Question

Column I		Column II
A	If $4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$ then $2x$ equals	P	1
B	The number of solutions of ${\log }_7{\log }_5(\sqrt{x+5}+\sqrt{x})=0$ is	Q	2
C	The number of values of $x$ such that the middle term of ${\log }_{3}2,{\log }_3\left(2^x-5\right),{\log }_3\left(2^x-\frac{7}{2}\right)$ is the average of the other two is	R	3
D	If $\alpha ,\beta$ are the roots of the equation $x^2-\left(3+2^{\sqrt{{\log }_{2}3}}-3^{\sqrt{{\log }_{3}2}}\right)x-2\left(3^{{\log }_{3}2}-2^{{\log }_{2}3}\right)=0$ then $2(\alpha +\beta )-\alpha \beta$ equals	S	4

• A. (A) R; (B) P ; (C) P; (D) P
• B. (A) R; (B) P ; (C) P; (D) Q
• C. (A) R; (B) P ; (C) P; (D) R
• D. (A) R; (B) P ; (C) P; (D) S

Answer 1

Answer: (D)

(A) $\left(4^x+\frac{4^x}{2}\right)=3^x\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)$ or $\frac{3}{2}{4}^x=\frac{4}{\sqrt{3}}{3}^x$ or ${\left(\frac{4}{3}\right)}^x=\frac{8}{3\sqrt{3}}={\left(\frac{4}{3}\right)}^{3/2}$
$\Rightarrow x=\frac{3}{2}\Rightarrow 2x=3$ Ans.
(B) ${\log }_{7}a=0\Rightarrow a=1$
$\Rightarrow {\log }_5(\sqrt{x+5}+\sqrt{x})=1\Rightarrow \sqrt{x+5}+\sqrt{x}=5\Rightarrow \sqrt{x+5}=5-\sqrt{x}$
$\Rightarrow x+5=(5-\sqrt{x}{)}^2=25-10\sqrt{x}+x\Rightarrow \sqrt{x}=2\Rightarrow x=4$
(C) If logarithm of 3 numbers are in A.P, then the number must be in G.P.
$\therefore {\left(2^x-5\right)}^2=2\left(2^2-\frac{7}{2}\right)\Rightarrow {\left(2^x\right)}^2-12\left(2^x\right)+32=0$
$\Rightarrow 2^x=4,8\Rightarrow x=2,3$
$\text{ But }{2}^x-5>0\Rightarrow x=3\text{ Ans. }]$
(D) $x^2-3x+2=0$ as $2^{\sqrt{{\log }_{2}3}}=3^{\sqrt{{\log }_{3}2}}$ Hence $\alpha +\beta =3;\alpha \beta =2\Rightarrow 2(\alpha +\beta )-\alpha \beta =2(3)-2=6-2=4$