Question

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

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

Answer 1

The correct option is B

[-1,1]

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