f(x)={cases} α cosx/π -2x & if & x≠π /2 \\ 3 & if & x=π /2 {cases} which is continuous at x=π /2 where α is a constant. What is {x→0}{lim}f(x) equal to

Correct Answer is : 6π6/π π6

Correct Answer is : 6π6/π π6

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UPSC NDA I 2015 Mathematics Paper

Question Numbers: 74-75

For the next two (2) items that follow:

Consider the function

Question : 75 of 120

Marks: +1, -0

f(x)=\begin{cases} \frac{\alpha cosx}{\pi -2x} & if & x≠\frac{\pi }{2} \\ 3 & if & x=\frac{\pi }{2} \end{cases}

x=\frac{\pi }{2}

\underset{x→0}{lim}f(x)

equal to

Solution:

Concept:

Use the given continuity condition at

x = \frac{\pi}{2}

to find the constant

\alpha

, then evaluate the limit as

x \to 0

Explanation:

Since

f(x)

is continuous at

x = \frac{\pi}{2}

, we have:

\displaystyle \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 3

For

x \neq \frac{\pi}{2}

f(x) = \frac{\alpha \cos x}{\pi - 2x}

. Evaluate the limit as

x \to \frac{\pi}{2}

using L'Hospital's rule or substitution.

Let

t = \frac{\pi}{2} - x

, then as

x \to \frac{\pi}{2}

t \to 0

. Also

\cos x = \cos\left(\frac{\pi}{2} - t\right) = \sin t

, and

\pi - 2x = \pi - 2\left(\frac{\pi}{2} - t\right) = 2t

Thus

\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\alpha \cos x}{\pi - 2x} = \lim_{t \to 0} \frac{\alpha \sin t}{2t} = \frac{\alpha}{2}

. Set equal to 3:

\frac{\alpha}{2} = 3 \implies \alpha = 6

Now find

\displaystyle \lim_{x \to 0} f(x)

. For

x \to 0

(not equal to

\frac{\pi}{2}

\displaystyle \lim_{x \to 0} \frac{\alpha \cos x}{\pi - 2x} = \frac{\alpha \cdot \cos 0}{\pi - 0} = \frac{\alpha}{\pi}

Substitute

\alpha = 6

\displaystyle \frac{6}{\pi}

Answer:

\displaystyle \frac{6}{\pi}

(Option D).

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