Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& \text{The minimum value of}{9}^3{27}^{\text{cos}2x}{81}^{\text{sin}2x}\text{is}& \text{(p)}& 1\\ \text{(b)}& \text{Number of solutions of the equation}co{s}^{7}x+si{n}^{4}x=1,xϵ[0,2\pi ]& \text{(q)}& 2\\ \text{(c)}& \text{Value of a for which the equation}{a}^2-2a+se{c}^2\pi (a+x)=0\text{has a solution}& \text{(r)}& 3\\ \text{(d)}& \text{If cos (Psin x) = sin (P cos x), then the minimum possible value of}\frac{4\sqrt{2}}{\pi }\text{P is}& \text{(s)}& 4\end{array}$

Which of the following is correct?

• A. A-P,B-R,C-S,D-Q
• B. A-S,B-R,C-P,D-Q
• C. A-R,B-R,C-P,D-Q
• D. A-R,B-S,C-P,D-Q

Answer 1

The correct option is D

A-R,B-S,C-P,D-Q

$\left(a\right)3^6{3}^{3cos2x}{3}^{4sin2x}=3^{6+3cos2x+4sin2x}$

$3cos2x+4sin2x=5sin(\alpha +2x)$

$\text{minimum value of}5sin(\alpha +2x)=5(-1)=-5$

$\therefore \text{minimun value of}{9}^3{27}^{cos2x}{81}^{sin2x}=3^{6-5}=3.$

$\left(b\right)co{s}^{7}x+si{n}^{4}x=1$

$co{s}^{7}x=1-si{n}^{4}x$

$\Rightarrow co{s}^{7}x=co{s}^{2}x(1+si{n}^{2}x)$

$\Rightarrow co{s}^{2}x(co{s}^{5}x-(1+si{n}^{2}x))=0$

$\therefore co{s}^{2}x=0\text{or}co{s}^{5}x=1+si{n}^{2}x$

$\therefore cosx=0\text{or}cosx=1$

$\therefore x=\frac{\pi }{2},\frac{3\pi }{2}\text{or}x=0,2\pi$

$\therefore$ four solutions

$\left(c\right)a^2-2a+1+ta{n}^2\pi (a+x)=0$

$(a-1{)}^2+{\left(tan\pi \right(a+x\left)\right)}^2=0$

only possible when a = 1 and

$tan\pi (a+x)=0$

$\therefore a=1andtan(\pi +\pi x)=0$

$\left(d\right)cos\left(Psinx\right)=sin\left(Pcosx\right)$

$\Rightarrow cos\left(Psinx\right)=cos(\frac{\pi }{2}-Pcosx)\Rightarrow Psinx=\frac{\pi }{2}-Pcosx$

$\Rightarrow P(sinx+cosx)=\frac{\pi }{2}$

$max(sinx+cosx)=\sqrt{2}$

$\therefore \text{minimum possible value of}P=\frac{\pi }{2\sqrt{2}}$

$\therefore \text{minimum positive value of}\frac{4\sqrt{2}}{\pi }P=2$