New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Limits Continuity and Differentiability question

2023 · 12 Apr · Shift 1 · Q41

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Limits Continuity and Differentiability
2023 · 12 Apr · Shift 1 · Q41

Limits Continuity and Differentiability question

2023 · 12 Apr · Shift 1 · Q41

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1

Let

[x]

be the greatest integer

\leq x

. Then the number of points in the interval

(-2,1)

, where the function

f(x)=|[x]|+\sqrt{x-[x]}

is discontinuous, is

\underline{\hspace{2cm}}

.

Numerical answer

More from Limits Continuity and Differentiability

If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to2023 · MCQ
Let $[x]$ denote the greatest integer function and $f (x) = max {1 + x + [x], 2 + x, x + 2 [x]}, 0 \leq x \leq 2$ . Let $m$ be the number of points in $f$ , where $f$ is not continuous and $n$ be the number of points in $f$ , where $f$ is not…2023 · MCQ
$t \to 0 lim (1^{\frac{1}{s i n ^{2} t}} + 2^{\frac{1}{s i n ^{2} t}} + ... + n^{\frac{1}{s i n ^{2} t}})^{s i n^{2} t}$ is equal to2023 · MCQ
Let $f (x) = {x^{2} sin (\frac{1}{x}) 0, x e 0, x = 0$ Then at $x = 0$ 2023 · MCQ
The set of all values of $a$ for which $x \to a lim ([x - 5] - [2 x + 2]) = 0$ , where [ $α$ ] denotes the greatest integer less than or equal to $α$ is equal to2023 · MCQ
The value of $n \to \infty lim \frac{1 + 2 - 3 + 4 + 5 - 6 + ..... + ( 3 n - 2 ) + ( 3 n - 1 ) - 3 n}{2 n ^{4} + 4 n + 3 - n ^{4} + 5 n + 4}$ is :2023 · MCQ
If the function $f (x) = ⎩ ⎨ ⎧ (1 + ∣ cos x ∣)^{\frac{λ}{∣ c o s x ∣}} μ e^{\frac{c o t 6 x}{c o t 4 x}},,, 0 < x < \frac{π}{2} x = \frac{π}{2} \frac{π}{2} < x < π$ …2023 · MCQ
Let $x = 2$ be a root of the equation $x^{2} + p x + q = 0$ and $f (x) = {\frac{1 - c o s ( x ^{2} - 4 p x + q ^{2} + 8 q + 16 )}{( x - 2 p ) ^{4}}, 0, x e 2 p x = 2 p$ Then $x \to 2 p^{+} lim [f (x)]$ …2023 · MCQ