Limits, Continuity and Differentiability
Let ,$f:[-1,2] \rightarrow \mathbf{R}$ be given by $f(x)=2 x^2+x+\left[x^2\right]-[x]$, where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points, where $f$ is not continuous, is :
If the function $f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R}$, is continuous at $x=0$, then $f(0)$ is equal to :
If the function
$f(x)= \begin{cases}\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ a \log _e 2 \log _e 3 & , x=0\end{cases}$
is continuous at $x=0$, then the value of $a^2$ is equal to
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function given by
$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x < 0 \\ \alpha, & x=0, \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0\end{cases}$
where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :
If for some $\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21$, then $\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\right\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :
$ f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x<0 \\\\ x^2+c x+2 ; & 0 \leq x \leq 1 \\\\ 2 x+1 ; & x>1\end{cases} $
If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :
Consider the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m+n$ is
$\lim _\limits{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$
Let $g(x)$ be a linear function and $f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.$, is continuous at $x=0$. If $f^{\prime}(1)=f(-1)$, then the value $g(3)$ is
Consider the function $f:(0,2) \rightarrow \mathbf{R}$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by
$g(x)=\left\{\begin{array}{ll} \min \lfloor f(t)\}, & 0<\mathrm{t} \leq x \text { and } 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{array} .\right. \text { Then, }$
$\text { If } \lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3} \text {, then } 2 \alpha-\beta \text { is equal to : }$
$ f(x)=\left\{\begin{array}{cc} \frac{\mathrm{a}\left(7 x-12-x^2\right)}{\mathrm{b}\left|x^2-7 x+12\right|} & , x<3 \\\\ 2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\\\ \mathrm{~b} & , x=3, \end{array}\right. $
where $[x]$ denotes the greatest integer less than or equal to $x$. If $\mathrm{S}$ denotes the set of all ordered pairs (a, b) such that $f(x)$ is continuous at $x=3$, then the number of elements in $\mathrm{S}$ is :
Let $f:(0, \pi) \rightarrow \mathbf{R}$ be a function given by $f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0< x<\frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{\mathrm{b}}{\mathrm{a}}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$
where $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to _________.
Explanation:
$\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=f\left(\frac{\pi}{2}\right)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x) \text { for continuity at } x=\frac{\pi}{2}$
$\begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}}\left(\frac{8}{7}\right)^{\left(\frac{\tan 8 x}{\tan 7 x}\right)} \quad \text { Let } x=\frac{\pi}{2}-h \\ & \Rightarrow \quad \lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1 \\ & \Rightarrow \quad a-8=1 \Rightarrow a=9 \end{aligned}$
$\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}, x=\frac{\pi}{2}+h$
$\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}=\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}$
$\begin{aligned} & =\lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} \\ & =e^{\frac{b}{9}}=1 \quad \Rightarrow b=0 \\ & \Rightarrow a^2+b^2=81+0=81 \end{aligned}$
If $\alpha=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$ and $\beta=\lim _\limits{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $\mathrm{a} x^2+\mathrm{b} x-\sqrt{\mathrm{e}}=0$, then $12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b})$ is equal to _________.
Explanation:
$\begin{aligned} \alpha & =\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\ & =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\ \beta & =\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}=\lim _{x \rightarrow 0} e^{(\sin x)\left(\frac{1}{2} \cot x\right)} \\ & =\lim _{x \rightarrow 0} e^{\frac{1}{2} \cos x}=e^{1 / 2} \end{aligned}$
$\begin{aligned} & \text { Product of roots }=\sqrt{e}=\frac{-\sqrt{e}}{a} \Rightarrow a=-1 \\ & \text { Sum of roots }=\frac{-b}{a}=1+\sqrt{e} \\ & \qquad=b=\sqrt{e}+1 \\ & \Rightarrow 12 \ln (a+b)=12 \ln (\sqrt{e}+1-1)=12 \ln \left(e^{1 / 2}\right)=6 \end{aligned}$
The value of $\lim _\limits{x \rightarrow 0} 2\left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \ldots . \sqrt[10]{\cos 10 x}}{x^2}\right)$ is __________.
Explanation:
$\mathop {\lim }\limits_{x \to 0} 2\left( {{{1 - \cos x{{(\cos 2x)}^{{1 \over 2}}}{{(\cos 3x)}^{{1 \over 3}}}\,...\,{{(\cos 10x)}^{{1 \over {10}}}}} \over {{x^2}}}} \right)$ $\left(\frac{0}{0} \text { form }\right)$
Using L' hospital
$2 \lim _\limits{x \rightarrow 0} \frac{\sin x(\cos 2 x)^{\frac{1}{2}} \ldots(\cos 10 x)^{\frac{1}{10}} \ldots(\sin 2 x)(\cos x)(\cos 3 x)^{\frac{1}{3}}+\ldots}{2 x}$
$\begin{aligned} \Rightarrow & \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\frac{\sin 2 x}{x}+\ldots+\frac{\sin 10 x}{x}\right) \\ \quad & =1+2+\ldots+10=55 \end{aligned}$
Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbf{R}$ be a function defined by $f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$. Let $\mathrm{S}$ be the set of all points in the interval $[0,8]$ at which $f$ is not continuous. Then $\sum_\limits{\text {aes }} a$ is equal to __________.
Explanation:
$\begin{aligned} f(x) & =\left[\frac{x}{3}+3\right]-[\sqrt{x}] \\ & =\left[\frac{x}{2}\right]-[\sqrt{x}]+3 \end{aligned}$
Critical points where $f(x)$ might change behaviours when $\frac{x}{2} \in$ integer and $\sqrt{x} \in$ integer
$\Rightarrow$ Critical points,
$\begin{aligned} & f(0)=3 \\ & f\left(0^{+}\right)=3 \\ & f\left(1^{-}\right)=3 \\ & f\left(1^{+}\right)=2 \\ & f\left(2^{-}\right)=2 \\ & f\left(2^{+}\right)=3 \\ & f\left(3^{+}\right)=f\left(3^{-}\right)=3=f\left(4^{+}\right)=f\left(4^{-}\right)=f\left(5^{+}\right)=f\left(5^{-}\right) \\ & f\left(6^{-}\right)=3 \\ & f\left(6^{+}\right)=4 \\ & f\left(7^{-}\right)=f\left(7^{+}\right)=4 \\ & f\left(8^{-}\right)=4 \\ & f(8)=5 \end{aligned}$
$\begin{aligned} & \Rightarrow f(x) \text { is not continuous at } \\ & x=1,2,6,8 \\ & \Rightarrow \sum_{a \in s} a=1+2+6+8 \\ & \quad=17 \end{aligned}$
Let $\mathrm{a}>0$ be a root of the equation $2 x^2+x-2=0$. If $\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}=\alpha+\beta \sqrt{17}$, where $\alpha, \beta \in Z$, then $\alpha+\beta$ is equal to _________.
Explanation:
$\because 2 x^2+x-2=0$ has two roots where $a=\frac{\sqrt{17}-1}{4}$ and another root is $\frac{-\sqrt{17}-1}{4}$
And $2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$
Now $\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}$
$\begin{aligned} & =\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin ^2\left(\frac{2+x-2 x^2}{2}\right)}{a^2\left(\frac{1}{a}-x\right)^2} \\ & =\lim _{x \rightarrow \frac{1}{a}} \frac{\left(x+\frac{4}{\sqrt{17}+1}\right)^2 32 \cdot\left(\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}{a^2\left(\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2} \\ & =2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\ & =2\left(\frac{4}{\sqrt{17}-1}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\ & =\frac{17 \times 4}{18-2 \sqrt{17}}=\frac{68}{9-\sqrt{17}} \\ & =17(9+\sqrt{17}) \\ & \alpha+\beta=170 \end{aligned}$
Let $f$ be a differentiable function in the interval $(0, \infty)$ such that $f(1)=1$ and $\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then $2 f(2)+3 f(3)$ is equal to _________.
Explanation:
$\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{(t-x)}=1 \quad\left(\frac{0}{0} \text { form }\right)$
$\begin{aligned} & \lim _{t \rightarrow x} \frac{2 \operatorname{tf}(x)-x^2 f^{\prime}(t)}{1}=1 \\ & \Rightarrow 2 x f(x)-x^2 f'(x)=1 \\ & \frac{d y}{d x}-\frac{2 x y}{x^2}=\frac{-1}{x^2} \\ & \Rightarrow \frac{d y}{d x}-\left(\frac{2}{x}\right) y=\frac{-1}{x^2} \\ & \Rightarrow \text { I.F. }=e^{\int \frac{-2}{x} d x}=e^{-2 \ln x}=x^{-2}=\frac{1}{x^2} \end{aligned}$
$\begin{aligned} \Rightarrow & y\left(\frac{1}{x^2}\right)=\int\left(\frac{-1}{x^2}\right)\left(\frac{1}{x^2}\right) d x+C \\ & \frac{y}{x^2}=\frac{1}{3 x^3}+C \text { at } x=1, y=1 \\ \Rightarrow & 1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3} \\ \Rightarrow & y=\frac{1}{3 x}+\frac{2}{3} x^2=f(x) \\ \Rightarrow & 2 f(2)+3 f(3)=24 \end{aligned}$
If $\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $8 \mathrm{~m}+12 \mathrm{n}$ is equal to _______.
Explanation:
$I=\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(+5)^{1 /}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}$
From: $\frac{0}{0}$, using $\mathrm{L}-\mathrm{H}$ rule
$\begin{aligned} & I=\lim _{x \rightarrow 1} \frac{\frac{1}{3} \times 5(5 x+1)^{-2 / 3}--(+5)}{\frac{1}{2} \times 2(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}} \\ & =\frac{\left(\left.\frac{5}{3}-\frac{1}{3} \right\rvert\, 6^{-2 / 3}\right.}{\frac{1}{2} 5^{-1 / 2}}=\frac{8}{3} \times \frac{5^{1 / 2}}{6^{2 / 3}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}} \\ & \Rightarrow m=8, n=3 \\ & \Rightarrow 8 m+12 n=100 \end{aligned}$
Explanation:
R = $ \begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) =\lim _{h \rightarrow 0} f(h) \end{gathered} $
$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\\\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\cos ^{-1}\left(1-\mathrm{h}^2\right)}{\mathrm{h}}\left(\frac{\sin ^{-1} 1}{1}\right)\end{aligned}$
Let $\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2$
$ \begin{aligned} & =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}} \\\\ & =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}} \\\\ & =\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}} \\\\ & \therefore \mathrm{R}=\frac{\pi}{\sqrt{2}} \end{aligned} $
Now finding left hand limit
$ \begin{aligned} & L=\lim _{x \rightarrow 0^{-}} f(x) =\lim _{h \rightarrow 0} f(-h) \end{aligned} $
$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^2\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)}\end{aligned}$
$\begin{aligned} & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^2\right)} \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^2+2 h}\right) \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right) \\\\ & \quad L=\frac{\pi}{4}\end{aligned}$
$\begin{aligned} & \frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\\\ & =16+2 \\\\ & =18\end{aligned}$
If $\lim _\limits{x \rightarrow 0} \frac{a x^2 e^x-b \log _e(1+x)+c x e^{-x}}{x^2 \sin x}=1$, then $16\left(a^2+b^2+c^2\right)$ is equal to ________.
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{a{x^2}\left( {1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ....} \right) - b\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - .....} \right) + cx\left( {1 - x + {{{x^2}} \over {x!}} - {{{x^3}} \over {3!}} + .....} \right)} \over {{x^3}\,.\,{{\sin x} \over x}}}$
$=\lim _\limits{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots}{x^3}=1$
$\begin{aligned} & \mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0 \\ & \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} \\ & \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \\ & 16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81 \end{aligned}$
If the function
$f(x)= \begin{cases}\frac{1}{|x|}, & |x| \geqslant 2 \\ \mathrm{a} x^2+2 \mathrm{~b}, & |x|<2\end{cases}$
is differentiable on $\mathbf{R}$, then $48(a+b)$ is equal to __________.
Explanation:
$\mathrm{f}(\mathrm{x})\left\{\begin{array}{c} \frac{1}{\mathrm{x}} ; \mathrm{x} \geq 2 \\ \mathrm{ax}^2+2 \mathrm{~b} ;-2<\mathrm{x}<2 \\ -\frac{1}{\mathrm{x}} ; \mathrm{x} \leq-2 \end{array}\right.$
Continuous at $\mathrm{x}=2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$
Continuous at $\mathrm{x}=-2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$
Since, it is differentiable at $\mathrm{x}=2$
$-\frac{1}{x^2}=2 \mathrm{ax}$
Differentiable at $x=2 \quad \Rightarrow \frac{-1}{4}=4 a \Rightarrow a=\frac{-1}{16}, b =\frac{3}{8}$
Let $f(x)=\sqrt{\lim _\limits{r \rightarrow x}\left\{\frac{2 r^2\left[(f(r))^2-f(x) f(r)\right]}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right\}}$ be differentiable in $(-\infty, 0) \cup(0, \infty)$ and $f(1)=1$. Then the value of ea, such that $f(a)=0$, is equal to _________.
Explanation:
$\begin{aligned} & f(1)=1, f(a)=0 \\ & {f^2}(x) = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}({f^2}(r) - f(x)f(r))} \over {{r^2} - {x^2}}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\ & = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}f(r)} \over {r + x}}{{(f(r) - f(x))} \over {r - x}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\ & f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}} \\ & y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}} \\ & \frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}} \end{aligned}$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\begin{aligned} & v=v+x \frac{d v}{d x}-\frac{x}{v} e^v \\ & \frac{d v}{d x}=\frac{e^v}{v} \Rightarrow e^{-v} v d v=d x \end{aligned}$
Integrating both side
$\begin{aligned} & \mathrm{e}^{\mathrm{v}}(\mathrm{x}+\mathrm{c})+1+\mathrm{v}=0 \\ & \mathrm{f}(1)=1 \Rightarrow \mathrm{x}=1, \mathrm{y}=1 \end{aligned}$
$\begin{aligned} & \Rightarrow c=-1-\frac{2}{e} \\ & e^v\left(-1-\frac{2}{e}+x\right)+1+v=0 \\ & e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0 \\ & x=a, y=0 \Rightarrow a=\frac{2}{e} \\ & a e=2 \end{aligned}$
$f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2$. Let $m$ be the number of
points in $[0,2]$, where $f$ is not continuous and $n$ be the number of points in
$(0,2)$, where $f$ is not differentiable. Then $(m+n)^{2}+2$ is equal to :
If $\lim_\limits{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17$, then $5 a^{2}+b^{2}$ is equal to
Let $f$ and $g$ be two functions defined by
$f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ |x-1|, & x \geq 0\end{array}\right.$ and $\mathrm{g}(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1, & x \geq 0\end{array}\right.$
Then $(g \circ f)(x)$ is :
Let $f(x)=\left[x^{2}-x\right]+|-x+[x]|$, where $x \in \mathbb{R}$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is :
If $\alpha > \beta > 0$ are the roots of the equation $a x^{2}+b x+1=0$, and $\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text { then } \mathrm{k} \text { is equal to }$ :
$\lim_\limits{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^{2}(3 x)\right.}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$ is equal to _____________.
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{\mathrm{n}}$ be $\mathrm{n}$ positive consecutive terms of an arithmetic progression. If $\mathrm{d} > 0$ is its common difference, then
$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)$ is
$f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$
$g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.$
and $h(x)=2[x]-f(x)$, where $[x]$ is the greatest integer $\leq x$. Then the
value of $\lim\limits_{x \rightarrow 1} g(h(x-1))$ is :
Suppose $f: \mathbb{R} \rightarrow(0, \infty)$ be a differentiable function such that $5 f(x+y)=f(x) \cdot f(y), \forall x, y \in \mathbb{R}$. If $f(3)=320$, then $\sum_\limits{n=0}^{5} f(n)$ is equal to :
Let $x=2$ be a root of the equation $x^2+px+q=0$ and $f(x) = \left\{ {\matrix{ {{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x \ne 2p} \cr {0,} & {x = 2p} \cr } } \right.$
Then $\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)]$, where $\left[ . \right]$ denotes greatest integer function, is
If the function $f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr \mu & , & {x = {\pi \over 2}} \cr e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr } } \right.$
is continuous at $x = {\pi \over 2}$, then $9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}$ is equal to
The value of $\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }}$ is :
The set of all values of $a$ for which $\mathop {\lim }\limits_{x \to a} ([x - 5] - [2x + 2]) = 0$, where [$\alpha$] denotes the greatest integer less than or equal to $\alpha$ is equal to
$\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}$ is equal to
Let $f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr 0 & {,\,x = 0} \cr } } \right.$
Then at $x=0$
Let $[x]$ be the greatest integer $\leq x$. Then the number of points in the interval $(-2,1)$, where the function $f(x)=|[x]|+\sqrt{x-[x]}$ is discontinuous, is ___________.
Explanation:
1. The greatest integer function $[x]$ is discontinuous at every integer, since it jumps from one integer to the next without taking any values in between. The absolute value does not affect where this function is continuous or discontinuous. So within the interval $(-2,1)$, $[x]$ is discontinuous at $-2, -1, 0$ and $1$. However, because the interval is open $(-2,1)$, the endpoints $-2$ and $1$ are not included.
2. The function $\sqrt{x-[x]}$ is the square root of the fractional part of $x$. The fractional part function $x-[x]$ is continuous everywhere, as it always takes a value between $0$ and $1$ (inclusive of $0$, exclusive of $1$). However, the square root function $\sqrt{x}$ is only defined for $x \geq 0$, and so $\sqrt{x-[x]}$ is discontinuous wherever $x-[x] < 0$. This happens exactly at the points where $x$ is a negative integer, as the fractional part of a negative integer is $1$ (considering that the "fractional part" is defined as the part of the number to the right of the decimal point, which for negative numbers works a bit differently). Within the interval $(-2,1)$, this is the case for $-2$ and $-1$. However, since the interval is open $(-2,1)$, the endpoint $-2$ is not included.
Now, let's analyze the discontinuities within the given interval $(-2,1)$:
At $x=-1$: $f(-1^{+})=1+0=1$ and $f(-1^{-})=2+1=3$. Since these two values are different, $f(x)$ is discontinuous at $x=-1$.
At $x=0$: $f(0^{+})=0+0=0$ and $f(0^{-})=1+1=2$. Again, these two values are different, so $f(x)$ is discontinuous at $x=0$.
At $x=1$: $f(1^{+})=1+0=1$ and $f(1^{-})=0+1=1$. These two values are the same, so $f(x)$ is continuous at $x=1$. However, this point is not within the open interval $(-2,1)$.
So within the interval $(-2,1)$, the function $f(x) = |[x]| + \sqrt{x-[x]}$ is discontinuous at the points $-1$ and $0$ (2 points in total).
Let $f:( - 2,2) \to R$ be defined by $f(x) = \left\{ {\matrix{ {x[x],} & { - 2 < x < 0} \cr {(x - 1)[x],} & {0 \le x \le 2} \cr } } \right.$ where $[x]$ denotes the greatest integer function. If m and n respectively are the number of points in $( - 2,2)$ at which $y = |f(x)|$ is not continuous and not differentiable, then $m + n$ is equal to ____________.
Explanation:
When $[x]$ is denotes greatest integer function
Clearly, $|f(x)|$ remains same.
Given that, $m$ and $n$ respectively are the number points in $(-2,2)$ at which $y=|f(x)|$ is not continuous and not differentiable
So, $m=1$ where $y=|f(x)|$ not continuous
and $n=3$ where $|f(x)|$ is not differentiable.
Thus, $m+n=4$
Let $\mathrm{k}$ and $\mathrm{m}$ be positive real numbers such that the function $f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.$ is differentiable for all $x > 0$. Then $\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}$ is equal to ____________.
Explanation:
$\because f(x)$ is differentiable at $x>0$
So, $f(x)$ is differentiable at $x=1$
$ \begin{gathered} f\left(1^{-}\right)=f(1)=f\left(1^{+}\right) \\\\ 3+k \sqrt{2}=m+k^2 ......(i) \end{gathered} $
$ \begin{aligned} & f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) \\\\ & 6(1)+\frac{k}{2 \sqrt{1+1}}=2 m(1) \\\\ & \Rightarrow 6+\frac{k}{2 \sqrt{2}}=2 m ......(ii) \end{aligned} $
Using (i) and (ii),
$ \begin{aligned} & 3+k \sqrt{2}=3+\frac{k}{4 \sqrt{2}}+k^2 \\\\ & \Rightarrow k^2+k\left[\frac{1}{4 \sqrt{2}}-\sqrt{2}\right]=0 \end{aligned} $
$ \Rightarrow k\left[k+\frac{1-8}{4 \sqrt{2}}\right]=0 \Rightarrow k=0, \frac{7}{4 \sqrt{2}} $
As the problem specifies k to be a positive real number, we can rule out k = 0. Hence, k = $\frac{7}{4 \sqrt{2}}$
$ \begin{aligned} & \text { for } k=\frac{7}{4 \sqrt{2}}, m=3+\frac{\frac{7}{4 \sqrt{2}}}{4 \sqrt{2}} \\\\ & =3+\frac{7}{32}=\frac{96+7}{32}=\frac{103}{32} \end{aligned} $
$ \text { So, } \frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=\frac{8 \times\left[2 \times \frac{103}{32} \times 8\right]}{6 \times \frac{1}{8}+\frac{7}{4 \sqrt{2}} \times 2 \sqrt{918}} $
$ =\frac{412}{\frac{3}{4}+\frac{7}{12}}=\frac{412}{\frac{9+7}{12}}=\frac{412 \times 12}{16}=309 $
Concept :
$f(x)$ is differentiable at $x=a$, if $f^{\prime}\left(a^{-}\right)=f'\left(a^{+}\right)$
Let $a \in \mathbb{Z}$ and $[\mathrm{t}]$ be the greatest integer $\leq \mathrm{t}$. Then the number of points, where the function $f(x)=[a+13 \sin x], x \in(0, \pi)$ is not differentiable, is __________.
Explanation:
Given that $ f(x) = [a + 13\sin(x)] $, where $[t]$ is the greatest integer function and $ x \in (0, \pi) $.
The function $[t]$ is not differentiable wherever $ t $ is an integer because at these points, the function has a jump discontinuity.
For $ f(x) $ to have a point of non-differentiability, the value inside the greatest integer function, i.e., $ a + 13\sin(x) $, should be an integer.
So, we need to find the values of $ x $ in the interval $ (0, \pi) $ for which $ a + 13\sin(x) $ is an integer.
Now, $ \sin(x) $ varies from 0 to 1 in the interval $ (0, \pi) $, so the maximum value of $ 13\sin(x) $ in this interval is 13.
For each integer value of $ 13\sin(x) $ between 0 and 13, we'll have a corresponding value of $ x $. There will be two such values of $ x $ for each value of $ 13\sin(x) $ (because of the periodic and symmetric nature of sine function over the interval $ (0, \pi) $), except for the maximum value, 13, which will have only one corresponding value of $ x $ (namely $ x = \frac{\pi}{2} $).
Thus, the total number of integer values of $ 13\sin(x) $ between 0 and 13 is 13. Excluding the maximum value, we have 12 integer values, each giving rise to two values of $ x $. Including the maximum value, which gives one value of $ x $, we have :
$ 12 \times 2 + 1 = 25 $
Thus, $ f(x) $ is not differentiable at 25 points in the interval $ (0, \pi) $.
$ \text { Let the function } f(x)=\left\{\begin{array}{cl} \frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x \neq 0 \\ 10 & ; \text { if } x=0 \end{array} \text { be continuous at } x=0 .\right. $
Then $\alpha$ is equal to
If $\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$, where $\alpha, \beta, \gamma \in \mathbf{R}$, then which of the following is NOT correct?
The number of points, where the function $f: \mathbf{R} \rightarrow \mathbf{R}$,
$f(x)=|x-1| \cos |x-2| \sin |x-1|+(x-3)\left|x^{2}-5 x+4\right|$, is NOT differentiable, is :
The function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by
$f(x)=\lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$ is continuous for all x in :
If for $\mathrm{p} \neq \mathrm{q} \neq 0$, the function $f(x)=\frac{\sqrt[7]{\mathrm{p}(729+x)}-3}{\sqrt[3]{729+\mathrm{q} x}-9}$ is continuous at $x=0$, then :