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Q. The equation $ {{\tan }^{4}}x-2{{\sec }^{2}}x+{{a}^{2}}=0 $ will have atleast one solution, if

Jharkhand CECEJharkhand CECE 2012

Solution:

$ {{\tan }^{4}}x-2{{\sec }^{2}}x+{{a}^{2}}=0 $
$ \Rightarrow $ $ {{\tan }^{4}}x-2(1+{{\tan }^{2}}x)+{{a}^{2}}=0 $
$ \Rightarrow $ $ {{\tan }^{4}}x-2{{\tan }^{2}}x+1=3-{{a}^{2}} $
$ \Rightarrow $ $ {{({{\tan }^{2}}x-1)}^{2}}=3-{{a}^{2}} $
$ \Rightarrow $ $ 3-{{a}^{2}}\ge 0 $
$ \Rightarrow $ $ {{a}^{2}}\le 3 $
$ \Rightarrow $ $ |a|\,\,\le \sqrt{3} $