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1

AIEEE 2004

The normal to the curve x = a(1 + cos $$\theta$$), $$y = a\sin \theta$$ at $$'\theta '$$ always passes through the fixed point
A
$$(a, a)$$
B
$$(0, a)$$
C
$$(0, 0)$$
D
$$(a, 0)$$

Explanation

$${{dx} \over {d\theta }} = - a\sin \theta$$ and $${{dy} \over {d\theta }} = a\cos \theta$$

$$\therefore$$ $${{dy} \over {dx}} = - \cot \theta .$$

$$\therefore$$ The slope of the normal at $$\theta$$ = $$- {1 \over { - \cot \theta }}$$$$= \tan \theta$$

$$\therefore$$ The equation of the normal at $$\theta$$ is

$$y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)$$

$$\Rightarrow$$ $$y - a\sin \theta =$$ $${{\sin \theta } \over {\cos \theta }}$$ $$\left( {x - a - a\cos \theta } \right)$$

$$\Rightarrow y\cos \theta - a\sin \theta \cos \theta$$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$= x\sin \theta - a\sin \theta - a\sin \theta \cos \theta$$

$$\Rightarrow x\sin \theta - y\cos \theta = a\sin \theta$$

$$\Rightarrow y = \left( {x - a} \right)\tan \theta$$

which always passes through $$(a, 0)$$
2

AIEEE 2004

If $$2a+3b+6c=0$$, then at least one root of the equation
$$a{x^2} + bx + c = 0$$ lies in the interval
A
$$(1, 3)$$
B
$$(1, 2)$$
C
$$(2, 3)$$
D
$$(0, 1)$$

Explanation

Let us define a function

$$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$$

Being polynomial, it is continuous and differentiable, also,

$$f\left( 0 \right) = 0\,$$ and $$\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$$

$$\Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0$$ (given)

$$\therefore$$ $$f\left( 0 \right) = f\left( 1 \right)$$

$$\therefore$$ $$f(x)$$ satisfies all conditions of Rolle's theorem

therefore $$f'\left( x \right) = 0$$ has a root in $$\left( {0,1} \right)$$

i.e. $$a{x^2} + bx + c = 0$$ has at lease one root in $$(0, 1)$$
3

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 $$(2,1)$$

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

$$\therefore$$ $$f\left( x \right) = {\left( {x - 1} \right)^3}$$
4

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 $$\left( {{9 \over 8},{9 \over 2}} \right)$$

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