Maths

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhateachedgeofthecuboidis2units.\\ \therefore Coordinatesoftheverticesare\\ A\left(2,0,0\right),B\left(2,2,0\right),C\left(0,2,0\right),D\left(0,2,2\right),E\left(0,0,2\right),F\left(2,0,2\right),G\left(2,2,2\right)andO\left(0,0,0\right).\end{array}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab)\frac{2b^2}{a}=10\Rightarrow b^2=5a$

Now,  $?\left(t\right)=\frac{5}{12}+t-t^2=\frac{8}{12}-{\left(t-\frac{1}{2}\right)}^2$
$?(\mathrm{t}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{4}{9}$

$\Rightarrow {\mathrm{a}}^2=81\mathrm{}$  (From (i) and (ii)

So,  $a^2+b^2=81+45=126$

$f\left(2\right)=8,f^{\mathrm{\text{'}}}\left(2\right)=5,f^{\mathrm{\text{'}}}\left(x\right)\ge 1,f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)\ge 4,\forall x\in \left(\mathrm{1,6}\right)$

Using LMVT

$f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)=\frac{f^{\mathrm{\text{'}}}\left(5\right)-f^{\mathrm{\text{'}}}\left(2\right)}{5-2}\ge 4\Rightarrow f^{\mathrm{\text{'}}}\left(5\right)\ge 17$

$f^{\mathrm{\text{'}}}\left(x\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\ge 1\Rightarrow f\left(5\right)\ge 11$

Therefore $f^{\mathrm{\text{'}}}\left(5\right)+f\left(5\right)\ge 28$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Letthegiven\text{\hspace{0.17em}points}areA\left(0,-1,-7\right),B\left(2,1,-9\right),C\left(6,5,-13\right)\\ AB=\sqrt[]{{\left(2-0\right)}^2+{\left(1+1\right)}^2+{\left(-9+7\right)}^2}=\sqrt[]{4+4+4}=\sqrt[]{12}=2\sqrt[]{3}\\ BC=\sqrt[]{{\left(6-2\right)}^2+{\left(5-1\right)}^2+{\left(-13+9\right)}^2}=\sqrt[]{16+16+16}=\sqrt[]{48}=4\sqrt[]{3}\\ AC=\sqrt[]{{\left(6-0\right)}^2+{\left(5+1\right)}^2+{\left(-13+7\right)}^2}=\sqrt[]{36+36+36}=\sqrt[]{108}=6\sqrt[]{3}\\ 2\sqrt[]{3}+4\sqrt[]{3}=6\sqrt[]{3}\\ i.e.,AB+BC=AC\\ \therefore AB:AC=2\sqrt[]{3}:6\sqrt[]{3}=1:3\\ Hence,\text{\hspace{0.17em}point}AdividesBandCin1:3externally.\end{array}$

${\int }_0^3?g\left(x\right)-f\left(x\right)={\int }_0^3?\left|\right|x-2|-2|dx-{\int }_0^3?|x-2|dx$

$=\left(\frac{1}{2}*2*2+1+\frac{1}{2}*1*1\right)-\left(\frac{1}{2}*2*2+\frac{1}{2}*1*1\right)$

$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$

P (E_n) = n/36 for n = 1, 2, 3, …., 8

$P\left(A\right)=\frac{Anypossiblesumof\left(1,2,3,.........,8\right)\left(=\alpha say\right)}{36}$

$\frac{\alpha }{36}\ge \frac{4}{5}$

$\therefore a\ge 29$

If one of the number from {1, 2, ….8} is left then total $\ge$ 29 by 3 ways

Similarly by leaving terms more 2 or 3 we get 16 more combinations

$\therefore$ Total number of different set a possible is 16 + 3

= 19

The direction of ratios of the lines,   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ , are $-3,2k,2and3k,1,-5$ respectively.

It is known that two lines with direction ratios,   $a_1, b_1, c_1 and a_2, b_2,c_2$ , are perpendicular, if  $a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\begin{array}{l}\therefore -3(3k)+2k*1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

$L_1:4x+3y+2=0$  

$L_2:3x-4y-11=0$               

Since circle C touches the line L2 at Q intersection point L1 and L2 is (1, -2)

P lies of L₁

$\therefore P\left(x,-\frac{1}{3}\left(2+4x\right)\right)$               

Now,

PQ = 5 ? (x – 1)² + ${\left(\frac{4x+2}{3}-2\right)}^2=25$  

$\Rightarrow x=4,-2$                     

$?$ The circle lies below the axis

y = -6

p (4, -6)

Now distance of P from 5x – 12 y + 51 = 0

$=\left|\frac{20+72+51}{13}\right|=\frac{143}{13}=11$                

$x^2-3x+p=0$

$\alpha ,\beta ,\gamma ,\delta$ in G.P.

$\alpha +\alpha r=3$

$x^2-6x+q=0$

$\alpha r^2+\alpha r^3=6$

$\left(2\right)÷\left(1\right)\Rightarrow r^2=2$

So,  $\frac{2q+p}{2q-p}=\frac{2r^5+r}{2r^5-r}=\frac{2r^4+1}{2r^4-1}=\frac{9}{7}$

$\left(1-x^2\right)dy=\left(xy+\left(x^3+2\right)\sqrt[]{1-x^2}\right)dx$

$\therefore \frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{x^3+3}{\sqrt[]{1-x^2}}$

$\therefore l.F.=e^{{\displaystyle \int \frac{X}{1-x^2}dx}}=\sqrt[]{1-x^2}$

$\therefore y\left(x\right)=\frac{x^4+12x}{4\sqrt[]{1-x^2}}$

$\therefore {\displaystyle \underset{\frac{-1}{2}}{\overset{\frac{1}{2}}{\int }}\sqrt[]{1-x^2}y\left(x\right)dx={\displaystyle \underset{\frac{-1}{2}}{\overset{\frac{1}{2}}{\int }}\left(\frac{x^4+12x}{4}\right)}dx}$

$\therefore k=\frac{1}{320}$

$\therefore k^{-1}=320$