Question

If \[A={{\sin }^{8}}\theta +{{\cos }^{14}}\theta \] then for all values of \[\theta \] the range of \[A\] is
(a) \[A > 1\]
(b) \[A\ge 1\]
(c) \[A< 1\]
(d) \[A\le 1\]

Answer 1

Hint: We solve this problem by using the given expression in the inequality of known identity.
We have the standard identity of trigonometric ratios as
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
So, we try to find the range of given trigonometric ratios in terms of standard identity to get the required range.
We use the condition that \[0\le \sin \theta \le 1,\forall \theta \] and \[0\le \cos \theta \le 1,\forall \theta \]to find the range of a given expression.

Complete step by step answer:
We are given that the equation of trigonometric ratios as
\[A={{\sin }^{8}}\theta +{{\cos }^{14}}\theta \]
We know that the range of sine ratio as
\[0\le \sin \theta \le 1,\forall \theta \]
Here, we can see that the range of sine function is greater than 0 and less than 1.
So, we can say that if the power of sine ratio is high then it will be the lowest.
We know that there will be only rational numbers between 0 and 1 so that if we increase the power of sine function more and more then it approaches 0 more and more.
By suing the above statements we can conclude that
\[\Rightarrow {{\sin }^{8}}\theta < {{\sin }^{2}}\theta ,\forall \theta ........equation(i)\]
Similarly we know that the range of cosine ratio as
\[0\le \cos \theta \le 1,\forall \theta \]
We know that there will be only rational numbers between 0 and 1 so that if we increase the power of cosine function more and more then it approaches 0 more and more.
By suing the above statements we can conclude that
\[\Rightarrow {{\cos }^{14}}\theta < {{\cos }^{2}}\theta ,\forall \theta ........equation(ii)\]
Now, by adding the equation (i) and equation (ii) then we get
\[\Rightarrow {{\sin }^{8}}\theta +{{\cos }^{14}}\theta < {{\sin }^{2}}\theta +{{\cos }^{2}}\theta ,\forall \theta \]
We know that the standard identity of trigonometric ratios as
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
\[\Rightarrow {{\sin }^{8}}\theta +{{\cos }^{14}}\theta < 1,\forall \theta \]
Now, by substituting the value that is \[A={{\sin }^{8}}\theta +{{\cos }^{14}}\theta \] in above equation we get
\[\Rightarrow A< 1,\forall \theta \]
Therefore we can conclude that the for all values of \[\theta \] we have \[A< 1\]
So, option (c) is the correct answer.

Note:
Students may make mistakes in taking the powers of the domain.
If we have the range of number as
\[0< x< 1\]
We know that there will be only a rational number between 0 and 1 so that if we increase the power of the number \[x\] more and more then it approaches 0 more and more.
By using the above statement we get
\[\Rightarrow {{x}^{n}}< {{x}^{p}},\forall n< p\]
This is possible only if the range is \[0< x< 1\] but if the range is \[0< x< k,k > 1\] the above statement will not hold because there will be integers between them.
But students may use this property in all cases where it leads to the wrong answers.