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# If $f(\theta ) = {\sin ^4}\theta + {\cos ^2}\theta$, then range of $f(\theta )$ isA) $\left[ {\dfrac{1}{2},1} \right]$B) $\left[ {\dfrac{1}{2},\dfrac{3}{4}} \right]$C) $\left[ {\dfrac{3}{4},1} \right]$D) None of these

Last updated date: 07th Sep 2024
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Hint: This problem comes under differential calculus of maximum and minimum. Here we are asked to find the range of the given function. The range is maximum and minimum value of the function. To find we need to solve the function by substituting the function with variable (t) thus differentiate the given function has to be solved with variable (t) and then find the maximum and minimum value of the function. Formulas are there to be used and complete step by step explanation.

Formula used: ${\cos ^2}\theta = 1 - {\sin ^2}\theta$
If ${\text{f(x) = }}{{\text{x}}^{\text{n}}}$ then $\dfrac{{{\text{df}}}}{{{\text{dt}}}}{\text{ = n}}{{\text{x}}^{{\text{n - 1}}}}$

Let the function be
$f(\theta ) = {\sin ^4}\theta + {\cos ^2}\theta$
Let us use the formula mentioned in formula used to rearrange the given function
That is, ${\cos ^2}\theta = 1 - {\sin ^2}\theta$
Then, the function will be
$\Rightarrow$$f(\theta ) = {\sin ^4}\theta + 1 - {\sin ^2}\theta$
Now Put ${\sin ^2}\theta = t$
Then the differentiate will be in the function of variable $t$,
Then the function will be,
$\Rightarrow$$f(t) = {t^2} - t + 1$
$\Rightarrow$$f\left( t \right)$ is quadratic equation with $D < 0$
Thus D means nature of the root, we can find the maximum and minimum value of with the nature of the roots.
Now we need to differentiate,
Differentiate ‘$f$’ with respect to ‘$t$’
$\Rightarrow$$f(t) = {t^2} - t + 1$
Use the differentiate formula mention on formula used, we get
$\Rightarrow$$\dfrac{{df}}{{dt}} = 2t - 1 + 0$
Solving we get,
$\Rightarrow$$\dfrac{{df}}{{dt}} = 2t - 1$
Now, the minimum value of the function
If $D < 0$, then the minimum value of $f(t)$, when $\dfrac{{df(t)}}{{dt}} = 0$
Then $\dfrac{{df(t)}}{{dt}} = 0 - - - - - - - - - - - (1)$
$\Rightarrow$$\dfrac{{df}}{{dt}} = 2t - 1 - - - - - - - - - - - - - - (2)$
Now comparing equation 1 & 2, we get
$\Rightarrow$$2t - 1 = 0$
Solving to $t$ we get,
$\Rightarrow$$2t = 1$
$\Rightarrow$$t = \dfrac{1}{2}$
Now the minimum value of $f(t)$ is substitute the value of $t$ in $f(t)$
$\Rightarrow$$f{(t)_{\min }} = {\left( {\dfrac{1}{2}} \right)^2} - \dfrac{1}{2} + 1$
Squaring the terms we get,
$\Rightarrow$$f{(t)_{\min }} = \dfrac{1}{4} - \dfrac{1}{2} + 1$
Simplifying the term,
$\Rightarrow$$f{(t)_{\min }} = \dfrac{1}{4} - \dfrac{2}{4} + \dfrac{4}{4}$
Add and subtract the terms we get,
$\Rightarrow$$f{(t)_{\min }} = \dfrac{3}{4}$
Now, the maximum value of the function
If $D < 0$, then the maximum value of $t = 1$, therefore the maximum value of $f(t)$
$\Rightarrow$$f{(t)_{\max }} = 1 - 1 + 1$
$\Rightarrow$$f{(t)_{\max }} = 1$
Thus we find the maximum and minimum value of the function, thus we find the range of the function
Range of $f\left( t \right) \in \left[ {\dfrac{3}{4},1} \right]$

$\therefore$ Thus the answer is option C) $\left[ {\dfrac{{\text{3}}}{{\text{4}}}{\text{,1}}} \right]$

Note: This kind of problem needs to know about differential calculus and maximum and minimum function which are able to solve this problem. The trigonometric identities and differential formula have a huge part to solve in it. We have to convert the function to an easy solvable method so that we can solve using identities. All this kind will be similar as it is but the difference only on derivatives and we should know to solve it.