Question

If the function $$f(x) = \begin{cases} 1 + \sin\frac{\pi}{2}, & -\infty < x \le 1 \\ ax + b, & 1 < x < 3 \\ 6 \tan\frac{x\pi}{12}, & 3 \le x < 6 \end{cases}$$ is continuous in $(-\infty, 6)$, then the values of $a$ and $b$ are respectively.

• A. 1, 1
• B. 2, 1
• C. 0, 2
• D. 2, 0

Hint:

Always ensure your trigonometric values are rock solid. Forgetting that $\sin(\frac{\pi}{2}) = 1$ or $\tan(\frac{\pi}{4}) = 1$ will derail an otherwise simple algebra problem!

Answer 1

The Correct Option is D

Step 1: What continuity needs.
For the pieces of this function to join smoothly, the left value and right value must match at each joining point. The joins are at $x=1$ and $x=3$.
Step 2: Match at x=1 (left side).
Just before $x=1$ the rule gives $1+\sin(\pi/2)=1+1=2$.
Step 3: Match at x=1 (right side).
Just after $x=1$ the rule is $ax+b$, which gives $a+b$. Setting the two equal: $a+b=2$. Call this Equation 1.
Step 4: Match at x=3 (left side).
Just before $x=3$ the rule $ax+b$ gives $3a+b$.
Step 5: Match at x=3 (right side).
Just after $x=3$ the rule gives $6\tan(3\pi/12)=6\tan(\pi/4)=6\cdot 1=6$. Setting equal: $3a+b=6$. Call this Equation 2.
Step 6: Solve the two equations.
Subtract Equation 1 from Equation 2: $2a=4$, so $a=2$. Then $2+b=2$ gives $b=0$. So $a=2,\ b=0$, which is option (D).
\[ \boxed{\,a=2,\ b=0\,} \]