Maths · Limits, continuity and differentiability
Assertion If =0 \) has two distinct positive real roots then number of non- diff
Assertion If \( f(x)=0 \) has two distinct positive real roots then number of non- differentiable points of \( \boldsymbol{y}=|\boldsymbol{f}(-|\boldsymbol{x}|)| \) is \( \mathbf{1} \) Reason Graph of \( \boldsymbol{y}=\boldsymbol{f}(|\boldsymbol{x}|) \) is symmetrical about y-axis
- A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- C. Assertion is correct but Reason is incorrect
- D. Assertion is incorrect but Reason is correct
Step-by-step solution
The assertion is correct because with two distinct positive roots of f(x)=0, f(-|x|) never zero (since argument is non-positive, never equals positive roots), so the only non-differentiable point arises from the absolute value of x at x=0, giving 1 such point. The reason is also correct because y=f(|x|) is symmetrical about y-axis, but this does not explain the assertion as the given function is |f(-|x|)|, not f(|x|).
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