Question

For $n\in I$, the line $x=n\pi +\dfrac{\pi }{2}$ does not intersect the graph of
        (a) $\cot \left( x+\pi \right)$
        (b) $\sin x$
        (c) $\cos \left( x-\pi \right)$
        (d) $\tan x$

Answer 1

Hint: Put the value of x from the given line in each function. If we get a value which is defined for a graph that means the line is intersecting that graph.

Complete step-by-step solution -
The equation of the given line is:
$x=n\pi +\dfrac{\pi }{2}........(1)$
If the line (1) intersects a graph, that graph and the line (1) both will have a common point. That means, if we put the value of x from the line in that graph we will get a value which is also defined for that graph.
Let us take the options one by one.

At first we have,
$\cot \left( x+\pi \right)$
We will put the value of x from (1).
$=\cot \left( n\pi +\dfrac{\pi }{2}+\pi \right)$
$=\cot \left( \left( n+1 \right)\pi +\dfrac{\pi }{2} \right)$
For any integer n, $\left( n+1 \right)\pi +\dfrac{\pi }{2}$ will always fall either in first coordinate or in third coordinate. We know that in the first coordinate and in the third coordinate cot is positive. Therefore,
$=\cot \dfrac{\pi }{2}$
$=0$
Therefore the line satisfies the graph.

The second option is:
$\sin x$
Now we will put the value of x from (1).
$=\sin \left( n\pi +\dfrac{\pi }{2} \right)$
For any integer n, $n\pi +\dfrac{\pi }{2}$ will always fall either in first coordinate or in third coordinate. We know that in first coordinate sine is positive and in third coordinate sine is negative. For odd integers sine will give negative value and of even integers sin will give positive value. Therefore,
$={{\left( -1 \right)}^{n}}\sin \dfrac{\pi }{2}$
$={{\left( -1 \right)}^{n}}$
Therefore the line satisfies the graph.

The third option is:
$\cos \left( x-\pi \right)$
Now we will put the value of x from (1).
$=\cos \left( n\pi +\dfrac{\pi }{2}-\pi \right)$
$=\cos \left( n\pi -\dfrac{\pi }{2} \right)$
For any integer n, $n\pi -\dfrac{\pi }{2}$ will always fall either in second coordinate or in fourth coordinate. We know that the second coordinate cosine is negative and the fourth coordinate cosine is positive. For odd integers cosine will give negative value and of even integers cosine will give positive value. Therefore,
$={{\left( -1 \right)}^{n}}\cos \dfrac{\pi }{2}$
$=0$
Therefore the line satisfies the graph.

The fourth option is:
$\tan x$
Now we will put the value of x from (1).
$=\tan \left( n\pi +\dfrac{\pi }{2} \right)$
For any integer n, $n\pi +\dfrac{\pi }{2}$ will always fall either in first coordinate or in third coordinate. We know that in first coordinate tan is positive and in third coordinate tan is positive. Therefore,
$=\tan \dfrac{\pi }{2}$
= undefined.
Therefore the line is not satisfying the graph.
Hence, the line $x=n\pi +\dfrac{\pi }{2}$, does not intersect the graph $\tan x$ .
Therefore, option (d) is correct.

Note: since, the given line is a vertical line and we know that the graph of sine and cosine are horizontal, a vertical line will always intersect the graph of sine and cosine. Therefore we can easily cancel out option (b) and (c). In this way we will be able to solve the problem quickly.