IIT-JEE 2010 · Paper 1 · Mathematics Q56

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MathematicsInteger+3Question 56 of 84

Let f be a real-valued differentiable function on $\mathbb R$ (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

Your answer

Answer and worked solution

Official answer: 9

Tangent's y-intercept. At $(x,y)$ the tangent meets the $y$ -axis at $y-xy'$ . So
$y-xy'=x^3\implies y'-\frac yx=-x^2\implies\left(\frac yx\right)'=-x.$

Solve. $\frac yx=-\frac{x^2}2+C$ ; $f(1)=1$ gives $C=\frac32$ . So $f(x)=\frac{3x-x^3}2$ .

Value. $f(-3)=\frac{-9+27}2=9$ .

Answer: 9

Topics:Linear first-order equations Tangents and normals

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IIT-JEE 2010 · Paper 1, Mathematics question 56