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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 45-46

Consider the following for the next items that follow:

A function is defined by

f(x)= \begin{vmatrix} x+1 & 2 & 3 \\ 2 & x+4 & 6 \\ 3 & 6 & x+9 \end{vmatrix}

Question : 46 of 120

Marks: +1, -0

The function attains local minimum value at :

Solution:

Concept:

A local minimum occurs where

f'(x) = 0

and

f''(x) > 0

, i.e., the function changes from decreasing to increasing.

Explanation:

Given

f(x) = \begin{vmatrix} x+1 & 2 & 3 \\ 2 & x+4 & 6 \\ 3 & 6 & x+9 \end{vmatrix}

Expand the determinant:

f(x) = (x+1)[(x+4)(x+9)-36] - 2[2(x+9)-18] + 3[12 - 3(x+4)]

Simplify each term:

(x+4)(x+9)-36 = x^2+13x+36-36 = x^2+13x

2(x+9)-18 = 2x+18-18 = 2x

12-3(x+4) = 12-3x-12 = -3x

Thus

f(x) = (x+1)(x^2+13x) - 2(2x) + 3(-3x)

= x^3+13x^2+x^2+13x -4x -9x

= x^3+14x^2

Differentiate:

f'(x) = 3x^2+28x = x(3x+28)

Set

f'(x)=0

x=0

3x+28=0 \Rightarrow x=-\frac{28}{3}

These are critical points.

Second derivative:

f''(x) = 6x+28

x=0

f''(0)=28 > 0

, so local minimum.

x=-\frac{28}{3}

f''(-\frac{28}{3}) = 6(-\frac{28}{3})+28 = -56+28 = -28 < 0

, so local maximum.

Answer:

x=0

(option C).

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