JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 9)
Let the functions f : R $$ \to $$ R and g : R $$ \to $$ R be defined as :
$$f(x) = \left\{ {\matrix{ {x + 2,} & {x < 0} \cr {{x^2},} & {x \ge 0} \cr } } \right.$$ and
$$g(x) = \left\{ {\matrix{ {{x^3},} & {x < 1} \cr {3x - 2,} & {x \ge 1} \cr } } \right.$$
Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :
$$f(x) = \left\{ {\matrix{ {x + 2,} & {x < 0} \cr {{x^2},} & {x \ge 0} \cr } } \right.$$ and
$$g(x) = \left\{ {\matrix{ {{x^3},} & {x < 1} \cr {3x - 2,} & {x \ge 1} \cr } } \right.$$
Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :
0
3
1
2
Explanation
$$fog(x) = \left\{ {\matrix{
{{x^3} + 2,} & {x \le 0} \cr
{{x^6},} & {0 \le x \le 1} \cr
{{{(3x - 2)}^2},} & {x \ge 1} \cr
} } \right.$$
$$ \because $$ fog(x) is discontinuous at x = 0 then non-differentiable at x = 0
Now,
at x = 1
$$RHD = \mathop {\lim }\limits_{h \to 0} {{f(1 + h) - f(1)} \over h} = \mathop {\lim }\limits_{h \to 0} {{{{(3(1 + h) - 2)}^2} - 1} \over h} = 6$$
$$LHD = \mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^6} - 1} \over { - h}} = 6$$
Number of points of non-differentiability = 1
$$ \because $$ fog(x) is discontinuous at x = 0 then non-differentiable at x = 0
Now,
at x = 1
$$RHD = \mathop {\lim }\limits_{h \to 0} {{f(1 + h) - f(1)} \over h} = \mathop {\lim }\limits_{h \to 0} {{{{(3(1 + h) - 2)}^2} - 1} \over h} = 6$$
$$LHD = \mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^6} - 1} \over { - h}} = 6$$
Number of points of non-differentiability = 1
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