Question

Find the value of a in order that \[f(x) = \sin x - \cos x - ax + b\] decreases for all real value of x.
A.\[a \ge \sqrt 2 \]
B.\[a < \sqrt 2 \]
C.\[a \ge 1\]
D.\[a < 1\]

Answer 1

Hint: First differentiate the given equation with respect to x. Then obtain the inequality for \[f'(x) \le 0\]. Now find for what value of x, \[\sin x + \cos x\] is maximum, then from the inequality \[a \ge \sin x + \cos x\] obtain the range of a by substituting \[\sqrt 2 \] for \[\sin x + \cos x\].

Formula used:
1. \[\dfrac{d}{{dx}}(ax) = a\dfrac{d}{{dx}}(x)\]
2. \[\dfrac{d}{{dx}}(x) = 1\]
3. \[\dfrac{d}{{dx}}(C) = 0,C = cons\tan t\]
4. \[\dfrac{d}{{dx}}(\sin x) = \cos x\]
5. \[\dfrac{d}{{dx}}(\cos x) = - \sin x\]

Complete step by step solution:
Differentiate the given equation with respect to x.
\[\begin{array}{I}f'(x) = \dfrac{d}{{dx}}\left[ {\sin x - \cos x - ax + b} \right]\\ = \cos x - ( - \sin x) - a\\ = \cos x + \sin x - a\end{array}\]
Since it is decreasing function so the derivative should be less than equal to zero.
Solve for \[f'(x) \le 0\].
\[\begin{array}{l}\cos x + \sin x - a \le 0\\\sin x + \cos x \le a\\a \ge \sin x + \cos x\end{array}\]
The maximum value of \[\sin x + \cos x{\rm{ }}\] is \[\sqrt 2 \] for \[x = \dfrac{\pi }{4}\].
Hence, we can conclude that \[a \ge \sqrt 2 \].

The correct option is A.

Note: Sometimes students unable to find the maximum value of \[\sin x + \cos x\], so to find the value they can do the rough work as just let \[f(x) = \sin x + \cos x\], then differentiate f(x) with respect to x, equate \[\cos x - \sin x\] to zero and solve to obtain x as \[\dfrac{\pi }{4}\] .