NIMCET 2023 — Mathematics PYQ

Step 1: Fundamental Condition for Same Side

For two given points $P(x_1,y_1)$ and $Q(x_2,y_2)$ to lie on the same side of a line $L(x,y)=ax+by+c=0$, the expressions obtained by substituting the coordinates of both points into the line equation must have the same sign.

Mathematically, this means:

L(x_1, y_1) \cdot L(x_2, y_2) > 0

Let the given line be:

$$L(x,y)=x+y-1=0$$

Step 2: Check the First Point $(3,2)$

Substitute the coordinates of the first point $P(3,2)$ into the line equation:

$$L(3,2)=3+2-1$$

$$L(3,2)=4$$

Since 4 > 0, the expression is positive. For the second point to be on the same side, its corresponding expression must also be strictly positive.

Step 3: Evaluate the Second Point $(\cos \theta ,\sin \theta )$

Substitute the coordinates of the second point $Q(\cos \theta ,\sin \theta )$ into the line equation:

$$L(\cos \theta ,\sin \theta )=\cos \theta +\sin \theta -1$$

Since L(3, 2) > 0, we must have:

\cos \theta + \sin \theta - 1 > 0

\cos \theta + \sin \theta > 1

Step 4: Solve the Trigonometric Inequality

To solve \cos \theta + \sin \theta > 1, we can divide the entire inequality by $\sqrt{1^2+1^2}=\sqrt{2}$:

\frac{1}{\sqrt{2}}\cos \theta + \frac{1}{\sqrt{2}}\sin \theta > \frac{1}{\sqrt{2}}

Using the trigonometric identity $\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$ and $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$, we can rewrite this as:

\sin \theta \cos\left(\frac{\pi}{4}\right) + \cos \theta \sin\left(\frac{\pi}{4}\right) > \frac{1}{\sqrt{2}}

Apply the identity $\sin (A+B)=\sin A\cos B+\cos A\sin B$:

\sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}}

Step 5: Determine the Interval for $\theta$

We know that \sin x > \frac{1}{\sqrt{2}} in the first and second quadrants when the angle lies between $\frac{\pi }{4}$ and $\frac{3\pi }{4}$. Therefore:

\frac{\pi}{4} < \theta + \frac{\pi}{4} < \frac{3\pi}{4}

Subtract $\frac{\pi }{4}$ from all parts of the inequality to isolate $\theta$:

\frac{\pi}{4} - \frac{\pi}{4} < \theta < \frac{3\pi}{4} - \frac{\pi}{4}

0 < \theta < \frac{2\pi}{4}

0 < \theta < \frac{\pi}{2}

Thus, the range of $\theta$ is $\left(0,\frac{\pi }{2}\right)$, which lies completely within the given domain interval of $(0,\pi )$.

Conclusion

The range of values of $\theta$ is $\left(0,\frac{\pi }{2}\right)$.

The correct option is B) $\left(0,\frac{\pi }{2}\right)$.