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1

MCQ (Single Correct Answer)

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

$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$

$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is

A

$$a-b$$

B

$$a+b$$

C

$$\sqrt {{a^2} + {b^2}} $$

D

$$\sqrt {{a^2} - {b^2}} $$

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 $$=a+b$$

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

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

$$=a+b$$

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

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Complex Numbers

Quadratic Equation and Inequalities

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Mathematical Reasoning

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Application of Derivatives

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Differential Equations