Q50 / 84 · JEE Advanced 2010 (Paper 1) · Mathematics
Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$, then the value of $f(-3)$ is equal to
Correct answer: 9
Solution
The equation of the tangent at $(x, y)$ to the given curve $y=f(x)$ is
$$
\begin{gathered}
Y-y=\frac{d y}{d x}(X-x) \\
Y \text {-intercept }=y-x \frac{d y}{d x}
\end{gathered}
$$
According to the question
$$
x^3=y-x \frac{d y}{d x}
$$
$$
\Rightarrow \quad \frac{d y}{d x}-\frac{y}{x}=-x^2
$$
which is linear in $x$.
$$
\text { IF }=e^{\int \frac{-1}{x} d x}=\frac{1}{x}
$$
$\therefore$ Required solution is
$$
\begin{aligned}
& y \cdot \frac{1}{x}=\int-x^2 \cdot \frac{1}{x} d x \\
& \Rightarrow \quad \frac{y}{x}=\frac{-x^2}{2}+c \\
& \Rightarrow \quad y=\frac{-x^3}{2}+c x \\
& \text { At } x=1, y=1 \text {, } \\
& 1=\frac{-1}{2}+c \\
& \Rightarrow \quad c=\frac{3}{2} \\
&
\end{aligned}
$$
Now, $\begin{aligned} f(-3) & =\frac{27}{2}+\frac{3}{2}(-3) \\ & =\frac{27-9}{2}=9\end{aligned}$
Hence, the correct answer is 9.