Ncert Solutions Maths class 11th

$f\left(x\right)=\frac{1-x}{1+x},0<x<1$ (on simplification)

$f\text{'}\left(x\right)=\frac{-2}{{\left(1+x\right)}^2}$

${\left(1-x\right)}^{2}f\text{'}\left(x\right)=-2{\left(f\left(x\right)\right)}^2$

Given equation

$\Rightarrow \frac{cosx}{1+sinx}=\frac{2\left|sinx\right|cosx}{\left|co{s}^{2}x-si{n}^{2}x\right|}$

$\Rightarrow \left|1-2si{n}^{2}x\right|=2\left|sinx\right|\left(1+sinx\right)$

Put sin x = t

then equation,

$\left|1-2t^2\right|=2\left|t\right|\left(1+t\right),t\in \left(-1,1\right)-\left\{\pm \frac{1}{\sqrt[]{2}}\right\}$

Case I : For $\ge \frac{1}{\sqrt[]{2}}$ $\frac{}{\text{exception 25:}}$

the equation has no solution

Case II : For $0\le t<\frac{1}{\sqrt[]{2}}$ $\frac{}{\text{exception 25:}}$

Equation t = $\frac{\sqrt[]{5}-1}{4}\Rightarrow x=18°$ $\frac{\text{exception 25:}}{}$

Case III : $For\frac{-1}{\sqrt[]{2}}<t<0$ $\frac{}{\text{exception 25:}}$

Equation t = $\frac{-1}{2}\Rightarrow x=-30°$ $\frac{}{}$

Case IV : For $T<-\frac{1}{\sqrt[]{2}}$ $\frac{}{\text{exception 25:}}$

Equation t = $\frac{\sqrt[]{5}+1}{4}\Rightarrow$ $\frac{\text{exception 25:}}{}$ x = 54°

Sum of solutions = 18° - 30° - 54° = 66°

Total number = 20 + 8 + 24 = 52

$x^2+y^2+{\left(x-1\right)}^2+y^2+x^2+{\left(y-1\right)}^2+{\left(x-1\right)}^2+{\left(y-1\right)}^2=18$

$\Rightarrow x^2+y^2-x-y-\frac{7}{2}=0$

$r^2=\frac{1}{4}+\frac{1}{4}+\frac{7}{2}=4$

$LHS={\displaystyle \sum _{r=1}^{15}r\left(r!\right)}$

$=\sum \left(r+1-1\right)r!={\displaystyle \sum _{r=1}^{15}\left(r+1\right)!-r!}$

$=\left(2!-1!\right)+\left(3!-2!\right)+....+\left(16!-15!\right)$

$\begin{array}{l}=16!-1\\ {=}^{16}{P}_{16}-1\end{array}$

Equation : 2x² – (6 + k)x + 4 + 3k = 0

D < 0 => 6  $-4\sqrt[]{2}<k<6+4\sqrt[]{2}$

Sum of integral values of K is 1+2+3+.+11 = 66

Side of square = a

Radius of circle = r

Given : 4a + 2pr = 36

S = a² + pr²

^{$=\left(\frac{{\pi }^2}{4}+\pi \right)r^2-9\pi r+81$}

^{$\frac{ds}{dr}=0\Rightarrow r=\frac{18}{\pi +4}$}

Let P (h, k) be the mid point of the chord x² – y² = 4

$\therefore$ its equation is xh – yk = h² – k²

$Or,y=\left(\frac{h}{x}\right)x+\frac{k^2-h^2}{k}$ if this line is tangent to y² = 8x then $\frac{k^2-h^2}{k}$ = $\frac{2}{h/k}=\frac{2k}{h}$

$h\left(k^2-h^2\right)=2k^2$              

$\therefore$ Required locus is 2y² = x (y² – x²)

$\Rightarrow x^3=y^2\left(x-2\right)$

  $ta{n}^{-1}2=\theta$

->tan q = 2

$sin\left(\theta -\alpha \right)=\frac{1}{\sqrt[]{5}}$

->4 – 2 tanq = 1 + 2 tan a tan a = $\frac{3}{4}$  

$\alpha =ta{n}^{-1}\frac{3}{4}$