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

The Correct Option is B

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)