AP EAMCET · Maths · Continuity and Differentiability
Assertion (A) : If \(y=f(x)=(|x|-|x-1|)^2\), then \(\left(\frac{d y}{d x}\right)_{x=1}=1\)
Reason (R) : If \(\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\) exist, then it is called derivative of \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=\mathrm{a}\).
Then
- A (A) is true, (R) is true, (R) is correct explanation to (A)
- B (A) is true, (R) is true, (R) is not the correct explanation to (A)
- C (A) is true, (R) is false
- D (A) is false, (R) is true
Answer & Solution
Correct Answer
(D) (A) is false, (R) is true
Step-by-step Solution
Detailed explanation
For \(x \ge 1\), \(f(x)=(x-(x-1))^2 = 1\). For \(0 Left-hand derivative at \(x=1\): \(f'(1^-) = \lim_{h \to 0^-} \frac{(2(1+h)-1)^2 - 1}{h} = \lim_{h \to 0^-} \frac{(1+2h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{4h+4h^2}{h} = 4\). Right-hand derivative at \(x=1\):…
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