JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 4)
The function
$$f(x) = \left| {{x^2} - 2x - 3} \right|\,.\,{e^{\left| {9{x^2} - 12x + 4} \right|}}$$ is not differentiable at exactly :
$$f(x) = \left| {{x^2} - 2x - 3} \right|\,.\,{e^{\left| {9{x^2} - 12x + 4} \right|}}$$ is not differentiable at exactly :
four points
three points
two points
one point
Explanation
$$f(x) = \left| {(x - 3)(x + 1)} \right|\,.\,{e^{{{(3x - 2)}^2}}}$$
$$f(x) = \left\{ {\matrix{ {(x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in (3,\infty )} \cr { - (x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in [ - 1,3]} \cr {(x - 3)\,.\,(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in ( - \infty , - 1)} \cr } } \right.$$
Clearly, non-differentiable at x = $$-$$1 & x = 3.
$$f(x) = \left\{ {\matrix{ {(x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in (3,\infty )} \cr { - (x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in [ - 1,3]} \cr {(x - 3)\,.\,(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in ( - \infty , - 1)} \cr } } \right.$$
Clearly, non-differentiable at x = $$-$$1 & x = 3.
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