Question

If $ta{n}^{2}x+secx-a=0$ has alteast one solution, then $a\in$ ___

• A. $(-\infty ,1]$
  
• B. $[-1,\infty ]$
• C. $[1,\infty )$
• D. $(-\infty ,-1]$

Answer 1

The correct option is B

$[-1,\infty ]$

$ta{n}^{2}x+secx-a=0(se{c}^{2}x-1)+secx-a=0=se{c}^{2}x+secx+\frac{1}{4}-1-\frac{1}{4}-a=0=(secx+\frac{1}{2}{)}^2=\frac{5}{4}+aNow,secx\ge 1or\le -1\Rightarrow secx+\frac{1}{2}\ge \frac{3}{2}orsecx+\frac{1}{2}\le -\frac{1}{2}\therefore (secx+\frac{1}{2}{)}^2\ge \frac{9}{4}or(secx+\frac{1}{2}{)}^2\ge \frac{1}{4}i.e.\frac{5}{4}+a\ge \frac{1}{4}\Rightarrow a\ge -1$