JEE Main 2019 (Online) 10th January Evening Slot | Limits, Continuity and Differentiability Question 145 | Mathematics

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Let f : ($$-$$1, 1) $$ \to $$ R be a function defined by f(x) = max $$\left\{ { - \left| x \right|, - \sqrt {1 - {x^2}} } \right\}.$$ If K be the set of all points at which f is not differentiable, then K has exactly -

one element

three elements

five elements

two elements

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

Let f : R $$ \to $$ R be a function such that f(x) = x³ + x²f'(1) + xf''(2) + f'''(3), x $$ \in $$ R. Then f(2) equals -

$$-$$ 2

$$-$$ 4

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

For each t $$ \in $$ R , let [t] be the greatest integer less than or equal to t

Then $$\mathop {\lim }\limits_{x \to 1 + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$$

equals $$-$$ 1

equals 1

equals 0

does not exist

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

Let $$f\left( x \right) = \left\{ {\matrix{ {\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr {8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr } } \right.$$

Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S

equals $$\left\{ { - 2, - 1,1,2} \right\}$$

equals $$\left\{ { - 2, - 1,0,1,2} \right\}$$

equals $$\left\{ { - 2,2} \right\}$$

is an empty set

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