Question

Let f(x) = min {1, 1 + x sin x}, 0 ≤ x ≤ 2π. If m is the number of points, where f is not differentiable, and n is the number of points, where f is not continuous, then the ordered pair (m , n) is equal to

• A. (2, 0)
• B. (1, 0)
• C. (1, 1)
• D. (2, 1)

Answer 1

Answer: (B)

(1, 0)

Answer 2

The correct answer is (B) : (1, 0)

$f(x) = min{1, 1 + xsinx}, 0 ≤x ≤2π$
$f(x) = \begin{cases} 1,0 \leq x < \pi \\\ 1 + x\sin x, \pi \leq x \leq 2\pi \end{cases}$
Now at x = $\pi$,
$\lim_{{x \to \pi^-}} \limits$ = 1 =$\lim_{{x \to \pi^-}} \limits$ $ƒ(x)$
∴ f(x) is continuous in [0, 2π]
Now, at x = π
L.H.D = $\lim_{{h \to 0}} \limits$ $\frac{ƒ( π - h ) - ƒ( π )}{-h}$ = 0
R.H.D = $\lim_{{h \to 0}} \limits$ $\frac{ƒ( π + h ) - ƒ( π )}{-h}$ = $1$ -$$\frac{ (π + h)sinh - 1}{h}
$= –π$
∴ f(x) is not differentiable at x = π
∴ (m , n) = (1, 0)