Differential Calculus Questions - JEE

$\lim_{x \to \infty} \sqrt[3]{(x+a)(x+b)(x+c)} - x =$

Define $f: [0, \pi] \to \mathbb{R}$ by $f(x) = \begin{cases} \tan^2 x + \sqrt{2\sin^2 x + 3\sin x + 4 - \sqrt{\sin^2 x + 6\sin x + 2}} & x \neq \pi/2 \\ k & x = \pi/2 \end{cases}$ is continuous at $x = \pi/2$, then $k$ is equal to:

The function $f(x) = [x] + \sqrt{\{x\}}$, where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function respectively, is discontinuous at

Let $f(x) = \begin{cases} \frac{\tan^2 [x]}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{[x]} \cot [x] & \text{for } x < 0 \end{cases}$ where $[x]$ is the step up function and $\{x\}$ is the fractional part function of $x$, then :

The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

The function, $f(x) = [x] - [[x]]$, where $[ ]$ denotes greatest integer function:

$f$ is a continuous function in $[a, b]$; $g$ is a continuous function in $[b, c]$. A function $h(x)$ is defined as: $h(x) = f(x)$ for $x \in [a, b]$ $= g(x)$ for $x \in [b, c]$ if $f(b) = g(b)$, then

Which of the following limits vanish?

$\lim_{x \to c} f(x)$ does not exist when:

Let $f(x) = |x - 1|([x] - [-x])$, then which of the following statement(s) is/are correct. (where $[.]$ denotes greatest integer function.)

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

If $y = f(x)$ defined parametrically by $x = 2t - |t - 1|$ and $y = 2t^2 + t|t|$, then:

$f(x)$ is defined for $x \geq 0$ and has a continuous derivative. It satisfies $f(0) = 1, f'(0) = 0$ and $(1 + f(x))f''(x) = 1 + x$. The values $f(1)$ can't take is/are:

Consider $f(x) = \int \left(t + \frac{1}{t}\right) dt$ and $g(x) = f'(x)$ for $x \in \left[-3, -\frac{1}{2}\right]$. If P is a point on the curve $y = g(x)$ such that the tangent to this curve at P is parallel to a chord joining the points $\left(\frac{1}{2}, g\left(\frac{1}{2}\right)\right)$ and $(3, g(3))$ of the curve, then the coordinates of the point P

P and Q are two points on a circle of centre C and radius $a$, the angle PCQ being 20 then the radius of the circle inscribed in the triangle CPQ is maximum when

Let $f$ be a function defined on $(-\pi/2, \pi/2)$ as follows: $f(x) = \begin{cases} \frac{2^{[1/n]} - [x] - \frac{[x]}{[n2-1]}}{x\tan x} & x \neq 0 \\ k & x = 0 \end{cases}$. The value of $k$ so that $f$ is continuous at $x = 0$ is:

The function $f(x) = [x] + \sqrt{\{x\}}$, where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function respectively, is discontinuous at

Define $f: [0, \pi] \to \mathbb{R}$ by $f(x) = \begin{cases} \tan^2 x + \sqrt{2\sin^2 x + 3\sin x + 4 - \sqrt{\sin^2 x + 6\sin x + 2}} & x \neq \pi/2 \\ k & x = \pi/2 \end{cases}$ is continuous at $x = \pi/2$, then $k$ is equal to:

Let $f: R \to R$ be defined by $f(x) = \begin{cases} x + 2x^2 \sin \frac{1}{x} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$ then

Let $f(x) = \lim_{n \to \infty} \frac{2x^{2n} \sin + x}{1 + x^{2n}}$ then which of the following alternative(s) is/are correct?

Assume that $\lim_{\theta \to 1} f(0)$ exists and $\frac{\theta^2 + 0 - 2}{\theta + 3} \leq \frac{f(0)}{\theta^2} \leq \frac{\theta^2 + 20 - 1}{\theta + 3}$ holds for certain interval containing the point $\theta = -1$ then $\lim_{\theta \to 1} f(0)$ and $\lim_{\theta \to 1} \frac{f(0)}{\theta^2}$ is :

Let $f(x) = \begin{cases} \frac{\tan^2 [x]}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{[x]} \cot [x] & \text{for } x < 0 \end{cases}$ where $[x]$ is the step up function and $\{x\}$ is the fractional part function of $x$, then :

$\lim_{x \to c} f(x)$ does not exist when:

The function, $f(x) = [x] - [[x]]$, where $[ ]$ denotes greatest integer function:

The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

$f$ is a continuous function in $[a, b]$; $g$ is a continuous function in $[b, c]$. A function $h(x)$ is defined as: $h(x) = f(x)$ for $x \in [a, b]$ $= g(x)$ for $x \in [b, c]$ if $f(b) = g(b)$, then

In which of the following cases the given equations has atleast one root in the indicated interval?

If $f(x) = \begin{cases} \frac{x \cdot \ln(\cos x)}{\ln(1+x^2)} & x \neq 0 \\ 0 & x = 0 \end{cases}$ then:

Which of the following limits vanish?

Let $f(x) = |x - 1|([x] - [-x])$, then which of the following statement(s) is/are correct. (where $[.]$ denotes greatest integer function.)

If $y = f(x)$ defined parametrically by $x = 2t - |t - 1|$ and $y = 2t^2 + t|t|$, then:

$\lim f(x)$ does not exist when (where $[x]$ denotes the greatest integer less than or equal to $x$)

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

A function is defined as $f(x) = [\tan x] + \sqrt{\tan x - [\tan x]}$ $0 \leq x 2 \\ 5x - 7 & \text{if } x \leq 2 \end{cases}$ then:

If $F(x) = f(x)g(x)$ and $f'(x)g'(x) = c$, then (where $f$ and $g$ are thrice differentiable)

$f(x)$ is defined for $x \geq 0$ and has a continuous derivative. It satisfies $f(0) = 1, f'(0) = 0$ and $(1 + f(x))f''(x) = 1 + x$. The values $f(1)$ can't take is/are:

P and Q are two points on a circle of centre C and radius $a$, the angle PCQ being 20 then the radius of the circle inscribed in the triangle CPQ is maximum when

Let a function $f$ be defined as $f(x) = \begin{cases} \frac{|x-1|}{x^2+1} & \text{if } x > -1 \\ x^2 & \text{if } x \leq -1 \end{cases}$. Then the number of critical point(s) on the graph of this function is/are: