4-2.Quadratic Equations and Inequations
hard

જો $\alpha $ અને $\beta $ એ દ્રીઘાત સમીકરણ ${x^2}\,\sin \,\theta  - x\,\left( {\sin \,\theta \cos \,\,\theta  + 1} \right) + \cos \,\theta  = 0\,\left( {0 < \theta  < {{45}^o}} \right)$ ના ઉકેલો હોય અને $\alpha  < \beta $ તો $\sum\limits_{n = 0}^\infty  {\left( {{\alpha ^n} + \frac{{{{\left( { - 1} \right)}^n}}}{{{\beta ^n}}}} \right)} $ = ......

A

$\frac{1}{{1 - \cos \,\theta }} - \frac{1}{{1 + \sin \,\theta \,}}$

B

$\frac{1}{{1 + \cos \,\theta }} + \frac{1}{{1 - \sin \,\theta \,}}$

C

$\frac{1}{{1 - \cos \,\theta }} + \frac{1}{{1 + \sin \,\theta \,}}$

D

$\frac{1}{{1 + \cos \,\theta }} - \frac{1}{{1 - \sin \,\theta \,}}$

(JEE MAIN-2019)

Solution

Using quadratic formula,

$x=\frac{(\cos \theta \sin \theta+1) \pm \sqrt{(\cos \theta \sin \theta+1)^{2}-4 \sin \theta \cos \theta}}{2 \sin \theta}$

$=\frac{(\cos \theta \sin \theta+1)^{2} \pm(\cos \theta \sin \theta-1)}{2 \sin \theta}$

$ = \cos \,\theta ,\,\cos ec\,\theta $

$\alpha  = \cos \,\theta ,\,\beta  = \cos ec\,\theta $

$\therefore \,\sum\limits_{n = 0}^\infty  {{\alpha ^n}}  + \frac{{{{( – 1)}^n}}}{{{\beta ^n}}}$

$ = \sum\limits_{n = 0}^\infty  {{{(\cos ec)}^n}}  + \sum\limits_{n = 0}^\infty  {{{( – \sin \theta )}^n}} $

$=\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta}$

$\therefore $ $(C)$ is the correct

Standard 11
Mathematics

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