JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer In a triangle \[ABC\] the value of \[\angle A\] is given by \[5\cos A+3=0\], then the equation  whose roots are \[\sin A\] and \[\tan A\] will be [Roorkee 1972]

    A) \[15{{x}^{2}}-8x+16=0\]

    B) \[15{{x}^{2}}+8x-16=0\]

    C) \[15{{x}^{2}}-8\sqrt{2}x+16=0\]

    D) \[15{{x}^{2}}-8x-16=0\]

    Correct Answer: B

    Solution :

    Given that \[5\cos A+3=0\]or \[\cos A=-\frac{3}{5}\] Let \[\alpha =\sin A\]and \[\beta =\tan A\], then the sum of roots \[=\alpha +\beta =\sin A+\tan A\] \[=\sin A+\frac{\sin A}{\cos A}=\frac{\sin A}{\cos A}(1+\cos A)\] \[=\frac{\sqrt{1-9/25}}{-3/5}\left( 1-\frac{3}{5} \right)=\frac{4}{-5}.\frac{5}{3}.\frac{2}{5}=\frac{8}{-15}\] and product of roots \[\alpha .\beta =\sin A\tan A=\frac{{{\sin }^{2}}A}{\cos A}\] \[=\frac{16/25}{-3/5}=-\frac{16}{25}\times \frac{5}{3}=-\frac{16}{15}\] Thus required equation  is \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\] Þ \[{{x}^{2}}+\frac{8x}{15}-\frac{16}{15}=0\] Þ \[15{{x}^{2}}+8x-16=0\]

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