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1

### IIT-JEE 2010 Paper 1 Offline

Numerical
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 $$f(-3)$$

## Explanation

The equation of the tangent at (x, y) to the given curve y = f(x) is

$$Y - y = {{dy} \over {dx}}(X - x)$$

Y-intercept $$= y - x{{dy} \over {dx}}$$

According to the question

$${x^3} = y - x{{dy} \over {dx}}$$

$$\Rightarrow {{dy} \over {dx}} - {y \over x} = - {x^2}$$

which is linear in x.

$$IF = {e^{\int {{{ - 1} \over x}dx} }} = {1 \over x}$$

$$\therefore$$ Required solution is

$$y\,.\,{1 \over x} = \int { - {x^2}\,.\,{1 \over x}dx}$$

$$\Rightarrow {y \over x} = {{ - {x^2}} \over 2} + c$$

$$\Rightarrow y = {{ - {x^3}} \over 2} + cx$$

At x = 1, y = 1,

$$1 = {{ - 1} \over 2} + c$$

$$\Rightarrow c = {3 \over 2}$$

Now, $$f( - 3) = {{27} \over 2} + {3 \over 2}( - 3)$$

$$= {{27 - 9} \over 2} = 9$$

2

### IIT-JEE 2010 Paper 2 Offline

Numerical
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 $$1$$.

3

### IIT-JEE 2009

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

4

### IIT-JEE 2009

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

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