JEE Main 2019 (Online) 10th April Morning Slot | Limits, Continuity and Differentiability Question 188 | Mathematics

If$$f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} & {,x > 0} \cr } } \right.$$
is continuous at x = 0, then the ordered pair (p, q) is equal to

$$\left( { - {3 \over 2}, - {1 \over 2}} \right)$$

$$\left( { - {1 \over 2},{3 \over 2}} \right)$$

$$\left( { - {3 \over 2}, {1 \over 2}} \right)$$

$$\left( { {5 \over 2}, {1 \over 2}} \right)$$

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

If $$f(x) = [x] - \left[ {{x \over 4}} \right]$$ ,x $$ \in $$ 4 , where [x] denotes the greatest integer function, then

Both $$\mathop {\lim }\limits_{x \to 4 - } f(x)$$ and $$\mathop {\lim }\limits_{x \to 4 + } f(x)$$ exist but are not equal

f is continuous at x = 4

$$\mathop {\lim }\limits_{x \to 4 + } f(x)$$ exists but $$\mathop {\lim }\limits_{x \to 4 - } f(x)$$ does not exist

$$\mathop {\lim }\limits_{x \to 4 - } f(x)$$ exists but $$\mathop {\lim }\limits_{x \to 4 + } f(x)$$ does not exist

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

If the function $$f(x) = \left\{ {\matrix{ {a|\pi - x| + 1,x \le 5} \cr {b|x - \pi | + 3,x > 5} \cr } } \right.$$
is continuous at x = 5, then the value of a – b is :-

$${2 \over {\pi - 5 }}$$

$${2 \over {5 - \pi }}$$

$${-2 \over {\pi + 5 }}$$

$${2 \over {\pi + 5 }}$$

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

Let ƒ(x) = 15 – |x – 10|; x $$ \in $$ R. Then the set of all values of x, at which the function, g(x) = ƒ(ƒ(x)) is not differentiable, is :

{10,15}

{5,10,15,20}

{10}

{5,10,15}