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1

AIEEE 2004

A function $$y=f(x)$$ has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$ If its graph passes through the point $$(2, 1)$$ and at that point the tangent to the graph is $$y = 3x - 5$$, then the function is
A
$${\left( {x + 1} \right)^2}$$
B
$${\left( {x - 1} \right)^3}$$
C
$${\left( {x + 1} \right)^3}$$
D
$${\left( {x - 1} \right)^2}$$

Explanation

$$f''\left( x \right) = 6\left( {x - 1} \right).$$ Inegrating,

we get $$f'\left( x \right) = 3{x^2} - 6x + c$$

Slope at $$\left( {2,1} \right) = f'\left( 2 \right) = c = 3$$

$$\left[ {\,\,} \right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\left. {\,\,} \right]$$

$$\therefore$$ $$f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}$$

Inegrating again, we get

$$f\left( x \right) = {\left( {x - 1} \right)^3} + D$$

The curve passes through $$(2,1)$$

$$\Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0$$

$$\therefore$$ $$f\left( x \right) = {\left( {x - 1} \right)^3}$$
2

AIEEE 2004

A point on the parabola $${y^2} = 18x$$ at which the ordinate increases at twice the rate of the abscissa is
A
$$\left( {{9 \over 8},{9 \over 2}} \right)$$
B
$$(2, -4)$$
C
$$\left( {{-9 \over 8},{9 \over 2}} \right)$$
D
$$(2, 4)$$

Explanation

$${y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}$$

Given $${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$$

Puting in $${y^2} = 18x \Rightarrow x = {9 \over 8}$$

$$\therefore$$ Required point is $$\left( {{9 \over 8},{9 \over 2}} \right)$$
3

AIEEE 2003

If the function $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1,$$ where $$a>0,$$ attains its maximum and minimum at $$p$$ and $$q$$ respectively such that $${p^2} = q$$ , then $$a$$ equals
A
$${1 \over 2}$$
B
$$3$$
C
$$1$$
D
$$2$$

Explanation

$$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$

$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$

$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$

For max. or min.

$$6{x^2} - 18ax + 12{a^2} = 0$$

$$\Rightarrow {x^2} - 3ax + 2{a^2} = 0$$

$$\Rightarrow x = a$$ or $$x=2a.$$

At $$x=a$$

$$f''(a) = 12a - 18a = -6a < 0$$

At $$x=a$$ maximum As f''($$a$$) < 0.

At $$x=2a$$

$$f''(a) = 24a - 18a = 6a > 0$$

At $$x=2a$$ minimum As f''(2$$a$$) > 0.

$$\therefore$$ $$p=a$$ and $$q=2a$$

As per question $${p^2} = q$$

$$\therefore$$ $${a^2} = 2a$$

$$\Rightarrow$$ $$a(a - 2) = 0$$

$$\therefore$$ $$a$$ = 0, 2

but $$a > 0,$$ therefore, $$a=2.$$
4

AIEEE 2002

If $$2a+3b+6c=0,$$ $$\left( {a,b,c \in R} \right)$$ then the quadratic equation $$a{x^2} + bx + c = 0$$ has
A
at least one root in $$\left[ {0,1} \right]$$
B
at least one root in $$\left[ {2,3} \right]$$
C
at least one root in $$\left[ {4,5} \right]$$
D
none of these

Explanation

Let $$f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0$$ and $$f(1)$$

$$= {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0$$

Also $$f(x)$$ is continuous and differentiable in $$\left[ {0,1} \right]$$ and

$$\left[ {0,1\left[ {.\,\,} \right.} \right.$$ So by Rolle's theorem $$f'\left( x \right) = 0.$$

i.e $$\,\,a{x^2} + bx + c = 0$$ has at least one root in $$\left[ {0,1} \right].$$

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