Question

Let $\alpha ,\beta ,\gamma$ be distinct real numbers such that
$a{\alpha }^2+b\alpha +c=\left(\sin \theta \right){\alpha }^2+\left(\cos \theta \right)\alpha a{\beta }^2+b\beta +c=\left(\sin \theta \right){\beta }^2+\left(\cos \theta \right)\beta a{\gamma }^2+b\gamma +c=\left(\sin \theta \right){\gamma }^2+\left(\cos \theta \right)\gamma$
where $a,b,c,\theta \in R$
Then which of the following is/are correct?

• A. The maximum value of $\frac{a^2+b^2}{a^2+3ab+5b^2}$ is $2$
• B. The minimum value of $\frac{a^2+b^2}{a^2+3ab+5b^2}$ is $\frac{2}{11}$
• C. If $\overrightarrow{V_1}=a\hat{i}+b\hat{j}+c\hat{k}$ makes an angle $\frac{\pi }{3}$ with $\overrightarrow{V_2}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}$, then the number of values of $\theta \in [0,2\pi ]$ is $3$.
• D. If $\overrightarrow{V_1}=a\hat{i}+b\hat{j}+c\hat{k}$ makes an angle $\frac{\pi }{3}$ with $\overrightarrow{V_2}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}$, then the number of values of $\theta \in [0,2\pi ]$ is $5$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The maximum value of $\frac{a^2+b^2}{a^2+3ab+5b^2}$ is $2$
B The minimum value of $\frac{a^2+b^2}{a^2+3ab+5b^2}$ is $\frac{2}{11}$
C If $\overrightarrow{V_1}=a\hat{i}+b\hat{j}+c\hat{k}$ makes an angle $\frac{\pi }{3}$ with $\overrightarrow{V_2}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}$, then the number of values of $\theta \in [0,2\pi ]$ is $3$.

Given : $a{\alpha }^2+b\alpha +c=\left(\sin \theta \right){\alpha }^2+\left(\cos \theta \right)\alpha a{\beta }^2+b\beta +c=\left(\sin \theta \right){\beta }^2+\left(\cos \theta \right)\beta a{\gamma }^2+b\gamma +c=\left(\sin \theta \right){\gamma }^2+\left(\cos \theta \right)\gamma$

Now, $(a-\sin \theta )x^2+(b-\cos \theta )x+c=0$
This equation has roots as $\alpha ,\beta ,\gamma$
As this is a quadratic having more than $2$ roots, so this is an identity
$(a-\sin \theta )=0,(b-\cos \theta )=0,c=0\Rightarrow a=\sin \theta ,b=\cos \theta$

Now, $E=\frac{a^2+b^2}{a^2+3ab+5b^2}$
$\Rightarrow E=\frac{1}{{\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta }\Rightarrow E=\frac{2}{1-\cos 2\theta +3\sin 2\theta +5(1+\cos 2\theta )}\Rightarrow E=\frac{2}{6+3\sin 2\theta +4\cos 2\theta }$
We know that,
$3\sin 2\theta +4\cos 2\theta \in [-5,5]$
$E\in [\frac{2}{11},2]$

$\overrightarrow{V_1}=a\hat{i}+b\hat{j}+c\hat{k}\Rightarrow \overrightarrow{V_1}=\cos \theta \hat{i}+\sin \theta \hat{j}\overrightarrow{V_2}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}\cos \frac{\pi }{3}=\frac{\overrightarrow{V_1}\cdot \overrightarrow{V_2}}{|\overrightarrow{V_1}||\overrightarrow{V_2}|}\Rightarrow \frac{1}{2}=\frac{\cos \theta +\sin \theta }{1\times 2}\Rightarrow \cos \theta +\sin \theta =1\Rightarrow \cos (\theta +\frac{\pi }{4})=\frac{1}{\sqrt{2}}\Rightarrow \sin (\theta +\frac{\pi }{4})=\frac{1}{\sqrt{2}}\Rightarrow \theta +\frac{\pi }{4}=\frac{\pi }{4},\frac{3\pi }{4},\frac{9\pi }{4}\therefore \theta =0,\frac{\pi }{2},2\pi$