Limits, Continuity & Differentiability Questions - JEE

If $y = \ln\\!\left\{\dfrac{x + \sqrt{a^2 + x^2}}{a}\right\}$, then the value of $\dfrac{dy}{dx}$ is:

Consider $f(x) = x \ln x$ and $g(x) = e^{2x}$. Let $a$ and $b$ be two values of $x$ satisfying $f(x) = g(x)$ with $a If \(h(x) = \frac{f(x)}{g(x)}$, then $h'(a)$ equals:

If the function $f(x)=\dfrac{\sin3x+\alpha\sin x-\beta\cos3x}{x^3}$, $x\in\mathbb{R}$, is continuous at $x=0$, then $f(0)$ is equal to:

$f(x)=|x^2-2x-3|\cdot e^{|9x^2-12x+4|}$ non-diff at:

Let $f: [1, 10] \to \mathbb{Q}$ be a continuous function and $f(1) = 10$, then $f(10)$ is equal to

Consider the function $$f(x)=\begin{cases}\dfrac{a(7x-12-x^2)}{b|x^2-7x+12|} & ,\; x<3\\2^{\sin\left(\frac{x-3}{x-[x]}\right)} & ,\; x>3\\b & ,\; x=3\end{cases}$$ where $[x]$ denotes the greatest integer $\le x$. If $S$ denotes the set of all ordered pairs $(a,b)$ such that $f(x)$ is continuous at $x=3$, then the number of elements in $S$ is:

Let $f(x) = \lim_{n \to \infty} \displaystyle\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1}) + \tan^3(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})}\right)$. Then $\lim_{x \to 0} \dfrac{e^x - e^{f(x)}}{x - f(x)}$ is equal to:

Multiplicative functional equation, $f'(1)=2024$. ODE:

Find $\lambda$ for $f''(0)$ to exist.

g(x)=|f(x)|+f(|x|). In (-2,2), g is:

$\lim_{x \to \infty} \dfrac{(2x^2 - 3x + 5)(3x-1)^{x/2}}{(3x^2 + 5x + 4)\sqrt{(3x+2)^x}}$ is equal to:

$f$ triangle function, $g=f(x+2)-f(x-2)$. $n+m$?

If the function $f(x)=\begin{cases}\dfrac{1}{|x|} & ,\; |x|\ge 2\\ax^2+2b & ,\; |x|<2\end{cases}$ is differentiable on $\mathbb{R}$, then $48(a+b)$ is equal to ______.

The value of $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{4(x-\pi)\cos^2 x}{\pi(\pi-2x)\tan\!\left(x-\dfrac{\pi}{2}\right)}$ is equal to:

Let $f(x) = \begin{cases} \left[1 + \ln(c^2 + c + 1)\tan^2(x-1)\right]^{\frac{1}{(\ln x)^2}}, & x \neq 1 \\ 3c, & x = 1 \end{cases}$ where $c \in R$. If $\lim_{x \to 1} f(x)$ exists but $f(x)$ is discontinuous at $x = 1$, then $c$ can take the value:

201. Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that for all $x, y \in \mathbb{R}$, $|f(x) - f(y)| \leq 6|x - y|^2$. If $f(3) = 6$, then $f(6)$ is equal to:

Consider the function $f:(0,2)\to\mathbb{R}$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $$g(x)=\begin{cases}\min\{f(t)\},\; 0<t\le x & ,\; 0<x\le 1\\\frac{3}{2}+x & ,\; 1<x<2\end{cases}$$ Then

The set of points where $ f(x) = \dfrac{x}{1+|x|} $ is differentiable, is

Let $f: R \to R$ defined by $f(x) = x^3 + 3x + 1$ and $g$ be the inverse of $f$, then the value of $g''(5)$ equals:

Same $f$ and $f_1(x)=|f(|x|)|$. Number of points in $[-2,10]$ where $f_1$ is NOT differentiable:

$f=\max\{1+x+[x],2+x,x+2[x]\}$ on $[0,2]$. $(m+n)^2+2=$

f piecewise. m+n = non-cont + non-diff in (-2,2):

200. If $y = 2\tan^{-1}\!\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$, then the value of $\dfrac{d^2y}{dx^2}$ at $x = 2$ is:

Let $f(x) = x|x|$ and $g(x) = \sin x$. Then $g \circ f$ is:

Let $\displaystyle\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2n}{(n^2+1)\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8n}{(n^2+4)\sqrt{n^4+16}}+\cdots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2n\cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}}\right)$ be $\dfrac{\pi}{k}$, using only the principal values of the inverse trigonometric functions. Then $k^2$ is equal to

If $x^n = x^2 + x + 1$, then $\lim_{n \to \infty} x^n$ (as $n \to \infty$, $x \to 1$) equals:

Let a be a positive real number. Let f : R →R and g : (a, ∞) →R be defined by f(x) = sin πx 12 , g(x) = 2 loge(√x −√a) loge(e √x −e √a) . Then the value of limx→a+ f(g(x)) is

If $y^2 = P(x)$ which is a polynomial of degree 3, then $2\dfrac{d}{dx}\left(y^3 \dfrac{d^2y}{dx^2}\right)$ equals

If limx→0 eax−cos(bx)−cxe−cx 2 1−cos 2x = 17, find 5a2 + b2.

$\lim_{x \to 0} \frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}$ is equal to

Let f(x) = ( sin x x ∈Z 0 x /∈Z and g(x) =      x2 + 1 x ̸= 0, 2 4 x = 0 5 x = 2 , then:

limx→1− √π− √ 2 sin−1 x √1−x is equal to:

Given $(2x)^{2y} = 4e^{2x-2y}$, then $\dfrac{dy}{dx}(1 + \log_e 2x)^2$ equals:

Limit check for piecewise f(x):

Let $ S = \{t \in R : f(x) = |x - \pi| \cdot (e^{|x|} - 1)\sin|x| $ is not differentiable at $t\}$. Then the set S is equal to:

If limx→1 x+x2+···+xn−n x−1 = 820, then n is equal to:

Find g(x) = −x4b where b = limx→∞( √ x2 + x + 1 − √ x2 + 1).

If limx→1 x2−ax+b x−1 = 5, then a + b is equal to:

Find $\displaystyle\lim_{x \to \pi/2} \frac{\left[1 - \tan\left(\dfrac{x}{2}\right)\right]\left[1 - \sin x\right]}{\left[1 + \tan\left(\dfrac{x}{2}\right)\right]\left[\pi - 2x\right]^3}$.

Number of solutions of f(x) + g(x) = 0?

limx→2 3x+33−x−12 3−x/2−31−x is equal to:

limx→0 sin2(π cos4 x) x4 is equal to:

If $f(0) = 1$ and $\displaystyle\lim_{t \to x} \frac{\sec x \cdot f(t) - f(x) \sec t}{t - 1} = \sec^2 x$. The value of $\dfrac{f(0)}{f'(0)}$, is:

The value of $\displaystyle\lim_{x \to 0^+} \dfrac{\displaystyle\int_0^{\arctan x} (\sin t^2)\, dt}{x \cos x - x}$ is equal to:

$\lim_{x \to 1^-} \dfrac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}}$ is equal to __________ (up to four decimal places).

Let [x] denote the greatest integer function. Find limx→0 tan(π sin2 x)+(|x|−sin |x|)[x]2 x2 :

If $ f $ is a real-valued differentiable function satisfying $ |f(x) - f(y)| \leq (x-y)^2 $, $ x, y \in R $ and $ f(0) = 0 $, then $ f(1) $ equals

Let $f: R \to R$ defined by $f(x) = x^3 + 3x + 1$ and $g$ be the inverse of $f$, then the value of $g''(5)$ equals:

$\lim_{h \to 0} \frac{2\left[\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) - \cos\left(\frac{\pi}{6}+h\right)\right]}{\sqrt{3}h(\sqrt{3}\cos h - \sin h)}$ is equal to

If $f(x) = \frac{\sin[x]}{[x]}, [x] \neq 0$$= 0, [x] = 0$where $[x]$ is the greatest integer function, then $\lim_{x \to 0} f(x)$ equals:

Limits, Continuity & Differentiability Questions (1044)