Question:
Which of the following functions is differentiable at \(x = 0\)?
निम्नलिखित में से कौन सा फलन \(x = 0\) पर अवकलनीय (differentiable) है?
Options:
- \(\cos(|x|) + |x|\)
- \(\cos(|x|) – |x|\)
- \(\sin(|x|) + |x|\)
- \(\sin(|x|) – |x|\)
- \(\cos(|x|) + |x|\)
- \(\cos(|x|) – |x|\)
- \(\sin(|x|) + |x|\)
- \(\sin(|x|) – |x|\)
Correct Answer: d)
Explanation:
Let \(f(x) = \sin(|x|) – |x|\).
– For \(x \ge 0\): \(f(x) = \sin x – x\). Right derivative: \(R f'(0) = \lim_{h \to 0^+} \frac{\sin h – h}{h} = 1 – 1 = 0\).
– For \(x < 0\): \(f(x) = \sin(-x) - (-x) = -\sin x + x\). Left derivative: \(L f'(0) = \lim_{h \to 0^-} \frac{-\sin h + h}{h} = -1 + 1 = 0\).
Since \(R f'(0) = L f'(0) = 0\), the function is differentiable at \(x = 0\).
For all other options, the left and right derivatives differ at \(x = 0\).
Hence, the correct option is (d).
\(f(x) = \sin(|x|) – |x|\) के लिए:
दायाँ अवकलज: \(R f'(0) = \lim_{h \to 0^+} \frac{\sin h – h}{h} = 0\)।
बायाँ अवकलज: \(L f'(0) = \lim_{h \to 0^-} \frac{-\sin h + h}{h} = 0\)।
चूँकि बायाँ और दायाँ अवकलज दोनों 0 के बराबर हैं, अतः फलन \(x = 0\) पर अवकलनीय है।
अतः सही विकल्प (d) है।