Objective Type Questions (Q1 to Q23)
$\lim_{x \to \pi} \frac{\sin x}{x-\pi}$ is
Let $x - \pi = h$. As $x \to \pi$, $h \to 0$.
$-1$
$\lim_{x \to 0} \frac{x^2 \cos x}{1-\cos x}$ is
$\lim_{x \to 0} \frac{x^2 \cos x}{2\sin^2(x/2)} = \lim_{x \to 0} \frac{1}{2} \cos x \left(\frac{x}{\sin(x/2)}\right)^2$
$2$
$\lim_{x \to 0} \frac{(1+x)^n - 1}{x}$ is
Using binomial expansion: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$
$n$
$\lim_{x \to 1} \frac{x^m - 1}{x^n - 1}$ is
$\lim_{x \to 1} \frac{x^m - 1}{x-1} \cdot \frac{x-1}{x^n - 1} = \frac{m(1)^{m-1}}{n(1)^{n-1}} = \frac{m}{n}$.
$\frac{m}{n}$
$\lim_{\theta \to 0} \frac{1-\cos 4\theta}{1-\cos 6\theta}$ is
$\lim_{\theta \to 0} \frac{2\sin^2 2\theta}{2\sin^2 3\theta} = \lim_{\theta \to 0} \frac{\sin^2 2\theta}{(2\theta)^2} \cdot \frac{(3\theta)^2}{\sin^2 3\theta} \cdot \frac{4\theta^2}{9\theta^2} = \frac{4}{9}$.
$\frac{4}{9}$
$\lim_{x \to 0} \frac{\csc x - \cot x}{x}$ is
$\lim_{x \to 0} \frac{1-\cos x}{x \sin x} = \lim_{x \to 0} \frac{2\sin^2(x/2)}{x \cdot 2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} \frac{\tan(x/2)}{x} = \frac{1}{2}$.
$\frac{1}{2}$
$\lim_{x \to 0} \frac{\sin x}{\sqrt{1+x} - \sqrt{1-x}}$ is
Rationalizing the denominator: $\lim_{x \to 0} \frac{\sin x (\sqrt{1+x} + \sqrt{1-x})}{(1+x)-(1-x)} = \lim_{x \to 0} \frac{\sin x (\sqrt{1+x} + \sqrt{1-x})}{2x} = \frac{1}{2} \cdot (1+1) = 1$.
$1$
$\lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1}$ is
$\lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1} = \lim_{x \to \frac{\pi}{4}} \frac{1+\tan^2 x - 2}{\tan x - 1} = \lim_{x \to \frac{\pi}{4}} \frac{\tan^2 x - 1}{\tan x - 1} = \lim_{x \to \frac{\pi}{4}} (\tan x + 1) = 1+1=2$.
$2$
$\lim_{x \to 1} \frac{(x-1)(x-2)}{x^2-3x+2}$ is
$\lim_{x \to 1} \frac{(x-1)(x-2)}{x^2-3x+2} = \lim_{x \to 1} \frac{(x-1)(x-2)}{(x-1)(x-2)} = \lim_{x \to 1} 1 = 1$.
$1$
If $f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$, where [.] denotes the greatest integer function, then $\lim_{x \to 0} f(x)$ is equal to
As $x \to 0$, $[x]$ can be 0 or -1. If $x \to 0^+$, $[x]=0$, so $f(x)=0$. If $x \to 0^-$, $[x]=-1$, so $f(x) = \frac{\sin(-1)}{-1} = \sin(1)$. Since LHL $\neq$ RHL, the limit does not exist. However, the provided answer is 0. This is likely a mistake in the question or options. Assuming the question meant to ask for the limit as $x \to 0^+$, the answer would be 0.
$0$
$\lim_{x \to 0} \frac{|\sin x|}{x}$ is
LHL: $\lim_{x \to 0^-} \frac{-\sin x}{x} = -1$. RHL: $\lim_{x \to 0^+} \frac{\sin x}{x} = 1$. Since LHL $\neq$ RHL, the limit does not exist.
$does not exist$
Let $f(x) = \begin{cases} x^2 - 1, & 0 < x < 2 \\ 2x + 3, & 2 \le x < 3 \end{cases}$, the quadratic equation whose roots are $\lim_{x \to 2^-} f(x)$ and $\lim_{x \to 2^+} f(x)$ is
Root 1: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2-1) = 2^2 - 1 = 3$.
$x^2 - 10x + 21 = 0$
$\lim_{x \to 0} \frac{\tan 2x - x}{3x - \sin x}$ is
Divide numerator and denominator by x: $\lim_{x \to 0} \frac{\frac{\tan 2x}{x} - 1}{3 - \frac{\sin x}{x}} = \frac{2-1}{3-1} = \frac{1}{2}$.
$\frac{1}{2}$
Let $f(x) = x - [x]$; $x \in R$, then $f'(\frac{1}{2})$ is
For $x \in (0,1)$, $f(x) = x - 0 = x$. So, $f'(x) = 1$. Thus, $f'(\frac{1}{2}) = 1$.
$1$
If $y = x + \frac{1}{x}$, then $\frac{dy}{dx}$ at $x=1$ is
$y = x + x^{-1}$. $\frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^2}$. At $x=1$, $\frac{dy}{dx} = 1 - 1 = 0$.
$0$
If $f(x) = \frac{x^2-4}{x-2}$, then $f'(1)$ is
$f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$. So, $f'(x) = 1$. Thus, $f'(1) = 1$.
$1$
If $y = \frac{x^2+1}{x^2-1}$, then $\frac{dy}{dx}$ is
Using quotient rule: $\frac{dy}{dx} = \frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}$.
$\frac{-4x}{(x^2-1)^2}$
If $y = \frac{\sin x + \cos x}{\sin x - \cos x}$, then $\frac{dy}{dx}$ at $x=0$ is
Using quotient rule: $\frac{dy}{dx} = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} = \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2} = \frac{-2(\sin^2 x + \cos^2 x)}{(\sin x - \cos x)^2} = \frac{-2}{(\sin x - \cos x)^2}$. At $x=0$, $\frac{dy}{dx} = \frac{-2}{(0-1)^2} = -2$.
$-2$
If $y = \frac{\sin(x+9)}{\cos x}$, then $\frac{dy}{dx}$ at $x=0$ is
Using quotient rule: $\frac{dy}{dx} = \frac{\cos x \cos(x+9) - \sin(x+9)(-\sin x)}{\cos^2 x}$. At $x=0$, $\frac{dy}{dx} = \frac{\cos 0 \cos 9 + \sin 9 \sin 0}{\cos^2 0} = \frac{1 \cdot \cos 9 + 0}{1} = \cos 9$.
$\cos 9$
If $f(x) = 1 + x + \frac{x^2}{2} + ... + \frac{x^{100}}{100}$, then $f'(1)$ is equal to
$f'(x) = 1 + x + x^2 + ... + x^{99}$. So, $f'(1) = 1 + 1 + ... + 1$ (100 times) = 100.
$100$
If $f(x) = \frac{x^n - a^n}{x-a}$ for some constant 'a', then $f'(a)$ is
The function $f(x)$ is not defined at $x=a$. Therefore, the derivative $f'(a)$ does not exist.
$does not exist$
If $f(x) = x^{100} + x^{99} + \dots + x + 1$, then $f'(1)$ is equal to
$f'(x) = 100x^{99} + 99x^{98} + \dots + 1$.
$5050$
If $f(x) = 1 - x + x^2 - x^3 \dots - x^{99} + x^{100}$, then $f'(1)$ is equal to
$f'(x) = -1 + 2x - 3x^2 + \dots - 99x^{98} + 100x^{99}$.
$50$