Maths

Given hyperbola : $\frac{k{x}^2}{6}-\frac{y^2}{6}=1$ so eccentricity e = $\sqrt[]{1+k}$ and directrices $x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}\Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}=1$

k = 2 therefore equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{6}=1$

hence it passes through the point $\left(\sqrt[]{5},-2\right)$

 $f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$

If f (x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$ ……. (i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$ ……. (ii)

From (i) and (ii) we get $b\in [-6,-2)\cup (2,6]$

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$\therefore$ Equation of circle is $x^2+y^2-4x+2y-13=0$   ……………. (i)

But, given  $x^2+y^2+2ax+2by+c=0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12

 $f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x\le 0\hfill \\ \hfill \left|x-4\right|,\hfill & \hfill x>0\hfill \end{array}andg\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x-4\right)}^2+b,\hfill & \hfill x\ge 0\hfill \end{array}$

$?$ f (x) and g (x) are continuous on R $\therefore$ a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (2) = g (2) + f (-1) = -11 + 3 = -8

Equation of tangent of slope m to y = x² is y = mx $\frac{1}{4}{m}^2$ - …………. (i)

Equation of tangent of slope m to y = - (x - 2)² is y = m (x – 2) + $\frac{1}{4}{m}^2$  …………… (ii)

If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4

therefore equation of tangent is y = 4x – 4

 $f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

$\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20*\frac{20*21}{2}-\frac{20*21*41}{6}=1330$

 $\frac{x^2}{{\left(\sqrt[]{\frac{5}{2}}\right)}^2}+\frac{y^2}{{\left(\sqrt[]{\frac{5}{3}}\right)}^2}=1$

Equation of tangent having slope m is

$y=mx\pm \sqrt[]{\frac{5}{3}{m}^2+\frac{5}{3}},$ which passes through (1, 3) and we get m₁ + m₂ = -4 and m₁m₂ = $\frac{44}{9}$

$\therefore$ Acute angle between the tangents is α  = tan^-1 $\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=ta{n}^{-1}\left(\frac{24}{7\sqrt[]{5}}\right)$

Since a is a odd natural number then $\left|{\displaystyle \underset{1}{\overset{3}{\int }}{y}^{a}dy}\right|=\frac{364}{3}\Rightarrow \left|{\left(\frac{y^{a+1}}{a+1}\right)}_1^3\right|=\frac{364}{3}\Rightarrow \frac{3^{a+1}}{a+1}=\frac{364}{3}$

a = 5

| (A + I) (adj A + I)| = 4 |A adj A + A + Adj A + I| = 4 | (A)I + A + adj A + I|= 4|A| = 1

|A + adj A| = 4

$A=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow adjA=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill -c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow \left|\begin{array}{cc}\hfill \left(a+d\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \left(a+d\right)\hfill \end{array}\right|=4\Rightarrow a+d=\pm 2$