If \(f(x)=\begin{cases}\dfrac{\sin x+\sin 5x}{\cos x+\cos 5x} & \text{if } x\neq -\dfrac{\pi}{4}\\ K & \text{if } x=-\dfrac{\pi}{4}\end{cases}\), then find the value of K if f(x) is continuous at \(x=-\dfrac{\pi}{4}\)
1
Using sum-to-product formulas, \(\sin x+\sin 5x=2\sin 3x\cos 2x\) and \(\cos x+\cos 5x=2\cos 3x\cos 2x\).
For \(x\neq -\pi/4\), dividing gives \(f(x)=\dfrac{2\sin 3x\cos 2x}{2\cos 3x\cos 2x}=\tan 3x\) (the factor \(\cos 2x\) cancels away from the point where it vanishes).
For continuity at \(x=-\pi/4\), K must equal \(\displaystyle\lim_{x\to -\pi/4}\tan 3x=\tan\left(-\dfrac{3\pi}{4}\right)\).
Since \(\tan\) has period \(\pi\), \(\tan\left(-\dfrac{3\pi}{4}\right)=\tan\left(-\dfrac{3\pi}{4}+\pi\right)=\tan\dfrac{\pi}{4}=1\).
Hence \(K=1\), which is confirmed directly: at \(x=-\pi/4\) the numerator and denominator of the original expression both vanish, and L'Hopital's rule on \(\dfrac{\cos x+5\cos5x}{-\sin x-5\sin5x}\) also gives the value 1.
Important Questions from Continuity of a function
A function is defined as follows: \[ f(x) = \begin{cases} -\dfrac{x}{\sqrt{x^2}}, & x \ne 0, \\[6pt] 0, & x = 0. \end{cases} \] Which one of the following is correct in respect of the above function?
f(x) = x + |x| is continuous for
Let f : A → R, where A = R\(0) is such that \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + \left| {\rm{x}} \right|}}{{\rm{x}}}\) . On which one of the following sets is f(x) continuous?
Consider the following statements for f(x) = e -|x| ;
1. The function is continuous at x = 0.
2. The function is differentiable at x = 0.
Which of the above statements is / are correct?
If the function \(\rm f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {a + bx,\;\;}&{x < 1}\\ {5,}&{x = 1}\\ {b - ax,}&{x > 1} \end{array}} \right.\) is continuous, then what is the value of (a + b)?
