Question

Find the range of f(x) =

${cos}^{2} x + {sin}^{4} x$

for every x is a real numbet​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \:f(x) = {cos}^{2}x + {sin}^{4}x$

can be rewritten as

$\rm \:f(x) = {cos}^{2}x + {sin}^{2}x \times {sin}^{2}x$

We know,

$\boxed{\sf{ \:\rm \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm \: f(x) = {cos}^{2}x + {sin}^{2}x(1 - {cos}^{2}x)$

$\rm \: f(x) = {cos}^{2}x + {sin}^{2}x - {sin}^{2}x {cos}^{2}x$

$\rm \: f(x) = 1 - {sin}^{2}x {cos}^{2}x$

$\rm \: f(x) = 1 - {(sinxcosx)}^{2}$

$\rm \: f(x) = 1 - {\bigg(\dfrac{2sinxcosx}{2} \bigg)}^{2}$

$\rm \: f(x) = 1 - {\bigg(\dfrac{sin2x}{2} \bigg)}^{2}$

$\rm \: f(x) = 1 - \frac{1}{4} {sin}^{2} 2x$

Now, we know that

$\rm \: - 1 \leqslant sin2x \leqslant 1$

$\rm \: 0 \leqslant {sin}^{2}2x \leqslant 1$

$\rm \: 0 \leqslant \frac{1}{4} {sin}^{2}2x \leqslant \frac{1}{4}$

$\rm \: 0 \geqslant - \: \frac{1}{4} {sin}^{2}2x \geqslant - \: \frac{1}{4}$

$\rm \: - \: \frac{1}{4} \leqslant - \frac{1}{4} \: sin^{2} 2x \: \leqslant 0$

Now, Adding 1 in each term, we get

$\rm \: 1- \: \frac{1}{4} \leqslant 1- \frac{1}{4} \: sin^{2} 2x \: \leqslant 0 + 1$

$\rm \: \: \frac{4 - 1}{4} \leqslant f(x) \: \leqslant 1$

$\rm \: \: \frac{3}{4} \leqslant f(x) \: \leqslant 1$

Hence,

$\rm\implies \:Range \: of \: f(x) \: \in \: \bigg[\dfrac{3}{4}, \: 1\bigg]$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosecx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered}$