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1

### 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 $$\left[ {0,1} \right].$$
2

### AIEEE 2002

The maximum distance from origin of a point on the curve
$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is
A
$$a-b$$
B
$$a+b$$
C
$$\sqrt {{a^2} + {b^2}}$$
D
$$\sqrt {{a^2} - {b^2}}$$

## Explanation

Distance of origin from $$\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}$$

$$= \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$$

$$\le \sqrt {{a^2} + {b^2} + 2ab}$$ $$\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$$

$$=a+b$$

$$\therefore$$ Maximum distance from origin $$=a+b$$

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