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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 : 45 of 120

Marks: +1, -0

The function is decreasing on :

Solution:

Concept:

A function

f(x)

is decreasing on an interval if its derivative

f'(x) \le 0

for all

x

in that interval.

Explanation:

First, evaluate the determinant to simplify

f(x)

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

Expanding along the first row:

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

Now differentiate:

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

For decreasing function,

f'(x) \le 0 \Rightarrow 3x^2 + 28x \le 0

Factor:

x(3x+28) \le 0

The product is non-positive when

-\frac{28}{3} \le x \le 0

Thus

f(x)

is decreasing on

\left[-\frac{28}{3}, 0\right]

Answer:

\left[-\frac{28}{3},0\right]

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