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UPSC NDA 2 2022 Mathematics Solved Paper

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Question : 85 of 120

Marks: +1, -0

What is

\lim\limits_{x\rightarrow\frac{\pi}{2}} \frac{4x-2\pi}{\cos x}

equal to ?

$-4$
$-2$
2
4

Solution:

Concept:

The limit is of the form

\frac{0}{0}

, so L'Hôpital's Rule applies.

Explanation:

Step 1: Check the limit form.

As

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

, numerator

4x-2\pi \to 0

and denominator

\cos x \to 0

.

Thus it is a

\frac{0}{0}

indeterminate form.

Step 2: Apply L'Hôpital's Rule.

Differentiate numerator:

\frac{d}{dx}(4x-2\pi)=4

.

Differentiate denominator:

\frac{d}{dx}(\cos x)=-\sin x

.

Step 3: Rewrite the limit:

\lim\limits_{x\rightarrow\frac{\pi}{2}} \frac{4x-2\pi}{\cos x} = \lim\limits_{x\rightarrow\frac{\pi}{2}} \frac{4}{-\sin x}

.

Step 4: Evaluate the limit.

At

x = \frac{\pi}{2}

,

\sin\frac{\pi}{2} = 1

.

So

\frac{4}{-\sin\frac{\pi}{2}} = \frac{4}{-1} = -4

.

Answer:

-4

(option A).

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