The minimum value of $3\cos\text{x}+4\sin\text{x}+8$ is:
- A5
- B9
- C7
- D3
Solution:
The given expression is $3\cos\text{x}+4\sin\text{x}+8$
Let $\text{y}=3\cos\text{x}+4\sin\text{x}+8$
$\Rightarrow\text{y}-8=3\cos\text{x}+4\sin\text{x}$
Minimum value of $\text{y}-8=-\sqrt{(3)^2+(4)^2}$
$\Rightarrow\text{y}-8=-\sqrt{9+16}=-5$
$\Rightarrow\text{y}=8-5=3$
so, the minimum value of the given expression is 3.
Hence, the correct option is (d).
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

If $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$ for some constant, a, then f'(a) is equal to:
Equation of the hyperbola with eccentricty $\frac{3}{2}$ and foci at $(\pm2,0)$ is:
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$
$\frac{\text{x}^2}{9}-\frac{\text{y}^2}{9}=\frac{4}{9}$
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
none of these.