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UPSC NDA II 2018 Mathematics Solved Paper

Question : 78 of 120

Marks: +1, -0

What is

\underset{x→\frac{\pi }{6}}{lim} \frac{2sin^{2}x + sinx - 1}{2sin^{2}x - 3sinx + 1}

equal to?

Solution:

Concept:

Factor the numerator and denominator as polynomials in

\sin x

, cancel the common factor

\left(2\sin x - 1\right)

, and evaluate the limit by substituting

x = \frac{\pi}{6}

Explanation:

We are to find

\displaystyle\lim_{x\to\frac{\pi}{6}}\frac{2\sin^{2}x + \sin x - 1}{2\sin^{2}x - 3\sin x + 1}

Factor the numerator:

2\sin^{2}x + \sin x - 1 = (2\sin x - 1)(\sin x + 1)

Factor the denominator:

2\sin^{2}x - 3\sin x + 1 = (2\sin x - 1)(\sin x - 1)

Thus, for

x \neq \frac{\pi}{6}

we can cancel the factor

(2\sin x - 1)

The limit becomes

\displaystyle\lim_{x\to\frac{\pi}{6}}\frac{\sin x + 1}{\sin x - 1}

Substitute

\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}

\frac{\frac{1}{2}+1}{\frac{1}{2}-1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3

Answer:

-3

(Option D).

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