Question

. solve this fast
please​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=sin^2(\theta)+cos^4(\theta)}$

$\sf{\implies\,y=\dfrac{1-cos(2\theta)}{2}+\bigg(\dfrac{1+cos(2\theta)}{2}\bigg)^2}$

$\sf{\implies\,y=\dfrac{1-cos(2\theta)}{2}+\dfrac{1+cos^2(2\theta)+2\,cos(2\theta)}{4}}$

$\sf{\implies\,y=\dfrac{2-2cos(2\theta)+1+cos^2(2\theta)+2\,cos(2\theta)}{4}}$

$\sf{\implies\,y=\dfrac{2+1+cos^2(2\theta)}{4}}$

$\sf{\implies\,y=\dfrac{3+cos^2(2\theta)}{4}}$

Now,

We know,

$\sf{0\le\,cos^2(2\theta)\le1}$

$\sf{\implies0+3\le\,3+cos^2(2\theta)\le1+3}$

$\sf{\implies3\le\,3+cos^2(2\theta)\le4}$

$\sf{\implies\dfrac{3}{4}\le\,\dfrac{3+cos^2(2\theta)}{4}\le\dfrac{4}{4}}$

$\sf{\implies\dfrac{3}{4}\le\,\dfrac{3+cos^2(2\theta)}{4}\le1}$

$\sf{\implies\dfrac{3}{4}\le\,y\le1}$

$\rm{\blue{\bold{Hence,\,\,\,range\,\,\,of\,\,\,the\,\,\,given\,\,\,function\,\,\,\in\,\bigg[\,\dfrac{3}{4}\,,\,1\,\bigg]}}}$