Limits, Continuity and Differentiability

Let $f:\mathbb R\to\mathbb R$ be given by

$$f(x) = \begin{cases}x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1, & x < 0;\\ x^2 - x + 1, & 0 \le x < 1;\\ \frac23x^3 - 4x^2 + 7x - \frac83, & 1 \le x < 3;\\ (x-2)\log_e(x-2) - x + \frac{10}3, & x \ge 3.\end{cases}$$

Then which of the following options is/are correct?

Let $f:\mathbb R\to\mathbb R$ be a function. We say that $f$ has

PROPERTY 1 if $\displaystyle\lim_{h\to0}\frac{f(h) - f(0)}{\sqrt{|h|}}$ exists and is finite, and

PROPERTY 2 if $\displaystyle\lim_{h\to0}\frac{f(h) - f(0)}{h^2}$ exists and is finite.

Then which of the following options is/are correct?

For $a\in\mathbb R$, $|a| > 1$, let

$$\lim_{n\to\infty}\left(\frac{1 + \sqrt[3]2 + \cdots + \sqrt[3]n}{n^{7/3}\left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)}\right) = 54.$$

Then the possible value(s) of $a$ is/are

For every twice differentiable function $f: \mathbb{R} \to [-2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statement(s) is (are) TRUE?

For any positive integer $n$, define $f_n: (0, \infty) \to \mathbb{R}$ as

$$f_n(x) = \sum_{j=1}^{n} \tan^{-1}\left(\frac{1}{1 + (x+j)(x+j-1)}\right) \quad \text{for all } x \in (0, \infty).$$

(Here, the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.)

Then, which of the following statement(s) is (are) TRUE?

Let $f: (0, \pi) \to \mathbb{R}$ be a twice differentiable function such that

$$\lim_{t \to x} \frac{f(x)\sin t - f(t)\sin x}{t - x} = \sin^2 x \quad \text{for all } x \in (0, \pi).$$

If $f\left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$, then which of the following statement(s) is (are) TRUE?

Let $f_1: \mathbb{R} \to \mathbb{R}$, $f_2: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}$, $f_3: \left(-1, e^{\pi/2} - 2\right) \to \mathbb{R}$ and $f_4: \mathbb{R} \to \mathbb{R}$ be functions defined by

(i) $f_1(x) = \sin\left(\sqrt{1 - e^{-x^2}}\right)$,

(ii) $f_2(x) = \begin{cases} \dfrac{|\sin x|}{\tan^{-1} x} & \text{if } x \ne 0 \\ 1 & \text{if } x = 0 \end{cases}$, where the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,

(iii) $f_3(x) = [\sin(\log_e(x + 2))]$, where, for $t \in \mathbb{R}$, $[t]$ denotes the greatest integer less than or equal to $t$,

(iv) $f_4(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \ne 0 \\ 0 & \text{if } x = 0 \end{cases}$.

The correct option is:

Let $[x]$ be the greatest integer less than or equals to $x$. Then, at which of the following point(s) the function $f(x) = x\cos(\pi(x + [x]))$ is discontinuous?

Let $f: \mathbb{R} \to (0, 1)$ be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval $(0, 1)$?

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 0$, $f\left(\frac{\pi}{2}\right) = 3$ and $f'(0) = 1$. If

$$g(x) = \int_x^{\frac{\pi}{2}} \left[f'(t)\operatorname{cosec} t - \cot t\operatorname{cosec} t\, f(t)\right] dt$$

for $x \in \left(0, \frac{\pi}{2}\right]$, then $\lim_{x \to 0} g(x) =$

Which of the following options is the only CORRECT combination?

Which of the following options is the only CORRECT combination?

Which of the following options is the only INCORRECT combination?

Let $f(x) = \dfrac{1 - x(1 + |1 - x|)}{|1 - x|}\cos\left(\dfrac{1}{1 - x}\right)$ for $x \ne 1$. Then

$\displaystyle\lim_{x\to\frac\pi2}\frac{\cot x-\cos x}{(\pi-2x)^3}$ equals :

$\displaystyle\lim_{x\to3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt2}$ is equal to :

The value of k for which the function
$$f(x)=\begin{cases}\left(\dfrac45\right)^{\frac{\tan4x}{\tan5x}}, & 0<x<\dfrac\pi2\\[2mm] k+\dfrac25, & x=\dfrac\pi2\end{cases}$$
is continuous at $x=\dfrac\pi2$, is :

If $\displaystyle\lim_{n\to\infty}\frac{1^a+2^a+\dots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\dots+(na+n)]}=\frac1{60}$ for some positive real number a, then a is equal to :

Let $f:(0,\infty)\to\mathbb R$ be a differentiable function such that $f'(x)=2-\dfrac{f(x)}{x}$ for all $x\in(0,\infty)$ and $f(1)\ne1$. Then

Let $\alpha,\beta\in\mathbb R$ be such that $\displaystyle\lim_{x\to0}\frac{x^2\sin(\beta x)}{\alpha x-\sin x}=1$. Then $6(\alpha+\beta)$ equals