Question

Observe the following lists

List - I	List - II
A) minimum value of $a^2{\sin }^{2}x+b^2\cos e{c}^{2}x$	1) $2\sqrt{ab}$
B) minimum value of acotx + b tanx	2) $(a+b{)}^2$
C) minimum value of $a^2{\sec }^{2}x+b^2\cos e{c}^{2}x$	3)$-\sqrt{a^2+b^2}$
D) minimum value of $a\cos x+b\sin x$	4) ab
	5) 2ab

• A. A - 5, B - 1, C - 2, D - 3.
• B. A - 5, B - 4, C - 1, D - 2.
• C. A - 4, B - 1, C - 2, D - 3.
• D. A - 4, B - 5, C - 1, D - 3.

Answer 1

The correct option is A A - 5, B - 1, C - 2, D - 3.

(i) $f\left(x\right)=a^{2}si{n}^{2}x+b^{2}cose{c}^{2}x$

Since, AM $\ge$ GM

$\Rightarrow \frac{a^{2}si{n}^{2}x+b^{2}cose{c}^{2}x}{2}\ge \sqrt{a^{2}si{n}^{2}x{b}^{2}cose{c}^{2}x}$

$\Rightarrow \frac{f\left(x\right)}{2}\ge ab$

$\Rightarrow f\left(x\right)\ge 2ab$

Minimum value of $f\left(x\right)$ is $2ab.$
(ii)$f\left(x\right)=acotx+btanx$

Since, AM $\ge$ GM

$\Rightarrow \frac{acotx+btanx}{2}\ge \sqrt{acotxbtanx}$

$\Rightarrow \frac{f\left(x\right)}{2}\ge \sqrt{ab}$

$\Rightarrow f\left(x\right)\ge 2\sqrt{ab}$

Minimum value of f(x) is $2\sqrt{ab}$.
(iii) $f\left(x\right)=a^{2}se{c}^{2}x+b^{2}cose{c}^{2}x$

$f^{\prime }\left(x\right)=2a^{2}se{c}^{2}xtanx-2b^{2}cose{c}^{2}xcotx$

For maxima or minima,

$f^{\prime }\left(x\right)=0$

$f^{\prime }\left(x\right)=2a^{2}se{c}^{2}xtanx-2b^{2}cose{c}^{2}xcotx=0$

$\Rightarrow ta{n}^{4}x=\frac{b^2}{a^2}$

$\Rightarrow ta{n}^{2}x=\frac{b}{a}$

$f^{\prime \prime }\left(x\right)>0$

Hence, minimum value of f(x) $=a^2(1+ta{n}^{2}x)+b^2(1+co{t}^{2}x)=(a+b{)}^2$
(iv) Minimum and maximum value of $acosx+bsinx$ is$-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$