Question

If the function \[f(x)=\dfrac{K\sin x+2\cos x}{\sin x+\cos x}\] is strictly increasing for all values of x, then
(A). K<1
(B). K>1
(C). K<2
(D). K>2

Answer 1

HINT: - For finding whether a function is increasing or decreasing at a particular value of x, we follow the following steps and get to the answer
 1. First of all, we take the derivative of the function that is given to us.
 2. Then we put the value of x at which we have to find whether the function is increasing or decreasing.
3. If the value of the derivative of the function at the entered value of x comes out to be positive, then we can say that the function is increasing and if the value of the derivative of the function at the entered value of x comes out to be negative, then we can say that the function is decreasing.

Complete step-by-step solution -
The most important formulae that would be used in solving this question are as follows
1. Quotient rule of differentiation:-
 \[\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\cdot \dfrac{du}{dx}-u\cdot \dfrac{dv}{dx}}{{{v}^{2}}}\]
2. \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] .
As mentioned in the question, we have to find the value of K for which the function that is given in the question is increasing for all values of x.
Now, as mentioned in the question, we have to first take the derivative of the function as follows
\[\begin{align}
  & f(x)=\dfrac{K\sin x+2\cos x}{\sin x+\cos x} \\
 & {{\left( \dfrac{u}{v} \right)}^{\prime }}=\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\cdot \dfrac{du}{dx}-u\cdot \dfrac{dv}{dx}}{{{v}^{2}}} \\
 & {f}'(x)=\dfrac{\left( K\cos x-2\sin x \right)\left( \sin x+\cos x \right)-\left( \cos x-\sin x \right)\left( K\sin x+2\cos x \right)}{{{\left( \sin x+\cos x \right)}^{2}}} \\
 & {f}'(x)=\dfrac{K\sin x\cdot \cos x-2{{\sin }^{2}}x+K{{\cos }^{2}}x-2\sin x\cdot \cos x-K\sin x\cdot \cos x-2{{\cos }^{2}}x+K{{\sin }^{2}}x+2\sin x\cdot \cos x}{{{\left( \sin x+\cos x \right)}^{2}}} \\
 & {f}'(x)=\dfrac{-2{{\sin }^{2}}x+K{{\cos }^{2}}x-2{{\cos }^{2}}x+K{{\sin }^{2}}x}{{{\left( \sin x+\cos x \right)}^{2}}} \\
 & {f}'(x)=\dfrac{K\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)-2\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}{{{\left( \sin x+\cos x \right)}^{2}}} \\
 & {f}'(x)=\dfrac{K-2}{{{\left( \sin x+\cos x \right)}^{2}}} \\
\end{align}\]
(Using the identities that are mentioned in the hint that is \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\])
Now, in the above expression, for f(x) to be increasing for all values of x, then the derivative of the function should be positive.
As the denominator is a square of a number, so, it cannot be negative; hence, we can say that we have to make only the numerator positive and for that we can write as follows
\[\begin{align}
  & \dfrac{K-2}{{{\left( \sin x+\cos x \right)}^{2}}}>0 \\
 & K-2>0 \\
 & K>2 \\
\end{align}\]
Hence, we have to get the value of K as mentioned above.

NOTE: - The students can make an error if they don’t know what the principle is behind the derivative test for finding the increasing or decreasing of a function and the principle that is involved here is that on taking the derivative, we are actually finding the slope of the function at a particular value of x when that value of x is entered into the function’s derivative.