Question

The number of solutions of the equation $\tan (\frac{x}{2}+\frac{\pi }{4})=1$ in the interval $x\in [-2\pi ,2\pi ]=$

• A. 5.00
• B. 05
• C. 05.0
• D. 5.0
• E. 5

Answer 1

Answer: (A), (B), (C), (D), (E)

Number of solutions $=$ Number of points of intersection of $y=\tan (\frac{x}{2}+\frac{\pi }{4})\&y=1$ in the interval $x\in [-2\pi ,2\pi ]$

Now, $y=\tan (\frac{x}{2}+\frac{\pi }{4})$ can be written as:
$y=\tan \left(\frac{1}{2}(x+\frac{\pi }{2})\right)$

Graph of $\tan x$
Now stretching this along the x-axis by $2$ times we get graph of $\tan \frac{x}{2}$ as:

Now, shifting this graph to the left by $\frac{\pi }{2}$ units we get the graph of $y=\tan \left(\frac{1}{2}(x+\frac{\pi }{2})\right)$
Drawing the $y=1$ we get the number of points of intersection in the interval $[-2\pi ,2\pi ]$ as $5$

So, number of solutions $=5$