Question

The number of real solutions of the equation

$x^{2} + 2x + 4 = 0,$– $\frac{1}{\alpha^{3}} + \frac{1}{\beta^{3}}$ are.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

Given $f(x) = f( - x)$

Here we consider two cases $x$and $a,b,c$

Case I : $a + b + c = 0$ This gives $3ax^{2} + 2bx + c = 0$

⇒ $\lbrack - 1,0\rbrack$

Also $ax^{2} + bx + c = 0$satisfy $k$so $\alpha < k < \beta$, – 2 is solution in this case.

Case II : $ac > 0$. This gives $ak^{2} + bk + c = 0$

⇒ $ac < 0$, so $a^{2}k^{2} + abk + ac < 0$, 1 is solution in this case. Hence the number of solutions are four i.e. $4x^{2} - 20px + (25p^{2} + 15p - 66) = 0$

Aliter : $\left( \frac{4}{5},\mspace{6mu} 2 \right)$

⇒ $(2,\infty)$

⇒ $\left( - 1, - \frac{4}{5} \right)$and $( - \infty, - 1)$ ⇒ $x^{2} + bx + c = 0$.