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UPSC NDA 2 2021 Mathematics Solved Paper

Question : 75 of 120

Marks: +1, -0

Let

f(x) = \begin{cases} 1+\frac{x}{2k}, & 0 < x < 2 \\ kx, & 2 \le x < 4 \end{cases}

\displaystyle\lim\limits_{x\to 2}

f(x) exists, then what is the value of k?

Solution:

Concept:

For a limit to exist at a point, the left-hand limit and right-hand limit must be equal.

Explanation:

The function is defined as

f(x) = \begin{cases} 1+\frac{x}{2k}, & 0 < x < 2 \\ kx, & 2 \le x < 4 \end{cases}

We compute the left-hand limit as

x\to 2^{-}

\displaystyle\lim_{x\to 2^{-}} f(x) = \lim_{x\to 2} \left(1+\frac{x}{2k}\right) = 1+\frac{2}{2k}=1+\frac{1}{k}

The right-hand limit as

x\to 2^{+}

\displaystyle\lim_{x\to 2^{+}} f(x) = \lim_{x\to 2} (kx) = 2k

Since the limit exists, we equate:

1+\frac{1}{k}=2k

Multiply by

k

k+1=2k^{2}

→

2k^{2}-k-1=0

Factor:

(k-1)(2k+1)=0

, so

k=1

k=-\frac{1}{2}

Among the given options (

-2,-1,0,1

), only

k=1

is listed.

Answer:

k=1

(Option D).

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