Question

If $ {{\cos }^{4}}x+{{\sin }^{2}}x-p=0,p\in R $ has real solutions, then
(a) $ p\le 1 $
(b) $ \dfrac{3}{4}\le p\le 1 $
(c) $ p\ge \dfrac{3}{4} $
(d) None of these

Answer 1

Hint: In order to solve this problem we need to know the standard identities as $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ and $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ $ 2{{\cos }^{2}}x-1=\cos 2x $ . Also, the range of $ \cos x $ is from -1 to +1. And the range of $ {{\cos }^{2}}x $ is -1. We need to simplify the equation and simplify to the point where we can substitute the conditions of range.

Complete step-by-step answer:
We have been given the equation of $ {{\cos }^{4}}x+{{\sin }^{2}}x-p=0 $ .
Let’s write the equation in terms of p.
 $ {{\cos }^{4}}x+{{\sin }^{2}}x=p.....................(i) $
We know the identity that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $
We can write the identity in terms of $ {{\cos }^{2}}x $ .
 $ {{\sin }^{2}}x=1-{{\cos }^{2}}x $ .
Substituting in equation (i), we get,
 $ p=1-{{\cos }^{2}}x+{{\left( {{\cos }^{2}}x \right)}^{2}} $
We can write $ {{\cos }^{2}}x $ as $ 2\times \dfrac{1}{2}{{\cos }^{2}}x $ .
Substituting we get,
 $ p=1-2\times \dfrac{1}{2}\times {{\cos }^{2}}x+{{\left( {{\cos }^{2}}x \right)}^{2}} $
We can also split 1 as the sum of $ \dfrac{3}{4}+{{\left( \dfrac{1}{2} \right)}^{2}} $ .
Substituting that we get,
 $ p=\dfrac{3}{4}+{{\left( \dfrac{1}{2} \right)}^{2}}-2\times \dfrac{1}{2}\times {{\cos }^{2}}x+{{\left( {{\cos }^{2}}x \right)}^{2}} $
As we can see that except first term all the other terms are in the form $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ .
Here $ a={{\cos }^{2}}x $ and $ b=\dfrac{1}{2} $ .
We can write the equation as $ p=\dfrac{3}{4}+{{\left( {{\cos }^{2}}x-\dfrac{1}{2} \right)}^{2}} $
Simplifying inside the bracket we get,
\[p=\dfrac{3}{4}+{{\left( \dfrac{2{{\cos }^{2}}x-1}{2} \right)}^{2}}\]
We know the identity that, $ 2{{\cos }^{2}}x-1=\cos 2x $
Substituting we get,
\[p=\dfrac{3}{4}+{{\left( \dfrac{\cos 2x}{2} \right)}^{2}}\]
Solving it further we get,
 $ p=\dfrac{3}{4}+\dfrac{{{\cos }^{2}}2x}{4} $
The range of $ \cos x $ is from -1 to +1.
By squatting it the range of $ {{\cos }^{2}}x $ becomes from 0 to 1.
Therefore, the maximum value that p can take is,
 $ {{p}_{\max }}=\dfrac{3}{4}+\dfrac{1}{4}=1 $ .
The minimum value that p will take is,
 $ {{p}_{\min }}=\dfrac{3}{4}+\dfrac{0}{4}=\dfrac{3}{4} $
Therefore, we can see that the range of p lies from $ \dfrac{3}{4} $ to 1.
So, the correct answer is “Option B”.

Note: The range of $ \cos x $ and $ \cos 2x $ is the same because the result will be the same and the values of x will be changed but the range will not. Also, squaring any term cancels all the terms because the square of any term cannot take negative values. As the terms cannot be negative, we can eliminate the option that gives a negative range. Hence, option (a) is eliminated.