Question

What is the value of k when $f\left( x \right) = \left[ {\dfrac{{k\sin x + 2cosx}}{{\sin x + \cos x}}} \right]$ when the function $x$when $f(x)$is increasing?
A. k < 1
B. k > 1
C. k < 2
D. k > 2

Answer 1

Hint: Given function is a rational function. First, find the first ordered derivative of the function. Then calculate for which value of k, the value of the first ordered derivative greater than zero.

Formula Used:
$\dfrac{d}{{dx}}\left( {\dfrac{{u(x)}}{{v(x)}}} \right) = \dfrac{{v(x) \times \dfrac{d}{{dx}}u(x) - u(x) \times \dfrac{d}{{dx}}v(x)}}{{{{\left( {v(x)} \right)}^2}}}$
$({\cos ^2}x + {\sin ^2}x) = 1$

Complete step by step solution:
Given:
$f\left( x \right) = \left[ {\dfrac{{k{\rm{ sin}}x + 2{\rm{ cos }}x}}{{\sin x + \cos x}}} \right]$
$f'(x) > 0$,
We get,
$f'(x) = \dfrac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}$
Differentiating both sides,
$f'(x) = \dfrac{{(K\cos x - 2\sin x)(\sin x + \cos x) - (\cos x - \sin x)(K\sin x + 2\cos x)}}{{{{(\sin x + \cos x)}^2}}}$
$f'(x) = \dfrac{{K\sin x\cos x + K{{\cos }^2}x + K{{\cos }^2}x - 2{{\sin }^2}x - 2\sin x\cos x - K\sin x\cos x - 2{{\cos }^2}x + K{{\sin }^2}x + 2\sin x\cos x}}{{{{(\sin x + \cos x)}^2}}}$
$f'(x) = \dfrac{{(K\cos x - 2\sin x)(\sin x + \cos x) - (\cos x - \sin x)(K\sin x + 2\cos x)}}{{{{(\sin x + \cos x)}^2}}}$
Further solving the above equation by expanding and removing the brackets,
$f'(x) = \dfrac{{K\sin x\cos x + K{{\cos }^2}x + K{{\cos }^2}x - 2{{\sin }^2}x - 2\sin x\cos x - K\sin x\cos x - 2{{\cos }^2}x + K{{\sin }^2}x + 2\sin x\cos x}}{{{{(\sin x + \cos x)}^2}}}$
Cancel out all the necessary terms to get the equation below,
$f'(x) = \dfrac{{K{{\cos }^2}x + K{{\sin }^2}x - 2{{\cos }^2}x - 2{{\sin }^2}x}}{{{{(\sin x + \cos x)}^2}}}$
Taking all the common terms we get,
$f'(x) = \dfrac{{K({{\cos }^2}x + {{\sin }^2}x) - 2({{\sin }^2}x + {{\cos }^2}x)}}{{{{(\sin x + \cos x)}^2}}}$
Using formula $({\cos ^2}x + {\sin ^2}x) = 1$, we get
$f'(x) = \dfrac{{K - 2}}{{{{(\sin x + \cos x)}^2}}}$
Now, as $f'(x) > 0$
$f'(x) = \dfrac{{K - 2}}{{{{(\sin x + \cos x)}^2}}} > 0$
Sending the denominator to RHS,
$K - 2 > 0$
$K > 2$

Option ‘D’ is correct

Note: A function is increasing in the interval $\left(a,b\right)$ if $f’\left(x\right)>0]$ in the interval $\left(a,b\right)$. So that we need to find the first ordered derivative of the given function. Then find the condition where $f’\left(x\right)>0]$ to calculate the value of k.