Ncert Solutions Maths class 12th

$f\left(x\right)=\left|5x-7\right|+\left[x^2+2x\right]$

$=\left|5x-7\right|+\left[{\left(x+1\right)}^2\right]-1$

f $\left(\frac{7}{4}\right)=0+4=4$

as both |5x – 7| and x² + 2x are increasing in nature after x = 7/5

f (2) = 3 + 8 = 11

$f{\left(\frac{7}{5}\right)}_{min}-4andf{\left(2\right)}_{max}=11$

Sum is 4 + 11 = 15

Let f (x) = (x -α) (x - β)

It is given that f (0) = p = p

and

$f\left(1\right)=\frac{1}{3}\Rightarrow \left(1-\alpha \right)\left(1-\beta \right)=\frac{1}{3}$

Now, let us assume that is the common root of f (x) = 0 and fofofof (x) = 0

fofofof (x) = 0

$f\left(-3\right)=\left(-3-\frac{7}{6}\right)\left(3-3\right)=25$

Line of shortest distance will be along ${\overline{b}}_1\times {\overline{b}}_2$ ${\stackrel{}{}}_{}{\stackrel{}{}}_{}$

where $\begin{array}{l}{\stackrel{}{}}_{}\stackrel{}{}\stackrel{}{}\\ \end{array}$ $\begin{array}{l}{\overline{b}}_1=\widehat{j}+\widehat{k}\\ and\\ {\overline{b}}_2=2\widehat{i}+2\widehat{j}+\widehat{k}\\ {\overline{b}}_1\times {\overline{b}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=-\widehat{i}+2\widehat{j}-2\widehat{k}\end{array}$

Angle between ${\overline{b}}_1\times {\overline{b}}_2$ ${\stackrel{}{}}_{}{\stackrel{}{}}_{}$ and plane P,

$\Rightarrow a^2=-\frac{25}{11}\left(notpossible\right)$

$li{m}_{n\to \infty }{\left(\frac{n+1}{n}\right)}^{k-1}\frac{1}{n}{\displaystyle {\sum }_{r=1}^n\left(k+\frac{r}{n}\right)}$

= $33{\displaystyle {\int }_0^1{x}^{k}dx}$

$f\left(x\right)=\{\begin{array}{l}\left|4x^2-8x+5\right|,if8x^2-6x+1\ge 0\\ \left[4x^2-8x+5\right],if8x^2-6x+1<0\end{array}$

$x=\frac{1}{4},\frac{2-\sqrt[]{2}}{2},\frac{1}{2}$

General term

${=}^{15}{C}_r{\left(t^2{x}^{\frac{1}{5}}\right)}^{15-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$    

for term independent on t

2 (15 – r) – r = 0

$\Rightarrow r=10$          

$\therefore T_{11}{=}^{15}{C}_{10}\times \left(1-x\right)$

Maximum value of x (1 – x) occur at

x =  $\frac{1}{2}$

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

Given : ${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2\&\overrightarrow{a}\cdot \overrightarrow{b}=3$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta =3$

$\left|\overrightarrow{a}\right|cos\theta =\frac{3}{\sqrt[]{6}}=\sqrt[]{\frac{9}{6}}=\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta =75$

$\left|\overrightarrow{a}\right|sin\theta =\sqrt[]{\frac{75}{6}}=\frac{5}{\sqrt[]{2}}$

$\left|\overrightarrow{a}\right|cos\theta =\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2=\frac{25}{2}+\frac{3}{2}=\frac{28}{2}=14$

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$               

y = 22x $-\frac{1939}{27}$  which is not tangent to the curve.

$3a^2-2a+1=22$ (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$        

$f\left(x\right)=4\left|2x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right],-20<x<20$ doubtful points for differentiability : $x=\frac{-3}{2},$

$f\left(x\right)=4\left(2x+3\right)+9\left(-1\right)-12\left(-2\right)-240=8x-213forx=\frac{-3}{2}+h$

= $-8x-230$ for x = $\frac{-3}{2}-h$

Not diff. at x = $\frac{-3}{2}$

other doubtful points : $x+\frac{1}{2}=integer$

$-20+\frac{1}{2}<x+\frac{1}{2}<20+\frac{1}{2}$

$x+\frac{1}{2}=-19,-18,....,19,20$

$x=-19.5,-18.5,-17.5,.......,18.5,19.5\to$ total 40 numbers.

No. of number = 19.5 – $\left(-19.5\right)+1=40\left(-1.5\right)included$

$-20<x<90\Rightarrow x=-19,-18,.....,18,15\to 39points$

No. of number = 19 (19) + 1 = 39

Total : 40 + 39 = 79