NIMCET 2021 — Mathematics PYQ

Analysis of the Equation 
The given equation is $3\cos \theta +4\sin \theta =k$. The range for $\theta$ is $0^{\circ }\le \theta \le 360^{\circ }$. It is also given that $|k|=5$, which implies $k=5$ or $k=-5$.
Transformation of the Equation 
1. The expression $a\cos \theta +b\sin \theta$ can be rewritten in the form $R\sin (\theta +\alpha )$ or $R\cos (\theta -\alpha )$, where $R=\sqrt{a^2+b^2}$. 
2. For the given equation, $a=3$ and $b=4$. 
3. The value of $R$ is calculated as $R=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. 
4. The equation can be rewritten as $5\left(\frac{3}{5}\cos \theta +\frac{4}{5}\sin \theta \right)=k$. 
5. Let $\cos \alpha =\frac{3}{5}$ and $\sin \alpha =\frac{4}{5}$ for some angle $\alpha$. 
6. Then, the equation becomes $5(\cos \alpha \cos \theta +\sin \alpha \sin \theta )=k$. 
7. Using the trigonometric identity $\cos (A-B)=\cos A\cos B+\sin A\sin B$, the equation is transformed to $5\cos (\theta -\alpha )=k$. 
Solving for k = 5 
1. When $k=5$, the equation becomes $5\cos (\theta -\alpha )=5$. 
2. This simplifies to $\cos (\theta -\alpha )=1$. 
3. The general solution for $\cos x=1$ is $x=2n\pi$, where $n$ is an integer. 
4. Therefore, $\theta -\alpha =2n\pi$. 
5. This implies $\theta =\alpha +2n\pi$. 
6. Since $0^{\circ }\le \theta \le 360^{\circ }$, and $\alpha$ is an acute angle (as \sin \alpha > 0 and \cos \alpha > 0), only one solution exists for $\theta$ in the given range, which is $\theta =\alpha$ (when $n=0$). 
Solving for k = -5 
1. When $k=-5$, the equation becomes $5\cos (\theta -\alpha )=-5$. 
2. This simplifies to $\cos (\theta -\alpha )=-1$. 
3. The general solution for $\cos x=-1$ is $x=(2n+1)\pi$, where $n$ is an integer. 
4. Therefore, $\theta -\alpha =(2n+1)\pi$. 
5. This implies $\theta =\alpha +(2n+1)\pi$. 
6. Since $0^{\circ }\le \theta \le 360^{\circ }$, only one solution exists for $\theta$ in the given range, which is $\theta =\alpha +\pi$ (when $n=0$). 
Conclusion on Distinct Solutions 
1. For $k=5$, one distinct solution is found in the range $0^{\circ }\le \theta \le 360^{\circ }$. 
2. For $k=-5$, one distinct solution is found in the range $0^{\circ }\le \theta \le 360^{\circ }$. 
3. These two solutions are distinct because $\alpha \ne \alpha +\pi$. 
4. Therefore, a total of $1+1=2$ distinct solutions exist. 
Final Answer 
The number of distinct solutions is $2$.