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UPSC NDA 2 2022 Mathematics Solved Paper

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Question : 73 of 120

Marks: +1, -0

What is the minimum value of the function f(x) = \log_{10}(x^{2} + 2x + 11) ?

0
1
2
10

Solution:

Concept:

The minimum value of a function occurs where its first derivative is zero and its second derivative is positive.

Explanation:

Given

f(x) = \log_{10}(x^{2} + 2x + 11)

.

Rewrite using natural logarithm:

f(x) = \frac{1}{\ln 10} \ln(x^{2} + 2x + 11)

.

Differentiate:

f'(x) = \frac{1}{\ln 10} \cdot \frac{2x + 2}{x^{2} + 2x + 11}

.

Set

f'(x) = 0 \implies 2x + 2 = 0 \implies x = -1

.

Now compute

f''(x)

.

f''(x) = \frac{1}{\ln 10} \cdot \frac{2(x^{2} + 2x + 11) - (2x + 2)^{2}}{(x^{2} + 2x + 11)^{2}}

.

Simplify numerator:

2x^{2} + 4x + 22 - (4x^{2} + 8x + 4) = -2x^{2} - 4x + 18

.

At

x = -1

:

f''(-1) = \frac{1}{\ln 10} \cdot \frac{-2(1) - 4(-1) + 18}{(1 - 2 + 11)^{2}} = \frac{1}{\ln 10} \cdot \frac{20}{100} = \frac{1}{5\ln 10} > 0

.

Thus

f(x)

has a minimum at

x = -1

.

Minimum value:

f(-1) = \log_{10}((-1)^{2} + 2(-1) + 11) = \log_{10}(10) = 1

.

Answer:

1

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