Maths

Tangents making angle $\frac{\pi }{4}$ $\frac{}{}$ with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$

So, these tangents are $\perp$ $$ . So ASB is a focal chord.

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\sqrt[]{3}}{4}{\mathcal{l}}^2=\frac{25}{2\sqrt[]{3}}$

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$

So, f(x) = x

Now, $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$

f(x) = αx – b

$=\frac{18x-4}{13}$

option (D) satisfies

$\left|x^2-9\right|=3$

$\Rightarrow x=\pm 2\sqrt[]{3},\pm \sqrt[]{6}$

Required area = A

$\frac{A}{2}={\displaystyle \underset{0}{\overset{\sqrt[]{6}}{\int }}\left(9-x^2-3\right)dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{9+y}-\sqrt[]{9-y}\right)dy}}$

$A=16\sqrt[]{6}+32\sqrt[]{3}-72=8\left[2\sqrt[]{6}+4\sqrt[]{3}-9\right]$

Note : No option in the question paper is correct.

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$

option (C) is incorrect, there will be minima.

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

$\Rightarrow \frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}+\frac{n}{b}{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=0$

 $\Rightarrow \frac{dy}{dx}=-\frac{b}{a}{\left(\frac{bx}{ay}\right)}^{n-1}$

$\frac{dy}{d{x}_{\left(a,b\right)}}=-\frac{b}{a}$

So line always touches the given curve.

$x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

=2

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$ $\frac{}{}\frac{}{{}^{}}\frac{}{{}^{}}$

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$

$\frac{25S}{36}=1+\frac{3/5}{1}$

$S=\frac{288}{125}$