Question

Column I		Column II
A	If $4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1}$ then $2 x$ 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 $\left(\frac{4}{3}\right)^x=\frac{8}{3 \sqrt{3}}=\left(\frac{4}{3}\right)^{3 / 2}$
$\Rightarrow x =\frac{3}{2} \Rightarrow 2 x =3$ 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.
$\therefore \left(2^x-5\right)^2=2\left(2^2-\frac{7}{2}\right) \Rightarrow \left(2^x\right)^2-12\left(2^x\right)+32=0$
$\Rightarrow 2^x=4,8 \Rightarrow x=2,3$
$\text { But } \left. 2^x-5>0 \Rightarrow x=3 \text { Ans. }\right]$
(D) $ x^2-3 x+2=0$ as $2^{\sqrt{\log _2 3}}=3^{\sqrt{\log _3 2}}$ Hence $\alpha+\beta=3 ; \alpha \beta=2 \Rightarrow 2(\alpha+\beta)-\alpha \beta=2(3)-2=6-2=4$