If \[f\left( x \right) = kx - sinx\] is monotonically increasing. Then find the value of \[k\].
A. \[k > 1\]
B. \[k > - 1\]
C. \[k < 1\]
D. \[k < - 1\]

Hint: In the given question, one monotonically increasing function is given. Differentiate the given function with respect to \[x\]. The derivative of a monotonically increasing function is always positive and we can find the value of \[k\].

Formula Used: Derivative formula:
\(\dfrac{d}{{dx}}\left( {sinx} \right) = cosx\)
\[\dfrac{d}{{dx}}\left( {kx} \right) = k\dfrac{d}{{dx}}\left( x \right)\]=k
The range of the function \[cosx\] is \[\left[ { - 1,1} \right]\].

Complete step by step solution:
The given monotonically increasing function is \[f\left( x \right) = kx - sinx\].
Let’s differentiate the given function with respect to \[x\].
\(f'\left( x \right) = \dfrac{d}{{dx}}\left( {kx - sinx} \right)\)
Apply the formulas \[\dfrac{d}{{dx}}\left( {sinx} \right) = cosx\] and \[\dfrac{d}{{dx}}\left( {kx} \right) = k\dfrac{d}{{dx}}\left( x \right)\]=k
\[f'\left( x \right) = k - cosx\]
Since the derivative of a monotonically increasing function is always positive.
Then,
\[f'\left( x \right) > 0\]
\[ \Rightarrow \]\[k - cosx > 0\]
\[ \Rightarrow \]\[k > cosx\]
\[ \Rightarrow \]\[k > \left[ { - 1,1} \right]\] [since, the range of \[cosx\] is \[\left[ { - 1,1} \right]\] ]
\[ \Rightarrow \]\[k > 1\]

Option ‘A’ is correct

Note:Students are often confused with the formula \[\dfrac{d}{{dx}}\left( {sinx} \right) = cosx\] and \[\dfrac{d}{{dx}}\left( {sinx} \right) = - cosx\] . But the correct formula is \[\dfrac{d}{{dx}}\left( {sinx} \right) = cosx\].