JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 14)
The number of points, where the function $$f: \mathbf{R} \rightarrow \mathbf{R}$$,
$$f(x)=|x-1| \cos |x-2| \sin |x-1|+(x-3)\left|x^{2}-5 x+4\right|$$, is NOT differentiable, is :
1
2
3
4
Explanation
$$f:R \to R$$.
$$f(x) = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|{x^2} - 5x + 4|$$
$$ = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|x - 1||x - 4|$$
$$ = |x - 1|[\cos |x - 2|\sin |x - 1| + (x - 3)|x - 4|]$$
Sharp edges at $$x = 1$$ and $$x = 4$$
$$\therefore$$ Non-differentiable at $$x = 1$$ and $$x = 4$$
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