JEE Mathematics 2024 Question with Solution

Given: $\cos(2\sin^{-1} x) = \frac{1}{9}$ and $x = \frac{m}{n}$ where $m, n$ are co-prime natural numbers.

Find: The line on which the point $(\alpha, \beta)$ lies, where $\alpha$ and $\beta$ are roots of $mx^2 - nx - m + n = 0$.

Use the identity

$$\cos(2\theta) = 1 - 2\sin^2\theta$$

with $\theta = \sin^{-1}x$. Then

$$\cos(2\sin^{-1}x) = 1 - 2x^2$$

So,

$$1 - 2x^2 = \frac{1}{9}$$ $$2x^2 = 1 - \frac{1}{9} = \frac{8}{9}$$ $$x^2 = \frac{4}{9}$$

Hence,

$$x = \pm \frac{2}{3}$$

Since $x = \frac{m}{n}$ with $m, n$ co-prime natural numbers, we take

$$x = \frac{2}{3}$$

Therefore,

$$m = 2, \quad n = 3$$

Substitute into the quadratic equation:

$$mx^2 - nx - m + n = 0$$ $$2x^2 - 3x - 2 + 3 = 0$$ $$2x^2 - 3x + 1 = 0$$

Now solve for the roots:

$$x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}$$

Thus,

$$\alpha = 1, \quad \beta = \frac{1}{2}$$

Now check the point $(1, \frac{1}{2})$ in the given lines. For

$$5x + 8y = 9$$

we get

$$5(1) + 8\left(\frac{1}{2}\right) = 5 + 4 = 9$$

So the point satisfies this line.

Therefore, the correct option is D, that is, $5x + 8y = 9$.

Given: $\cos(2\sin^{-1} x) = \frac{1}{9}$.

Find: Which option is satisfied by $(\alpha, \beta)$.

First obtain $x$ quickly from

$$1 - 2x^2 = \frac{1}{9}$$

which gives

$$x = \frac{2}{3}$$

so $m = 2$ and $n = 3$.

Then the quadratic becomes

$$2x^2 - 3x + 1 = 0$$

whose roots are $1$ and $\frac{1}{2}$. Since $\alpha > \beta$,

$$(\alpha, \beta) = \left(1, \frac{1}{2}\right)$$

Now substitute this point in the options. The expression

$$5x + 8y$$

at $\left(1, \frac{1}{2}\right)$ becomes

$$5(1) + 8\left(\frac{1}{2}\right) = 9$$

Hence the point lies on $5x + 8y = 9$.

Therefore, the correct option is D.