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Continuity & Differentiability · Question 42 of 125

Question 42

MediumOriginal question

For which value of kk is f(x)={ksin⁡x,x≤π2cos⁡x+1,x>π2f(x) = \begin{cases} k\sin x, & x \le \frac{\pi}{2} \\ \cos x + 1, & x > \frac{\pi}{2} \end{cases} continuous at x=π2x = \frac{\pi}{2}?

Options for question 42

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. The pieces must meet at x=π2x = \frac{\pi}{2}: k=1k = 1.

  2. So k=1k = 1.

Worth noting

0 takes cos⁡π2+1\cos \frac{\pi}{2} + 1 as 0, forgetting the +1+1.

Concept
A constant that makes a function continuous
Typical time
90 s
Your time
—
The rule this testsChoosing a constant for continuity
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Ten questions from Continuity & Differentiability, at your own pace.

Practise this chapter →

Ten questions from Continuity & Differentiability, at your own pace.