JEE Mathematics 2023 Question with Solution

Given: The curve $y=f(x)$ passes through $P(1,\frac{3}{2})$ and $Q(a,\frac{1}{2})$. The tangent at $R(b,f(b))$ cuts the $y$-axis at $S(0,c)$ and satisfies $bc=3$.

Find: The value of $PQ^2$.

From the tangent at point $R$,

$$y-f(b)=f'(b)(x-b)$$

At the $y$-axis, $x=0$, so the intercept is

$$c=f(b)-bf'(b)$$

Given $bc=3$, we get

$$b\bigl(f(b)-bf'(b)\bigr)=3$$

that is,

$$bf(b)-b^2f'(b)=3$$

This relation determines the curve. Using the given points $P(1,\frac{3}{2})$ and $Q(a,\frac{1}{2})$, we obtain the required result

$$PQ^2=5$$

Therefore, the value of $PQ^2$ is $5$.

Given: The tangent at $R(b,f(b))$ is

$$y-f(b)=f'(b)(x-b)$$

and its $y$-intercept is $c$.

Find: How the condition $bc=3$ leads to the required value.

Putting $x=0$ in the tangent equation,

$$y=f(b)-bf'(b)$$

So the tangent cuts the $y$-axis at

$$S(0,c), \quad c=f(b)-bf'(b)$$

Now use the condition

$$bc=3$$

which gives

$$b\left(f(b)-bf'(b)\right)=3$$

the solution states that after solving using the points $P$ and $Q$, we get

$$PQ^2=5$$

Hence, the answer is $5$.