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Question

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$

\left( a \right){\text{ - 5}} \\

\left( b \right){\text{ 5}} \\

\left( c \right){\text{ 25}} \\

\left( d \right){\text{ None of these}} \\

$

Answer

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Hint-Use the concept of maxima and minima.Highest and lowest point is generally the maxima and minima of a graph.

Here we have to find the maximum value of ${\text{3cos}}\theta {\text{ + 4sin}}\theta $

So let ${\text{f}}\left( \theta \right) = {\text{3cos}}\theta {\text{ + 4sin}}\theta $

Now our first derivative ${{\text{f}}^1}\left( \theta \right) = - 3\sin \theta + 4\cos \theta $

Now double differentiating it we get ${{\text{f}}^{11}}\left( \theta \right) = - 3\cos \theta - 4\sin \theta $

In order to find max and min value we have to make ${{\text{f}}^1}\left( \theta \right) = 0$

Hence ${\text{ - 3sin}}\theta {\text{ + 4cos}}\theta {\text{ = 0}}$

On solving above we get ${\text{tan}}\theta {\text{ = }}\dfrac{4}{3}$

As we know that ${\text{tan}}\theta {\text{ = }}\dfrac{P}{H}$

Hence our ${\text{sin}}\theta {\text{ = }}\dfrac{4}{5}{\text{ and cos}}\theta {\text{ = }}\dfrac{3}{5}$

Now for this value of ${\text{sin}}\theta {\text{ and cos}}\theta $, the value of double derivative of $f(\theta ) $should be less than zero as we have to find maximum value of the expression.

${{\text{f}}^{11}}\left( \theta \right) < 0$

$ - 3\cos \theta - 4\sin \theta < 0$

Putting the values

$ - 3\left( {\dfrac{3}{5}} \right) - 4\left( {\dfrac{4}{5}} \right) < 0$

So max value of

${\text{3cos}}\theta {\text{ + 4sin}}\theta = 3 \times \left( {\dfrac{3}{5}} \right) + 4 \times \left( {\dfrac{4}{5}} \right)$

$ = \dfrac{{25}}{5}$Which is equal to 5

Hence option (b) is the right answer.

Note- Whenever we face such a problem the key concept that we need to use is that we always put the first derivative equal to 0 to obtain the values. Now double differentiate and cross verify that whether the value obtained corresponds to maximum or minimum for the function. This helps in reaching the right answer.

Here we have to find the maximum value of ${\text{3cos}}\theta {\text{ + 4sin}}\theta $

So let ${\text{f}}\left( \theta \right) = {\text{3cos}}\theta {\text{ + 4sin}}\theta $

Now our first derivative ${{\text{f}}^1}\left( \theta \right) = - 3\sin \theta + 4\cos \theta $

Now double differentiating it we get ${{\text{f}}^{11}}\left( \theta \right) = - 3\cos \theta - 4\sin \theta $

In order to find max and min value we have to make ${{\text{f}}^1}\left( \theta \right) = 0$

Hence ${\text{ - 3sin}}\theta {\text{ + 4cos}}\theta {\text{ = 0}}$

On solving above we get ${\text{tan}}\theta {\text{ = }}\dfrac{4}{3}$

As we know that ${\text{tan}}\theta {\text{ = }}\dfrac{P}{H}$

Hence our ${\text{sin}}\theta {\text{ = }}\dfrac{4}{5}{\text{ and cos}}\theta {\text{ = }}\dfrac{3}{5}$

Now for this value of ${\text{sin}}\theta {\text{ and cos}}\theta $, the value of double derivative of $f(\theta ) $should be less than zero as we have to find maximum value of the expression.

${{\text{f}}^{11}}\left( \theta \right) < 0$

$ - 3\cos \theta - 4\sin \theta < 0$

Putting the values

$ - 3\left( {\dfrac{3}{5}} \right) - 4\left( {\dfrac{4}{5}} \right) < 0$

So max value of

${\text{3cos}}\theta {\text{ + 4sin}}\theta = 3 \times \left( {\dfrac{3}{5}} \right) + 4 \times \left( {\dfrac{4}{5}} \right)$

$ = \dfrac{{25}}{5}$Which is equal to 5

Hence option (b) is the right answer.

Note- Whenever we face such a problem the key concept that we need to use is that we always put the first derivative equal to 0 to obtain the values. Now double differentiate and cross verify that whether the value obtained corresponds to maximum or minimum for the function. This helps in reaching the right answer.