Question

The number of values of the triplet (a,b,c) for which $a{\cos }^{2}x+b{\sin }^{2}x+c=0$ is satisified by all real x,is

• A. 0
• B. 2
• C. 3
• D. infinite

Answer 1

The correct option is D infinite

Given

$a{\cos }^2\theta +b{\sin }^2\theta +c=0$ ---(i)

Since ${\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow {\cos }^2\theta +{\sin }^2\theta -1=0$ --(ii)

Comparing (i) & (ii), we get:

$\frac{a}{1}=\frac{b}{1}=\frac{c}{-1}$

$\Rightarrow a=b=-c$

$\Rightarrow (a,b,c)=(a,a,-a)$

$\therefore$ Infinite Solutions

Hence the answer is D