JEE Mathematics 2023 Question with Solution

Given: An arc $PQ$ subtends a right angle at the centre $O$, so $\angle POQ = 90^\circ$. Also, $R$ is the midpoint of arc $PQ$. We have $\overrightarrow{OP}=\overrightarrow{u}$, $\overrightarrow{OR}=\overrightarrow{v}$ and $\overrightarrow{OQ}=\alpha \overrightarrow{u}+\beta \overrightarrow{v}$.

Find: The quadratic equation whose roots are $\alpha$ and $\beta^2$.

Let $\overrightarrow{u}=\hat{i}$ and $\overrightarrow{OQ}=\hat{j}$. Since $R$ is the midpoint of arc $PQ$, $\overrightarrow{OR}$ bisects the right angle $\angle POQ$. Therefore,

$$\overrightarrow{v}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$$

Given

$$\hat{j}=\alpha \hat{i}+\beta \left(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}\right)$$

So,

$$\hat{j}=\left(\alpha+\frac{\beta}{\sqrt{2}}\right)\hat{i}+\frac{\beta}{\sqrt{2}}\hat{j}$$

Comparing coefficients of $\hat{i}$ and $\hat{j}$,

$$\alpha+\frac{\beta}{\sqrt{2}}=0$$

and

$$\frac{\beta}{\sqrt{2}}=1$$

From the second equation,

$$\beta=\sqrt{2}$$

Substituting into the first equation,

$$\alpha+1=0$$

Hence,

$$\alpha=-1$$

Now,

$$\beta^2=2$$

So the required roots are $-1$ and $2$. Hence the quadratic equation is

$$x^2-(\text{sum of roots})x+(\text{product of roots})=0$$

That is,

$$x^2-(-1+2)x+(-1)(2)=0$$ $$x^2-x-2=0$$

Therefore, the correct option is C.

Using the obtained values $\alpha=-1$ and $\beta=\sqrt{2}$, we get $\beta^2=2$.

If a quadratic has roots $r_1$ and $r_2$, then its equation is

$$x^2-(r_1+r_2)x+r_1r_2=0$$

Here,

$$r_1=\alpha=-1, \qquad r_2=\beta^2=2$$

So,

$$r_1+r_2=-1+2=1$$

and

$$r_1r_2=(-1)(2)=-2$$

Therefore,

$$x^2-x-2=0$$

Thus the correct option is C.