differentiable at $x=1$

(D) $f(x)$ is everywhere differentiable

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Continuity and Differentiability

f (x) = {\int_{0}^{x} (5 + ∣1 - t ∣) dt, 5 x + 1, x > 2 x \leq 2

, then

Answer: (C)

For

x > 2

the modulus splits at

t = 1

f (x) = \int_{0}^{1} (6 - t) d t + \int_{1}^{x} (4 + t) d t = \frac{11}{2} + (4 x + \frac{x ^{2}}{2} - \frac{9}{2}) = \frac{x ^{2}}{2} + 4 x + 1

x = 2

the lower branch gives

5 (2) + 1 = 11

and this expression gives

2 + 8 + 1 = 11

, so

f

is continuous there. The left derivative is

5

, while the right derivative is

x + 4

evaluated at

x = 2

, namely

6

, so the two do not agree. On

x \leq 2

the function is the straight line

5 x + 1

, which is differentiable at

x = 1

. So

f

is continuous but not differentiable at

x = 2

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option