Question

If x $\in$ R and the numbers $5^{(1-x)}+5^{(x+1)}$, a/2, $({25}^x+{25}^{-x})$ form an A. P. then a must lie in the interval ............ .

• A. $[12,\infty ]$
• B. $[-5,5]$
• C. $[-12,12]$
• D. $[8,\infty ]$

Answer 1

The correct option is A $[12,\infty ]$

Since $5^{1+x}+5^{1-x},a/2,{25}^x+{25}^{-x}$ are in A. P., we have
$2a/2=5^{1+x}+5^{1-x}+{25}^x+{25}^{-x},$
Now put $5^x=t$ so that $t>0,$ we then have
$a=5t+5/t^2+1/t^2=+ive$ ...(1)
$=(t^2+1/t^2)+5(t+1/t)\ge 2+5.2=12$
because $A.M.\ge G.M.$
Also when $a\ge 12$ then (1) becomes
$a={(t+\frac{1}{t})}^2-2+5(t+\frac{1}{t})$
or $y^2+5y-(a+2)=0$
where $y=t+\frac{1}{t}\ge 2asA.M.\ge G.M.$
$\therefore y=\frac{-5\pm \sqrt{25+4a+8}}{2}$
$y=\frac{-5\pm \sqrt{33+4a}}{2}\ge \frac{-5\pm \sqrt{33+48}}{2}$
$=\frac{-5\pm 9}{2}=\frac{9-5}{2}=2\therefore y\ge 2$
$\therefore y=t+\frac{1}{t}=5^x+5^{-x}\ge 2$
which we know is true and gives $a\in [12,\infty )$