11th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-1

  • question_answer If \[\alpha +\beta =\mathbf{9}{{\mathbf{0}}^{{}^\circ }}\], then the maximum value of sin \[\alpha \].sin \[\beta \]. is

    A) 1                                             

    B) \[\frac{1}{2}\]              

    C) \[\frac{3}{2}\]                          

    D) \[\frac{-1}{2}\]

    Correct Answer: B

    Solution :

    [b] \[\alpha +\beta ={{90}^{{}^\circ }}\] \[\because sin\alpha .sin\beta =\frac{1}{2}.2.sin\alpha .sin\beta \] \[=\frac{1}{2}[cos\left( \alpha -\beta  \right)-cos\left( \beta +\alpha  \right)\] \[=\frac{1}{2}\left[ cos\left( \alpha -\beta  \right)-cos{{90}^{{}^\circ }} \right]\] \[=\frac{1}{2}[cos\left( \alpha -\beta  \right)\] \[=\frac{1}{2}\times 1=\frac{1}{2}\] \[\therefore \]Maximum value of \[cos\left( \alpha -\beta  \right)=1\] Hence, option [b] is correct.


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