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1

### JEE Main 2015 (Offline)

Let $$f(x)$$ be a polynomial of degree four having extreme values
at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :
A
$$0$$
B
$$4$$
C
$$-8$$
D
$$-4$$

## Explanation

$$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$$

So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$

Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$

Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$

As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$

Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$

By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get

$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$

$$\Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$

and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$

$$\Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$

$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$
2

### JEE Main 2014 (Offline)

If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$
A
$$f'\left( c \right) = g'\left( c \right)$$
B
$$f'\left( c \right) = 2g'\left( c \right)$$
C
$$2f'\left( c \right) = g'\left( c \right)$$
D
$$2f'\left( c \right) = 3g'\left( c \right)$$

## Explanation

Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$

and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that

$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$

and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$

Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$
3

### JEE Main 2013 (Offline)

The intercepts on $$x$$-axis made by tangents to the curve,
$$y = \int\limits_0^x {\left| t \right|dt,x \in R,}$$ which are parallel to the line $$y=2x$$, are equal to :
A
$$\pm 1$$
B
$$\pm 2$$
C
$$\pm 3$$
D
$$\pm 4$$

## Explanation

Since, $$y = \int\limits_0^x {\left| t \right|} dt,x \in R$$

therefore $${{dy} \over {dx}} = \left| x \right|$$

But from $$y = 2x,{{dy} \over {dx}} = 2$$

$$\Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$

Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$

$$\therefore$$ equation of tangent is

$$y - 2 = 2\left( {x - 2} \right)$$ or $$y + 2 = 2\left( {x + 2} \right)$$

$$\Rightarrow$$ $$x$$-intercept $$= \pm 1.$$
4

### AIEEE 2012

A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is :
A
$$-{1 \over 4}$$
B
$$-4$$
C
$$-2$$
D
$$-{1 \over 2}$$

## Explanation

Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$ having

slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$

Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its

equation is

$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$

where $$m$$ is the slope of the line $$PQ$$.

Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$\Rightarrow 0 - 2 = m\left( {x - 1} \right)$$

$$\Rightarrow - 2 = m\left( {x - 1} \right)$$

$$\Rightarrow x - 1 = {{ - 2} \over m}$$

$$\Rightarrow x = {{ - 2} \over m} + 1$$

Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$

Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$\Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b

and $$OQ = y = 2 - m$$

Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$

( As Area of $$\Delta = {1 \over 2} \times$$ base $$\,\, \times \,\,$$ height )

$$= {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$

$$= 2 - {m \over 2} - {2 \over m}$$ Let Area $$= f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$

Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$

Put $$f'\left( m \right) = 0$$

$$\Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$

Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$

$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$

$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$

Area will be least at $$m=-2$$

Hence, slope of $$PQ$$ is $$-2.$$

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