If - - are the roots of x2-px-1-c-0- then the value of -2-2 -1-2-2 -c-2-2 -1-2-2 -c is

The correct option is B 1x^2 p(x+1) c = 0 ⇒ x^2 px p c = 0 Sum and product of the roots is, α +β = p, αβ = (p+c) (α +1)(β +1) = αβ +(α +β )+1 = 1 c Now ...

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Standard XII

Mathematics

Range

If α, β are t...

Question

If $α, β$ are the roots of $x^{2} - p (x + 1) - c = 0$ , then the value of $\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c}$ is

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Solution

The correct option is B $1$
$x^{2} - p (x + 1) - c = 0 \Rightarrow x^{2} - p x - p - c = 0$
Sum and product of the roots is,
$α + β = p, α β = - (p + c)$
$(α + 1) (β + 1) = α β + (α + β) + 1 = 1 - c$
Now given expression
$\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} = \frac{(α + 1)^{2}}{(α + 1)^{2} - (1 - c)} + \frac{(β + 1)^{2}}{(β + 1)^{2} - (1 - c)}$
Putting the value $1 - c = (α + 1) (β + 1)$ , we get
$= \frac{α + 1}{α - β} + \frac{β + 1}{β - α} = \frac{α + 1 - β - 1}{α - β} = 1$

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If α,β are the roots of x2−p(x+1)−c=0, then the value of α2+2α+1α2+2α+c+β2+2β+1β2+2β+c is

If $α, β$ are the roots of $x^{2} - p (x + 1) - c = 0$ , then the value of $\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c}$ is