Continuity Questions - JEE

Given $f(x) = 5 - |x - 2|$, graph of $y = f(x)$ is as shown. So, $f(x)$ is maximum at $x = 2$, $\alpha = 2$. Given $g(x) = |x + 1|$, graph of $y = g(x)$ is as shown. So, $g(x)$ is minimum at $x = -1$, $\beta = -1$. Therefore, find $\displaystyle\lim_{x \to -\alpha\beta} \frac{(x-1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$.

Evaluate $\lim_{x \to 0} \dfrac{\sqrt[3]{8+x} - \sqrt[3]{8+x^2-x^2}}{\sqrt[3]{8+x} - \sqrt[3]{8+x^2-x^2}}$Evaluate $\lim_{x \to 0} \dfrac{\sqrt{8+x} - \sqrt[3]{8+x^2} - x^2}{\sqrt[3]{8+x} - \sqrt[3]{8+x^2-x^2}}$

If $a_1 = 1$ and $a_n = n(1 + a_{n-1})$ $\forall\, n \geq 2$, and $L = \lim_{n \to \infty}\left(1 + \dfrac{1}{a_1}\right)\left(1 + \dfrac{1}{a_2}\right)\cdots\left(1 + \dfrac{1}{a_n}\right)$, then

$f(x+y)+f(x-y)=2f(x)\ \forall x,y\in\mathbb{R}$, $f(0)=0$ and $f$ is differentiable. Which are correct?

If $x = 3\tan t$ and $y = 3\sec t$, then the value of $\dfrac{d^2y}{dx^2}$ at $t = \dfrac{\pi}{4}$, is:

If $(\cos x)^y = (\sin y)^x$, then $\dfrac{dy}{dx}$ equals:

Let $g$ be the inverse of $f$. If $f(x)=x^2+3x-3$ (for appropriate domain) and $g(7)=1$, find the value of $g'(7)$. (Express as lowest fraction; if $p/q$, give $p+q$; answer 2 from key means $g'(7)=1/5$ giving $p+q=6$... or $g'(7)=2$).

Let $f$ be differentiable at $x=0$ and $f'(0)=1$. Then $\lim_{h\to 0}\dfrac{f(h)-f(-2h)}{h}=$

$\lim_{x \to 0} \dfrac{e^{x^2} - \cos x}{\sin^2 x}$

Let A = \[ A = \lim_{x \to 0} \frac{2^{\tan x} - 2^{\sin x}}{x^3} \] Find the value of $e^{4A}$.

$ \dfrac{d^2x}{dy^2} $ equals

If $\lim_{x\to 0}\dfrac{1-\cos\left(1-\cos\frac{x}{2}\right)}{2^m x^n}$ equals the left hand derivative of $e^{-|x|}$ at $x=0$, find |n+2m| divisible by?

If $x^2 + y^2 + \sin y = 4$, then the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2, 0)$ is

If $t = e^{x^2}$ and $y = t^2 - 1$ then $\left(\dfrac{dy}{dx}\right)_{x=1}$ is

Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable with $|f(x)-f(y)|\le|x-y|^3$ for all $x,y\in\mathbb{R}$. If $f(10)=100$, then $f(20)=$

Graph of $f(x)$ shown (piecewise linear on [0,5] with values f(0)=1, f(1)=1, f(2)=2, f(3)=2, f(4)=2, f(5)=2 approximately). Which are correct?

Given that $f(x) \geq 0$ and continuous $\forall x \in \mathbb{R}$, and $A = \int_{\pi/4}^{\beta} f(x)\, dx = \left(\beta \sin\beta + \dfrac{\pi}{4}\cos\beta + \sqrt{2}\right)\beta$, $\beta > \dfrac{\pi}{4}$. Find $f\left(\dfrac{\pi}{2}\right)$.

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

If $y = \left[x + \sqrt{x^2-1}\right]^{15} + \left[x - \sqrt{x^2-1}\right]^{15}$, then $(x^2-1)\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx}$ is equal to

Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$, $x \in R$. If $f(x)$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value at $\beta$, then $\lim_{x \to -\alpha\beta} \dfrac{(x-1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ is equal to

Let K be the set of all real values of x where the function $ f(x) = \sin|x| - |x| + 2(x - \pi)\cos|x| $ is not differentiable. Then the set K is equal to:

Let $f(x)$ be a differentiable function in $[-1, \infty)$ and $f(0) = 1$ such that $\lim_{t \to x+1} \frac{t^2 f(x+1) - (x+1)^2 f(t)}{f(t) - f(x+1)} = 1$. Find the value of $\lim_{x \to 1} \frac{\ln(f(x)) - \ln 2}{x-1}$.

Which statements are correct?(A) ∃ f:[0,1]→ℝ discontinuous everywhere with |f| continuous everywhere(B) F=f·g, f diff at x=a, f(a)=0, g continuous at x=a ⟹ F diff at x=a(C) Rf'(a)=2, Lf'(a)=3 ⟹ f non-diff at x=a but always continuous(D) f(a) and f(b) have opposite signs ⟹ ∃ solution of f(x)=0 in (a,b) if f continuous on [a,b]

If the function $f$ defined on $\left(\dfrac{\pi}{6},\,\dfrac{\pi}{3}\right)$ by \[f(x) = \begin{cases} \dfrac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \neq \dfrac{\pi}{4} \\ k, & x = \dfrac{\pi}{4} \end{cases}\] is continuous, then $k$ is equal to:

Without expansion or using L'Hospital's rule, prove that $\lim_{\theta \to 0} \dfrac{3\theta - \sin 3\theta}{\theta^3} = \dfrac{1}{6}$.

If $y = (x^2+2x)(3x^4+4x^3)$, find the number of zeros of $y'$ in $(0,\infty)$.

Match List-I (functions) with List-II (their derivatives w.r.t. $x$). List-I: (P) $\sin^{-1}(3x-4x^3)$ for $x\in(-1/2,1/2)$ (Q) $\cos^{-1}(4x^3-3x)$ for $x\in(1/2,1)$ (R) $\tan^{-1}\dfrac{3x-x^3}{1-3x^2}$ for $|x| (S) $\sin^{-1}(2x\sqrt{1-x^2})$ for $x\in(1/\sqrt{2},1)$ List-II: (1) $\dfrac{3}{\sqrt{1-x^2}}$ (2) $\dfrac{-3}{\sqrt{1-x^2}}$ (3) $\dfrac{3}{1+x^2}$ (4) $\dfrac{-2}{\sqrt{1-x^2}}$

If $f(x) = g(x)|(x-1)(x-2)\cdots(x-10)| - 2$ is derivable for all $x \in R$, where $g(x) = ax^9 + bx^6 + cx^3 + d,\; a, b, c, d \in R$, then $f'(-1)$ is equal to:

Find $\displaystyle\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2}\, x}$.

Evaluate: \[\lim_{x \to \frac{\pi}{2}} \frac{4(x-\pi)\sin^2\left(\frac{\pi}{2}-x\right)}{-2\pi\left(\frac{\pi}{2}-x\right)^2 \cdot \frac{\tan\left(x-\frac{\pi}{2}\right)}{\left(x-\frac{\pi}{2}\right)}}\]

Given the function $ f(x) = \dfrac{1}{1-x} $. The points of discontinuity of the composite function, $ y = f(f(x)) $ are at $ x = 0 $

If $x\log_a(\log_a x) - x^2 + y^2 = 4\,(y > 0)$, then $\frac{dy}{dx}$ at $x = e$ is equal to:

For $ x > 1 $, if $ (2x)^{2y} = 4e^{2x-2y} $, then $ (1 + \log_e 2x)^2 \dfrac{dy}{dx} $ is equal to:

Let f, g: R → R be two functions defined by \[f(x) = \begin{cases} x\sin\left(\dfrac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}\] and $g(x) = xf(x)$.Statement 1: f is a continuous function at x = 0.Statement 2: g is a differentiable function at x = 0.

If $f(x) = x^n$, then the value of $f(1) - \dfrac{f'(1)}{1!} + \dfrac{f''(1)}{2!} - \dfrac{f'''(1)}{3!} + \cdots + \dfrac{(-1)^n f^{(n)}(1)}{n!}$ is:

The function $ f : R \sim \{0\} \to R $ given by \[ f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1} \] can be made continuous at $ x = 0 $ by defining $ f(0) $ as

If $f(x) = \begin{vmatrix} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & 0\end{vmatrix}$, then which of the following are correct?

$f(x)=\begin{cases}x^2\cos(1/x) & x0\end{cases}$. Which are correct?

The value of $\lim_{n \to \infty} \left(\dfrac{n!}{n^n}\right)^{\frac{1}{n}}$ is equal to

If for $x \in \left(0, \dfrac{1}{4}\right)$, the derivative of $\tan^{-1}\left(\dfrac{6x\sqrt{x}}{1-9x^3}\right)$ is $\sqrt{x} \cdot g(x)$, then $g(x)$ equals

$g(x)=\begin{cases}x+b & x can be made differentiable at \(x=0$ if:

lim $x \to -\frac{1}{\sqrt{2}}$ \left( \sin(\cos^{-1} x) - x \right) is equal to

If $\lim_{x \to c} f(x) \cdot g(x)$ exists then both $\lim_{x \to c} f(x)$ and $\lim_{x \to c} g(x)$ exist.State whether this statement is true or false.

If $ \lim_{x \to 0} \dfrac{10 - \displaystyle\sum_{k=1}^{10}(\cos kx)}{x^2} = \dfrac{a}{b} $ where a and b are co-prime, then the value of $ (a + b) $ is equal to:

If $x^{2x} - 2x^x \cot y - 1 = 0$, then $\dfrac{dy}{dx}$ at $\left(1, \dfrac{\pi}{2}\right)$ is:

$\lim_{x \to 0} \dfrac{\sin(\pi\cos^2 x)}{x^2}$

If $x_1=\sqrt{3}$ and $x_{n+1}=\dfrac{x_n}{1+\sqrt{1+x_n^2}}$ for all $n\in\mathbb{N}$, then $\displaystyle\lim_{n\to\infty} 2^n x_n$ is equal to

The function $ f(x) = \left[ x^2 \left[ \dfrac{1}{x^2} \right] \right] $, $ x \neq 0 $ is ( [x] represents the greatest integer $ \leq x $)

$\lim_{x \to 0} \frac{2x^2 - \log(1+x)}{x^2}$ is equal to

Let $f$ be a differentiable function such that $\displaystyle\lim_{x \to 1} \frac{f(1+x^3-x)-f(x)}{\sin(x-1)} = \displaystyle\lim_{x \to 0} \frac{f(1-x)-f(1)}{x} + 10$, then $f'(1)$ is equal to:

Continuity Questions (1086)