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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Mathematics
Chapter
Differential Equations

JEE Advanced 2010 · Paper 1 · Mathematics · Differential Equations

Let f be a real-valued differentiable function on R (the set of all...

Let ff be a real-valued differentiable function on RR (the set of all real numbers) such that f(1)=1f(1)=1. If the yy-intercept of the tangent at any point P(x,y)P(x, y) on the curve y=f(x)y=f(x) is equal to the cube of the abscissa of PP, then the value of f(−3)f(-3) is equal to

Correct answer

0

Step-by-step solution

The equation of the tangent at (x,y)(x, y) to the given curve y=f(x)y=f(x) is Y−y=dydx(X−x)Y-intercept =y−xdydx\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 x³=y-x dydx⇒dydx−yx=−\frac{d y}{d x} \Rightarrow \quad \frac{d y}{d x}-\frac{y}{x}=-x^2 which is linear in xx.  IF =\text{ IF }=e^{∫−1x\int \frac{-1}{x}d x}=1x∴\frac{1}{x}\therefore Required solution is y⋅1x=∫−x2⋅1xdx⇒yx=−x22+c⇒y=−x32+cx At x=1,y=1, 1=−12+c⇒c=32\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, f(−3)=272+32(−3)=27−92=9\begin{aligned} f(-3) & =\frac{27}{2}+\frac{3}{2}(-3) \\ & =\frac{27-9}{2}=9\end{aligned}
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