
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
164.4k+ views
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.
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.
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