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If $f(x)=\cos^2x+\sec^2x$,then

$\begin{array}{1 1}(A)\;f(x) < 1&(B)\;f(x) =1\\(C)\;2 < f(x) < 1&(D)\;f(x) \geq 2\end{array} $

1 Answer

Toolbox:
  • $\cos^2x=1-\sin^2x$
  • $\sec^2x=1+\tan^2x$
$f(x)=\cos^2x+\sec^2x$
$\cos^2x=1-\sin^2x$
$\sec^2x=1+\tan^2x$
$f(x)=1-\sin^2x+\tan^2x+1$
$\Rightarrow 2+\large\frac{\sin^2x}{\cos^2x}$$-\sin^2x$
$\Rightarrow 2+\large\frac{\sin^2x-\sin^2x\cos^2x}{\cos^2x}$
$\Rightarrow 2+\large\frac{\sin^2x(1-\cos^2x)}{\cos^2x}$
$\Rightarrow 2+\large\frac{\sin^4x}{\cos^2x}$
Here $\large\frac{\sin^4x}{\cos^2x}$$ \geq 0$
$\therefore f(x) \geq 2$
Hence (D) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 
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