Ncert Solutions Maths class 12th

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

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$           

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$           

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$           

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$  

Each element of ordered pair (i, j) is either present in A or in B.

              So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$  

              = 20 (1 + 2 + 3 + … + 10) = 1100

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$?8x^2?2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)           

6x = 5 = 0             $x=\frac{5}{6}$  

P (atmost 2H)

$P\left(OH,5T\right)+P\left(1H,4T\right)+P\left(2H,3T\right)$  

$=\frac{1}{6^5}\left(1+25+250\right)=\frac{276}{6^5}=\frac{4^6}{6^4}$  

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=3\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{u}$

$\overrightarrow{b}*\left(\overrightarrow{c}*\overrightarrow{a}\right)=\overrightarrow{c}-2\overrightarrow{a}=\overrightarrow{v}$

$\overrightarrow{c}*\left(\overrightarrow{b}*\overrightarrow{a}\right)=3\overrightarrow{b}-2\overrightarrow{a}=\overrightarrow{w}$

$\therefore \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{w}$

so vectors

$\overrightarrow{u},\overrightarrow{v}and\overrightarrow{w}$

are coplanar, hence their Scalar triple product will be zero.

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$  

$?$  Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$  

0  $\lambda =7$  

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$\left(2,-\frac{1}{2},2\right)$  

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$  

and  $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$     

clearly 

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$  

So, a : b : c = 8 : 5 : -4

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$ a

=> 2x – z = 1

option (B) satisfies.

Let AB   $\equiv x-2y+1=0$

AC $\equiv x-2y+1=0$  

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)