Soln:

$a+\frac{1}{a}={(\sqrt{a}-\frac{1}{\sqrt{a}})}^2+2$

For minimum value of$a+\frac{1}{a}$,we want minimum value of${(\sqrt{a}-\frac{1}{\sqrt{a}})}^2$that will be zero.

So, minimum value of$a+\frac{1}{a}$will be 2. Hence option (b)

2nd method:- By, $A.M\ge G.M\frac{a+\frac{1}{a}}{2}\ge \sqrt{(a\times \frac{1}{a}}thusa+\frac{1}{a}>2(i.e.,)(a+\frac{1}{a})$min will be 2 .  option (b).

Given equation:$x^2-3x+2=0$
with roots p, q

When the roots become 2p and 2q,
the equation becomes:
${\left(\frac{x}{2}\right)}^2-3\left(\frac{x}{2}\right)+2=0,x^2-6x+8=0$ 
When the roots become
$(2p+1)\&(2q+1)$
the equation becomes:

$(x-1{)}^2-6(x-1)+8=0,x^2-8x+15=0$.
Hence option (D)

$f\left(x\right)=x^8+6x^7-5x^4-3x^2+2x-5$

Check the number of positive roots= no. of sign changes in f(x) = 3

Check the number of negative roots = no. of sign changes in f(-x)

$f(-x)=x^8-6x^7-5x^4-3x^2-2x-5$. No. of sign changes = 1

Now check for zero as a root.$f\left(0\right)=-5\ne 0$

So, maximum number of real roots = 3 + 1 = 4. Hence option (a)

a and b are the roots of the equation

$a+b=(\alpha -4),ab=\alpha$

$5ab-a^2-b^2$

$5ab-(a^2+b^2)$

$5ab-\left[\right(a+b{)}^2-2ab]$

$5\alpha -\left[\right(\alpha -4{)}^2-2\alpha ]$

$5\alpha -[{\alpha }^2-10\alpha +16]$

$-{\alpha }^2+15\alpha -16=-({\alpha }^2+15\alpha -16)=-({\alpha }^2+16\alpha -\alpha -16)=-\left[\alpha \right(\alpha +16)-1(\alpha +16\left)\right]=-(\alpha +16)(\alpha -1)$

So,the maximum value$=\frac{D}{4a}=\frac{[{15}^2-4(-1\left)\right(-16\left)\right]}{4\times 1}=\frac{161}{4}.$

Let the two digit number be ab.

$a\times b=12$

When 9 is added to the number

$10a+b+9=10b+a$

$9a-9b+9=0$

$a-b+1=0$

$a-\frac{12}{a}+1=0$

$a^2+a-12=0$

$(a+4)(a-3)=0$

a = -4 is not accepted. So, a = 3, b = 4

The number is 34 .

Let y be the given expression, then

$y=\sqrt{(6+y)}$

$y^2=6+y$

$y^2-y-6=0$

(y – 3)(y + 2) = 0

y = 3, y = -2

So, the value of the given expression is 3.

Assume$\alpha$be the 3rd root.

Sum of the roots, i.e.,$3+5+\alpha =k^2$

Product of the roots, i.e.$3\times 5\times \alpha =5k$

So,$8+\alpha =9{\alpha }^2$

Then,$\alpha =1or-\frac{8}{9}$

Hence the 3rd root is 1. 

$\begin{array}{cc}CASE& \text{No. of ways}\\ \text{2 SCIENTISTS, 1 BUREAUCRAT}& {}^3{C}_2{\times }^3{C}_1=3\times 3=9\\ \text{1 SCIENTIST, 2 BUREAUCRAT}& {}^3{C}_1{\times }^3{C}_2=3\times 3=9\\ \text{2 OUT OF 6 (3 SCIENTIST, 3 BUREAUCRAT),}& {}^6{C}_2\times 1=151\\ WHOISBOTH& \end{array}$

So, total number of ways = 9 + 9 + 15 = 33. Hence option (e)

At least 1 orange can be selected in$2^5-1=31$ways.

At least 1 apple can be selected in$2^4-1=15$ways.

Total no. of selections possible$=31\times 15=465$

Majority of fourth year means out of 7, minimum 4 should be from fourth year.

Considering different possibilities:

$\begin{array}{ccccc}\text{No.from 4th year}& \text{Ways of choosing 4th year}& \text{No.from 3rd year}& \text{Ways of choosing 3rd year}& \text{wys of choosing commitee}\\ \text{4}& \text{8C4=70}& \text{3}& \text{6C3=20}& \text{70}\times \text{20 =1400}\\ \text{5}& \text{8C5=56}& \text{2}& \text{6C2=15}& \text{56}\times \text{15 =840}\\ \text{6}& \text{8C6=28}& \text{1}& \text{6C1=6}& \text{28}\times \text{6 =168}\\ \text{7}& \text{8C7=8}& \text{0}& \text{6C0=1}& \text{8}\times \text{1=8}\\ \text{Total}& & & & \text{2416}\end{array}$

So, total no. of ways = 2416.