Question

The following question contains statements given in two columns that have to be matched. The statements in column – I are labeled as A, B, C, and D while the statements in column – II one labeled as p, q, r, and s. Any given statement in column – I can have correct matching with one statement in the column – II.

Column – I	Column – II
f (x)	Range
(A) \[\dfrac{{{\cos }^{2}}x+\cos x+2}{{{\cos }^{2}}x+\cos x+1}\]	(p) \[\left( 0,\dfrac{7}{3} \right]\]
(B) \[\left| \dfrac{\left( \sqrt{\cos x}-\sqrt{\sin x} \right)\left( \sqrt{\cos x}+\sqrt{\sin x} \right)}{3\left( \cos x+\sin x \right)} \right|\]	(q) \[\left[ \dfrac{4}{3},\dfrac{7}{3} \right]\]
(C) \[\dfrac{7}{3\left( {{x}^{6}}+3{{x}^{4}}+3{{x}^{2}}+1 \right)}\]	(r) \[\left[ 0,\dfrac{1}{3} \right]\]
(D) \[{{\log }_{8}}\left( {{x}^{2}}+2x+2 \right)\]	(s) \[\left[ 0,\infty \right)\]

Answer 1

Hint: Consider each expression one – by – one and convert them in such a form in which we will not be required to use the derivative method to find the range. Use the proper algebraic identities to simplify the expression. Take care of the domains of f (x) such that the value of f (x) does not get undefined.

Complete step by step answer:
Here, we have been provided with four expressions and we have to match them with their ranges. So, let us check them one – by – one.
(A) Here, we have the expression: - \[\dfrac{{{\cos }^{2}}x+\cos x+2}{{{\cos }^{2}}x+\cos x+1}\]. Let us assume its value as E.
\[\Rightarrow E=\dfrac{{{\cos }^{2}}x+\cos x+2}{{{\cos }^{2}}x+\cos x+1}\]
The above expression can be written as: -
\[\Rightarrow E=\dfrac{\left( {{\cos }^{2}}x+\cos x+1 \right)+1}{{{\cos }^{2}}x+\cos x+1}\]
Breaking the terms, we get,
\[\Rightarrow E=1+\dfrac{1}{{{\cos }^{2}}x+\cos x+1}\]
Using the completing square method we can write,
\[\begin{align}
  & \Rightarrow E=1+\dfrac{1}{{{\cos }^{2}}x+2\times \cos x\times \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}+1-{{\left( \dfrac{1}{2} \right)}^{2}}} \\
 & \Rightarrow E=1+\dfrac{1}{{{\left( \cos x+\dfrac{1}{2} \right)}^{2}}+1-\dfrac{1}{4}} \\
 & \Rightarrow E=1+\dfrac{1}{{{\left( \cos x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}} \\
\end{align}\]
Now, the above expression will have a maximum when the denominator will be minimum. This minimum will be obtained when \[{{\left( \cos x+\dfrac{1}{2} \right)}^{2}}\] will be 0. This will happen at \[x=\dfrac{2\pi }{3}\], where \[\cos x=-\dfrac{1}{2}\]. So, we have,
\[\begin{align}
  & \Rightarrow {{E}_{\max }}=1+\dfrac{1}{0+\dfrac{3}{4}} \\
 & \Rightarrow {{E}_{\max }}=1+\dfrac{4}{3} \\
 & \Rightarrow {{E}_{\max }}=\dfrac{7}{3} \\
\end{align}\]
Now, we know that for the minimum value of the expression we must have the denominator at its maximum. Since, we know that the maximum value of \[\cos x\] is equal to 1. So, we have,
\[\begin{align}
  & \Rightarrow {{E}_{\min }}=1+\dfrac{1}{{{\left( 1+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}} \\
 & \Rightarrow {{E}_{\min }}=1+\dfrac{1}{{{\left( \dfrac{3}{2} \right)}^{2}}+\dfrac{3}{4}} \\
 & \Rightarrow {{E}_{\min }}=1+\dfrac{1}{\dfrac{9}{4}+\dfrac{3}{4}} \\
\end{align}\]
\[\Rightarrow {{E}_{\min }}=1+\dfrac{4}{12}\]
\[\begin{align}
  & \Rightarrow {{E}_{\min }}=1+\dfrac{1}{3} \\
 & \Rightarrow {{E}_{\min }}=\dfrac{4}{3} \\
\end{align}\]
So, the range of E is given as: - \[E\in \left[ \dfrac{4}{3},\dfrac{7}{3} \right]\], i.e. option (q).
(B) Here, we have the expression: - \[\left| \dfrac{\left( \sqrt{\cos x}-\sqrt{\sin x} \right)\left( \sqrt{\cos x}+\sqrt{\sin x} \right)}{3\left( \cos x+\sin x \right)} \right|\].
Let us assume its value as F.
\[\Rightarrow F=\left| \dfrac{\left( \sqrt{\cos x}-\sqrt{\sin x} \right)\left( \sqrt{\cos x}+\sqrt{\sin x} \right)}{3\left( \cos x+\sin x \right)} \right|\]
Applying the algebraic identity \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\], in the numerator, we get,
\[\begin{align}
  & \Rightarrow F=\left| \dfrac{\sqrt{{{\cos }^{2}}x}-\sqrt{{{\sin }^{2}}x}}{3\left( \cos x+\sin x \right)} \right| \\
 & \Rightarrow F=\left| \dfrac{\cos x-\sin x}{3\left( \cos x+\sin x \right)} \right| \\
\end{align}\]
Dividing numerator and denominator with \[\cos x\], we get,
\[\begin{align}
  & \Rightarrow F=\left| \dfrac{1-\dfrac{\sin x}{\cos x}}{3\left( 1+\dfrac{\sin x}{\cos x} \right)} \right| \\
 & \Rightarrow F=\left| \dfrac{1-\tan x}{3\left( 1+\tan x \right)} \right| \\
\end{align}\]
We know that, \[\left( \dfrac{1-\tan x}{1+\tan x} \right)=\tan \left( \dfrac{\pi }{4}-x \right)\], so we get,
\[\Rightarrow F=\left| \dfrac{1}{3}\tan \left( \dfrac{\pi }{4}-x \right) \right|\] - (i)
Hence, \[x\in \left[ 0,\dfrac{\pi }{2} \right]\]. So, let us come to the expression (i). Here, the minimum value of F will be obtained if \[\tan \left( \dfrac{\pi }{4}-x \right)\] will be minimum. We know that the minimum value of tangent of any angle is 0 and here it will occur at \[x=\dfrac{\pi }{4}\]. So, we have,
\[\begin{align}
  & \Rightarrow {{F}_{\min }}=\left| \dfrac{1}{3}\tan 0 \right| \\
 & \Rightarrow {{F}_{\min }}=0 \\
\end{align}\]
Now, the maximum of tangent of any angle is \[\infty \] which occurs at an angle \[\dfrac{\pi }{2}\] but here in F we cannot obtain \[\dfrac{\pi }{2}\] because \[x\in \left[ 0,\dfrac{\pi }{2} \right]\]. We know that \[\tan \dfrac{\pi }{4}=1\] and if we will subtract any value of angle, lying in first quadrant, from \[\dfrac{\pi }{4}\] then the value of tangent will decrease. So, the maxima will occur at x = 0. Therefore, we have,
\[\begin{align}
  & \Rightarrow {{F}_{\max }}=\left| \dfrac{1}{3}\tan \dfrac{\pi }{4} \right| \\
 & \Rightarrow {{F}_{\max }}=\left| \dfrac{1}{3}\times 1 \right| \\
 & \Rightarrow {{F}_{\max }}=\dfrac{1}{3} \\
\end{align}\]
So, the range of F is given as: - \[F\in \left[ 0,\dfrac{1}{3} \right]\], i.e. option (r).
(C) Here, we have the expression: - \[\dfrac{7}{3\left( {{x}^{6}}+3{{x}^{4}}+3{{x}^{2}}+1 \right)}\]. Let us assume its value as G.
\[\Rightarrow G=\dfrac{7}{3\left( {{x}^{6}}+3{{x}^{4}}+3{{x}^{2}}+1 \right)}\]
Let us focus on the expression in the denominator. So, we have, \[{{x}^{6}}+3{{x}^{4}}+3{{x}^{2}}+1\]. This can be written as: -
\[\Rightarrow {{x}^{6}}+3{{x}^{4}}+3{{x}^{2}}+1={{\left( {{x}^{2}} \right)}^{3}}+{{1}^{3}}+3\times {{\left( {{x}^{2}} \right)}^{2}}\times 1+3\times {{x}^{2}}\times {{1}^{2}}\]
Using the algebraic identity: - \[{{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}={{\left( a+b \right)}^{3}}\], we get,
\[\Rightarrow {{x}^{6}}+3{{x}^{4}}+3{{x}^{2}}+1={{\left( {{x}^{2}}+1 \right)}^{3}}\]
Therefore, the expression becomes,
\[\Rightarrow G=\dfrac{7}{3{{\left( {{x}^{2}}+1 \right)}^{3}}}\]
Now, for G to have maximum, the denominator must be minimum. We can see that \[\left( {{x}^{2}}+1 \right)\] will be minimum for x = 0 because \[{{x}^{2}}\] cannot be negative. So, we have,
\[\begin{align}
  & \Rightarrow {{G}_{\max }}=\dfrac{7}{3{{\left( 0+1 \right)}^{3}}} \\
 & \Rightarrow {{G}_{\max }}=\dfrac{7}{3} \\
\end{align}\]
Further, in the expression of G we can see that as we keep on increasing the value of denominator, the value of G will keep on decreasing. Here, we can keep on increasing x up to \[\infty \]. So, we have,
\[\begin{align}
  & \Rightarrow {{G}_{\min }}=\dfrac{7}{3{{\left( {{\infty }^{2}}+1 \right)}^{3}}} \\
 & \Rightarrow {{G}_{\min }}=\dfrac{7}{3\times \infty } \\
 & \Rightarrow {{G}_{\min }}=0 \\
\end{align}\]
So, the range of G is given as: - \[\left( 0,\dfrac{7}{3} \right]\], i.e. option (p).
(D) Here, we have the expression: - \[{{\log }_{8}}\left( {{x}^{2}}+2x+2 \right)\]. Let us assume its value as H.
\[\Rightarrow H={{\log }_{8}}\left( {{x}^{2}}+2x+2 \right)\]
The argument of the log in the above expression can be written as: -
\[\begin{align}
  & \Rightarrow H={{\log }_{8}}\left( \left( {{x}^{2}}+2x+1 \right)+1 \right) \\
 & \Rightarrow H={{\log }_{8}}\left( {{\left( x+1 \right)}^{2}}+1 \right) \\
\end{align}\]
Now, the value of H will be minimum when (x + 1) will be 0. So, we have,
\[\begin{align}
  & \Rightarrow {{H}_{\min }}={{\log }_{8}}\left( {{0}^{2}}+1 \right) \\
 & \Rightarrow {{H}_{\min }}={{\log }_{8}}\left( 1 \right) \\
 & \Rightarrow {{H}_{\min }}=0 \\
\end{align}\]
Further we can see that if we will keep on increasing the value of x then the value of H will keep on increasing. Here, we can keep on increasing x up to \[\infty \]. So, we have,
\[\begin{align}
  & \Rightarrow {{H}_{\max }}={{\log }_{8}}\left( {{\infty }^{2}}+1 \right) \\
 & \Rightarrow {{H}_{\max }}={{\log }_{8}}\infty \\
\end{align}\]
\[\Rightarrow {{H}_{\max }}=\infty \]
So, the range of H is given as: - \[H\in \left[ 0,\infty \right)\], i.e. option (s).
Hence, the solution can be given as: -
A \[\to \] (q)
B \[\to \] (r)
C \[\to \] (p)
D \[\to \] (s)

Note:
One may note that here you should never try to use differentiation to get the maximum and minimum values of the expressions. You may use it in the expression given in option (D) but in other options, the differentiation method will make the question more difficult. In options (C) and (D) you can see that we have kept increasing the values of x up to \[\infty \], you can also keep decreasing it up to (-\[\infty \]) because it will not alter the answer as at the end when we will square the expression like x and (x + 1) in corresponding options, (-\[\infty \]) will become \[\infty \]. So, overall it will not change the minima or maxima that we have found.