Maths

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$

$={\left(Sinx\right)}_0^{\pi /2}=1$
$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$             

$P{A}^2+P{B}^2=co{s}^2\theta +{\left(sin\theta -3\right)}^2+co{s}^2\theta +{\left(sin\theta +6\right)}^2$

= 2 cos² θ + 2 sin² θ + 6 sin q + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin θ = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

Given

$\left(x\right)=\{\begin{array}{l}2sin\left(\frac{-\pi x}{2}\right),x<-1\\ \left|a{x}^2+x+b\right|,-1\le x\le 1\\ sin\pi x,x>1\end{array}$

If f (x) is continuous for all $x\in R$ then it should be continuous at x = 1 & x = -1

At x = -1, L.H.L = R.H.L. Þ 2 = |a + b - 1|

=>a + b – 3 = 0  OR  a + b + 1 = 0 . (i)

=>a + b + 1 = 0 . (ii)

           (i) & (ii), a + b =-1

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So  $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$    

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$  

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

=> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here $\begin{array}{cc}\hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill \end{array}\begin{array}{cc}\hfill \begin{array}{l}\\ \end{array}\hfill & \hfill \hfill \end{array}$

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right|\begin{array}{cc}\hfill \begin{array}{l}=\left(42-22\right)-2\left(14+11\right)-3\left(-4-6\right)\\ \end{array}\hfill & \hfill =20-50+30=0\hfill \end{array}$

For infinite solution ${}_{}{}_{}{}_{}$

20a – 8b – 4c = 0 5a = 2b + c

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

Given $\frac{dy}{dx}=\frac{x{y}^2+y}{x}=y^2+\frac{y}{x}$

OR $\frac{dy}{dx}-\frac{y}{x}=y^{2}OR\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{x}.\frac{1}{y}=1.........\left(i\right)$

=> $\frac{x}{y}=-\frac{x^2}{2}+c..........\left(ii\right)$      

Since curve intersect x + 2y = 4 at x = -2 then y = 3 so

From (ii) $\frac{-2}{3}=-2+cORc=2-\frac{2}{3}=\frac{4}{3}$

put x = 3, then $\frac{3}{y}=\frac{-9}{2}+\frac{4}{3}=\frac{-19}{6}$

$\Rightarrow y=\frac{-18}{19}$  

Given f (k) = $\{\begin{array}{l}k+1,kisodd\\ k,kiseven\end{array}$  

$?g:A\to A$ such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ * 1 = 10⁵

${\displaystyle \sum _{n=1}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r-1}{2}$

$Now{\displaystyle \sum _{r=\left(3,5,7,.....\right)}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}={\displaystyle \sum _{r=\left(3,5,7.....\right)}^{\infty }\left(\frac{\frac{{\left(r-1\right)}^2}{4}+3r-3+10}{r!}\right)}={\displaystyle \sum _{r=\left(3,5,7,......\right)}^{\infty }\left(\frac{r^2+10r+29}{4.r!}\right)}$

$=\frac{1}{8}\{41e-\frac{19}{e}-80)=\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$