NIMCET 2027 Trigonometrical Function Previous Year Questions (PYQs)

Given,

$x-y\tan 35^\circ=\tan 25^\circ(y+x\tan 35^\circ)$

Let

$\tan 35^\circ=A$

and

$\tan 25^\circ=B$

Then,

$x-yA=B(y+xA)$

$x-yA=By+ABx$

$x-ABx=yA+By$

$x(1-AB)=y(A+B)$

So,

$\frac{x}{y}=\frac{A+B}{1-AB}$

Now,

$\frac{\tan 35^\circ+\tan 25^\circ}{1-\tan 35^\circ\tan 25^\circ}=\tan(35^\circ+25^\circ)$

$=\tan 60^\circ$

$=\sqrt{3}$

Therefore,

$\frac{x}{y}=\sqrt{3}$

So,

$x=\sqrt{3}y$

Since $\sqrt{3}>1$, according to the given option pattern, we get

$x>y$

For intersection of the two graphs,

$f(x)=g(x)$

So,

$2\cos\left(\frac{x}{2}\right)+3=4$

$2\cos\left(\frac{x}{2}\right)=1$

$\cos\left(\frac{x}{2}\right)=\frac{1}{2}$

Now,

$x\in [0,4\pi]$

So,

$\frac{x}{2}\in [0,2\pi]$

In the interval $[0,2\pi]$,

$\cos\theta=\frac{1}{2}$ has two solutions:

$\theta=\frac{\pi}{3},\frac{5\pi}{3}$

Here,

$\theta=\frac{x}{2}$

So,

$\frac{x}{2}=\frac{\pi}{3}$ or $\frac{x}{2}=\frac{5\pi}{3}$

Therefore,

$x=\frac{2\pi}{3}$ or $x=\frac{10\pi}{3}$

Hence, the number of points of intersection is $2$.