Question

If $\alpha \beta$ are the zeros of the polynomial $f(x)=x^{2}+px+q$, then polynomial having $\frac{1}{\alpha} and \frac{1}{\beta}$ is its zero is (a) $x^{2}+qx+p$ (b) $x^{2}-px+q$ (c) $qx^{2}+px+1$ (d) $px^{2}+qx+1$

Answer 1

SOLUTION :

The correct option is (c) : qx² + px +1.

Given : α  and β are the zeroes of the  polynomial f(x) = x² + px + q

On comparing with ax² + bx + c,

a = 1, b= p, c = q

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = - p/1  

α + β  = - p ………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = q/1  

αβ = q ……………………(2)

Given : 1/α and  1/ β are the zeroes of the required polynomial .

Sum of zeroes =  1/α + 1/ β  

1/α + 1/ β = (α + β) /αβ

1/α + 1/ β = - p / q ……………(3)

[From eq 1 & 2 ]

Product of the zeroes = 1/α × 1/ β = 1/αβ

1/α × 1/ β = 1/ q ………….(4)

[From eq  2 ]

Then, the required quadratic polynomial is :  

[x² –(sum of the zeroes)x + (product of the zeroes)] = 0  

[x² –(1/α +1/ β)x + (1/α β)] = 0

[ x² - (-p/q) x + (1/q)] = 0

[From eq 3 & 4]

[ (qx² + px +1)/q] = 0

qx² + px +1 = 0

Hence, the required polynomial is = qx² + px +1

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

For any arbitrary motion in space, which of the following relations are true: a. $V_{average}= (\frac{1}{2})[v(t_1)+v(t_2)]$ b. $V_{average}=\frac {[r(t_2)-r(t_1)]} {(t_2-t_1)}$ c. v (t) = v (0) + at d. r (t) = r (0) + v (0) t + (1/2) at² e. $a_{average}=\frac {[v(t_2)-v(t_1)]} {(t_2-t_1)}$ (The 'average' stands for average of the quantity over the time interval $t_1 to t_2$)