What should be the value of \[\lambda \] such that the function defined below is continuous at \[{\text{x}} = \frac{\pi }{2}?\] \[{\text{f}}\left( {\text{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\lambda \cos {\text{x}}}}{{\frac{\pi }{2} – {\text{x}}}}}&{{\text{if x}} \ne \frac{\pi }{2}} \\ 1&{{\text{if x}} = \frac{\pi }{2}} \end{array}} \right.\] A. 0 B. \[\frac{2}{\pi }\] C. 1 D. \[\frac{\pi }{2}\]

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”[rac{2}{pi
” option3=”1″ option4=”\[\frac{\pi }{2}\]” correct=”option3″]

The function is continuous at $x=\frac{\pi}{2}$ if the following equality holds:

$$\lim_{x\to \frac{\pi}{2}}f(x)=f\left(\frac{\pi}{2}\right)$$

In this case, we have:

$$\lim_{x\to \frac{\pi}{2}}f(x)=\lim_{x\to \frac{\pi}{2}}\frac{\lambda \cos x}{\frac{\pi}{2}-x}=\lambda$$

and:

$$f\left(\frac{\pi}{2}\right)=1$$

Therefore, for the function to be continuous at $x=\frac{\pi}{2}$, we need to have $\lambda=1$.

The other options are incorrect because they do not make the function continuous at $x=\frac{\pi}{2}$. For example, if $\lambda=0$, then the function would be equal to $0$ for all values of $x$, including $x=\frac{\pi}{2}$. This would mean that the function is not continuous at $x=\frac{\pi}{2}$, since the left-hand and right-hand limits of the function would be different.

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