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If $y = \left[ {\dfrac{{\sin \left( {x + a} \right)}}{{\sin \left( {x + b} \right)}}} \right],a \ne b$, then $y$ is
1. minimum at $x = 0$
2. maxima at $x = 0$
3. neither minima nor maxima at $x = 0$
4. None of the above

Answer
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Hint: In this question, given that the value of $y = \left[ {\dfrac{{\sin \left( {x + a} \right)}}{{\sin \left( {x + b} \right)}}} \right]$ where $a \ne b$ and we have to find the minimum and maximum value of $y$ in terms of $x$. First, differentiate the function with respect to $x$and put first derivative equal to zero and get the maxima or minima.

Formula used:
Quotient rule –
$\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) - f\left( x \right)\dfrac{d}{{dx}}g\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$

Complete step by step solution: 
Given that,
$y = \left[ {\dfrac{{\sin \left( {x + a} \right)}}{{\sin \left( {x + b} \right)}}} \right] - - - - - (1)$
Differentiate equation (1) with respect to $x$,
$\dfrac{{dy}}{{dx}} = \dfrac{{\left[ {\sin \left( {x + b} \right)\cos \left( {x + a} \right) - \sin \left( {x + a} \right)\cos \left( {x + b} \right)} \right]}}{{\left[ {{{\sin }^2}\left( {x + b} \right)} \right]}}$
$\dfrac{{dy}}{{dx}} = \dfrac{{\sin \left( {b - a} \right)}}{{{{\sin }^2}\left( {x + b} \right)}}$
For maximum and minimum values
$\dfrac{{dy}}{{dx}} = 0$
$\dfrac{{\sin \left( {b - a} \right)}}{{{{\sin }^2}\left( {x + b} \right)}} = 0$
$\sin \left( {b - a} \right) = 0$
Here, the value does not depend on $x$
It implies that $y$ is neither minima nor maxima at $x = 0$
Hence, Option (3) is the correct answer..

Note: The key concept involved in solving this problem is the good knowledge of maxima and minima. Students must remember that maxima and minima are the highest and lowest values of a function within a given set of ranges. The maximum value of the function under the entire range is known as the absolute maxima, and the minimum value is known as the absolute minima.