JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 20)
The number of points, at which the function
f(x) = | 2x + 1 | $$-$$ 3| x + 2 | + | x2 + x $$-$$ 2 |, x$$\in$$R is not differentiable, is __________.
f(x) = | 2x + 1 | $$-$$ 3| x + 2 | + | x2 + x $$-$$ 2 |, x$$\in$$R is not differentiable, is __________.
Answer
2
Explanation
$$f(x) = |2x + 1| - 3|x + 2| + |{x^2} + x - 2|$$
$$f(x) = \left\{ {\matrix{ {{x^2} - 7;} & {x > 1} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr {{x^2} + 2x + 3;} & {x < - 2} \cr } } \right.$$
$$ \therefore $$ $$f'(x) = \left\{ {\matrix{ {2x;} & {x > 1} \cr {2x - 3;} & { - {1 \over 2} < x < 1} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr {2x + 2;} & {x < - 2} \cr } } \right.$$
Check at 1, $$-$$2 and $${{ - 1} \over 2}$$
Non. differentiable at x = 1 and $${{ - 1} \over 2}$$
$$f(x) = \left\{ {\matrix{ {{x^2} - 7;} & {x > 1} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr {{x^2} + 2x + 3;} & {x < - 2} \cr } } \right.$$
$$ \therefore $$ $$f'(x) = \left\{ {\matrix{ {2x;} & {x > 1} \cr {2x - 3;} & { - {1 \over 2} < x < 1} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr {2x + 2;} & {x < - 2} \cr } } \right.$$
Check at 1, $$-$$2 and $${{ - 1} \over 2}$$
Non. differentiable at x = 1 and $${{ - 1} \over 2}$$
Comments (0)
