Answer:
We have, \[\begin{matrix} x\ne 0 \\ x=0 \\ \end{matrix}\] Since, f (x) is continuous at x = 0. \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f\,(x)=f\,(0)\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\cos }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1} \right)=a\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-{{\sin }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1} \right)=a\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-\,2{{\sin }^{2}}x}{\sqrt{{{x}^{2}}+1-1}} \right)=a\] \[\Rightarrow \]\[-\,2\,\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}x}{\sqrt{{{x}^{2}}+1-1}} \right)\times \frac{\sqrt{{{x}^{2}}+1}+1}{\sqrt{{{x}^{2}}+1}+1}=a\] \[\Rightarrow \]\[-\,2\,\,\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin x}{x} \right)}^{2}}\times \underset{x\to 0}{\mathop{\lim }}\,\sqrt{{{x}^{2}}+1}+1=a\] \[\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right]\] \[\Rightarrow \] \[-\,2\,{{(1)}^{2}}\times (\sqrt{(10+1})+1)=a\] \[\Rightarrow \] \[-\,2\times 2=a\] \[\Rightarrow \] \[a=-\,4\]
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