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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 18-22

Direction: For the next five (5) items that follow:

Consider the function:

f(x) = \left| x - 1 \right| + x^{2}

Where

x \text{ in } R

Question : 21 of 120

Marks: +1, -0

x=\frac{1}{2}

and x = 1? (No need to solve)

Solution:

Concept:

The area between a curve

y = f(x)

and the

x

-axis from

x = a

x = b

\int\limits_{a}^{b} f(x) \, dx

, where

f(x) \ge 0

in the interval.

Explanation:

Given

f(x) = |x - 1| + x^{2}

Rewrite the modulus: for

x < 1

|x - 1| = 1 - x

; for

x \ge 1

|x - 1| = x - 1

Thus, on the interval

\left[\frac{1}{2}, 1\right]

, we have

x < 1

, so

f(x) = (1 - x) + x^{2} = x^{2} - x + 1

The required area is

\int\limits_{\frac{1}{2}}^{1} (x^{2} - x + 1) \, dx

Compute the integral:

\int\limits_{\frac{1}{2}}^{1} x^{2} dx = \left[\frac{x^{3}}{3}\right]_{\frac{1}{2}}^{1} = \frac{1}{3}\left(1 - \frac{1}{8}\right) = \frac{7}{24}

\int\limits_{\frac{1}{2}}^{1} x \, dx = \left[\frac{x^{2}}{2}\right]_{\frac{1}{2}}^{1} = \frac{1}{2}\left(1 - \frac{1}{4}\right) = \frac{3}{8}

\int\limits_{\frac{1}{2}}^{1} 1 \, dx = \left[x\right]_{\frac{1}{2}}^{1} = 1 - \frac{1}{2} = \frac{1}{2}

Therefore, area

= \frac{7}{24} - \frac{3}{8} + \frac{1}{2} = \frac{7}{24} - \frac{9}{24} + \frac{12}{24} = \frac{10}{24} = \frac{5}{12}

square units.

Answer:

\frac{5}{12}

square unit (Option A).

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