Question

The set of values of $x$ for which $2^{2x^2-10x+3}+6^{x^2-5x+1}\ge 3^{2x^2-10x+3}$ holds good

• A. $(-\infty ,-\left(\frac{5+\sqrt{21}}{2}\right)]$
• B. $(-\infty ,\infty )$
• C. $[\frac{5-\sqrt{21}}{2},\frac{5+\sqrt{21}}{2}]$
• D. $[-\frac{5+\sqrt{21}}{2},\frac{5+\sqrt{21}}{2}]$

Answer 1

The correct option is C $[\frac{5-\sqrt{21}}{2},\frac{5+\sqrt{21}}{2}]$

$2^{2x^2-10x+3}+6^{x^2-5x+1}\ge 3^{2x^2-10x+3}$
$2.2^{2(x^2-5x+1)}+(3.2{)}^{x^2-5x+1}-{3.3}^{2(x^2-5x+1)}\ge 0$
Factorize
$2.2^{2(x^2-5x+1)}-2\times (3.2{)}^{x^2-5x+1}+3\times (3.2{)}^{x^2-5x+1}-{3.3}^{2(x^2-5x+1)}\ge 0$
$2.2^{(x^2-5x+1)}(2^{(x^2-5x+1)}-3^{(x^2-5x+1)})+3\left(3{)}^{x^2-5x+1}\right(2^{(x^2-5x+1)}-3^{(x^2-5x+1)})\ge 0$
$(2^{(x^2-5x+1)}-3^{(x^2-5x+1)})({2.2}^{(x^2-5x+1)}+3.3^{(x^2-5x+1)})\ge 0$
Now solving the equation by taking log
$[\frac{5-\sqrt{21}}{2},\frac{5+\sqrt{21}}{2}]$