Ncert Solutions Maths class 11th

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

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

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.

  $f\left(x\right)=min\left\{1,1+xsinx\right\},0\le x\le 2\pi$

$f\left(x\right)=\{\begin{array}{l}1,0\le x<\pi \\ 1+xsinx,\pi \le x\le 2\pi \end{array}$

Now at x = π,

$\therefore f\left(x\right)$ is not differentiable at x = π

$\therefore$ (m, n) = (1, 0)

$A={\displaystyle {\sum }_{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}}$

$B={\displaystyle {\sum }_{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}}$

$A=\frac{11}{15},B=\frac{-9}{15}$

$\therefore \frac{A}{B}=\frac{-11}{9}$

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

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

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

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

So line always touches the given curve.

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$ …… (i)

Now,  ${\left(-2\right)}^3\equiv -1mod\left(7\right)$

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$ ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$

$\therefore$ Remainder = 5

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

(i) f(x)=sin x cos x

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{sin(x+h)cos(x+h)?sinxcosx}{h}$

$=\underset{h?0}{lim}\frac{1}{2h}*[2sin(x+h)cos(x+h)?2sinxcosx]$

$=\underset{h?0}{lim}\frac{1}{2h}[sin2(x+h)?sin2x]$

$=\underset{h?0}{lim}\frac{1}{2h}\left[2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)?2x}{2}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[cos\left(2x+h\right)sinh\right]$

$=\underset{h?0}{lim}cos(2x+h)*\underset{h?0}{lim}\frac{sinh}{h}$

$=cos(2x+0)$

$=cos2x$

$\mathrm{(ii)\; f}(x)=secx$

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[sec(x+h)?secx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{?2sin\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{(?1sin\left(\frac{2x+h}{2}\right)}{cos(x+h)cosx}*\underset{h?0}{lim}(?1)\frac{sinh/2}{h/2}$

$=\frac{sinx}{cosx?cosx}*1$

$=tanx?secx.$

(iii) Given f(x)=5 sec x+4 cosx.

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$\begin{array}{l}\\ =\underset{h?0}{lim}\frac{5sec(x+h)+4cos(x+h)?[5secx+4cosx]}{h}.\end{array}$

$=\underset{h?0}{lim}\frac{5}{h}[sec(x+h)?secx]+\underset{h?0}{lim}\frac{4}{h}[cos(x+h)?cosx]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h?x}{2}\right)\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)(cosx)}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{2x+h}{2}\right)sin\frac{h}{2}\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]?4\underset{h?0}{lim}sin\left(\frac{2x+h}{2}\right)\underset{h?0}{lim}\frac{sinh/2}{h/2}$

$=\frac{sin\left(\frac{2x+0}{2}\right)}{cos(x+0)cosx}*1?4sin\left(\frac{2x}{2}\right)$

$=5\frac{sinx}{cosx}?\frac{1}{cosx}?4sinx$

$=5tanx?secx?4sinx$

$\mathrm{(iv)\; Given\; f}(x)=cosecx$

$f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[cosec(x+h)?cosecx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{sin(x+h)}?\frac{1}{sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{sinx?sin(x+h)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)]}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{2x+h}{2}\right)sin(?h/2)]}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{cos\left(\frac{2x+h}{2}\right)}{sin(x+h)sinx}*\frac{(?1)sin(2)}{h/2})$

$=\frac{cos\left(\frac{2x+0}{2}\right)}{sin(x+0)sinx}*(?1)$

$=?\frac{cosx}{sinx}*\frac{1}{sinx}$

$=?cotx?cosecx$

(v) Given,f(x)=3 cot x+5cosecx.

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$ $=\frac{2}{cos(x+0)cosx}*1+7sin\left(\frac{2x+0}{2}\right)*(?1)$

$=\underset{h?0}{lim}3cot(x+h)+5cosec(x+h)?[3cotx+5cosx)$

${}_{h?0}\frac{3}{h}[cot(x+h)?cotx]+\underset{h?0}{lim}$

$=\text{?}\frac{5}{h}[cosec(x+h)?cosecx]$

$=\underset{h?0}{lim}\frac{3}{h}\left[\frac{cos(x+h)}{sin(x+h)}?\frac{cosx}{sinx}\right]+\underset{h?0}{lim}\frac{5}{h}\left[\frac{1}{sin(x+h)}?\frac{1}{sinx}\right]$

$=\underset{h?0}{lim}\frac{3}{h}\left[\frac{cos(x+h)sinx?cosxsin(x+h)}{sin(x+h)sinx}\right]+\underset{h?0}{lim}\frac{5}{h}\left[\frac{sinx?sin(x+h)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{3}{h}[\frac{sin(x?(x+h))}{sin(x+h)sinx}+\underset{h?0}{lim}\frac{5}{h}[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{sin(x+h)sinx}$

$=\underset{h?0}{lim}\frac{3}{sin(x+h)sinx}*\underset{h?0}{lim}(?1)\frac{sinh}{h}+\underset{h?0}{lim}\frac{5}{h}[\frac{2cos\left(\frac{2x+h}{2}\right)sin(?h/2)}{sin(x+h)sinx}$