Question

If the roots of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p}$ are $\frac{r + p - x}{q}$ and the roots of the equation $\frac{3x}{p + q + r} = 0$ are $x = p + q + r$, then value of p will be.

• A. $x = p - q + r$
• B. $x = \frac{p + q}{q + r}$
• C. $x = \frac{p}{q} + r$
• D. None of these

Answer 1

Answer: (B)

$x = \frac{p + q}{q + r}$

Answer 2

$\mathbf{a}\mathbf{x}^{\mathbf{2}}\mathbf{+ bx + c = 0}$ are the roots of $a_{n}x^{n} + a_{n - 1}x^{n - 1} + .... + a_{1}x = 0$.

So, $x = \alpha$and $na_{n}x^{n - 1} + (n - 1)a_{n - 1}x^{n - 2} + .... + a_{1} = 0$

Again $\alpha$ are the roots of $\alpha$ then

$\leq 2$ and $P(0) = 0,$

Now $P(1) = 1$

⇒ $P'(x) > 0\ \forall x \in (0,1)$

⇒ $S = 0$.