IIT-JEE 2010 Paper 1 Offline | Application of Derivatives Question 26 | Mathematics

IIT-JEE 2010 Paper 1 Offline

Numerical

-0

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 find the value of $$f(-3)$$

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IIT-JEE 2010 Paper 2 Offline

Numerical

-0

Let $$f$$ be a function defined on $$R$$ (the set of all real numbers)
such that $$f'\left( x \right) = 2010\left( {x - 2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}$$ for all $$x \in $$$$R$$

If $$g$$ is a function defined on $$R$$ with values in the interval $$\left( {0,\infty } \right)$$ such that $$$f\left( x \right) = ln\,\left( {g\left( x \right)} \right),\,\,for\,\,all\,\,x \in R$$$
then the number of points in $$R$$ at which $$g$$ has a local maximum is $$1$$.

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IIT-JEE 2009 Paper 2 Offline

Numerical

-1

Let $$p(x)$$ be a polynomial of degree $$4$$ having extremum at

$$x = 1,2$$ and $$\mathop {\lim }\limits_{x \to 0} \left( {1 + {{p\left( x \right)} \over {{x^2}}}} \right) = 2$$.

Then the value of $$p (2)$$ is

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IIT-JEE 2009 Paper 2 Offline

Numerical

-0

The maximum value of the function $$f(x) = 2{x^3} - 15{x^2} + 36x - 48$$ on the set $$A = \{ x|{x^2} + 20 \le 9x|\} $$ is __________.

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