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UPSC NDA 2 2016 Mathematics Solved Paper

Question Numbers: 21-22

Consider the following for the next two (02) items that follow:

Question : 21 of 120

Marks: +1, -0

Let

\mathrm{a}x^3 + \mathrm{b}x^2 + \mathrm{c}x + \mathrm{d} = \begin{vmatrix} \mathrm{x}+1 & 2\mathrm{x} & 3\mathrm{x} \\ 2\mathrm{x}+3 & \mathrm{x}+1 & \mathrm{x} \\ 2-\mathrm{x} & 3\mathrm{x}+4 & 5\mathrm{x}-1 \end{vmatrix}

What is the value of c?

Solution:

Concept:

The given polynomial equation equals a determinant. We find

c

by expanding the determinant and comparing coefficients.

Explanation:

Let

A = \begin{vmatrix} x+1 & 2x & 3x \\ 2x+3 & x+1 & x \\ 2-x & 3x+4 & 5x-1 \end{vmatrix}

Apply row operations to simplify:

R_2 \to R_2 - 2R_1

and

R_3 \to R_3 + R_1

A = \begin{vmatrix} x+1 & 2x & 3x \\ 1 & -3x+1 & -5x \\ 3 & 5x+4 & 8x-1 \end{vmatrix}

Apply

R_3 \to R_3 - 3R_2

A = \begin{vmatrix} x+1 & 2x & 3x \\ 1 & -3x+1 & -5x \\ 0 & 14x+1 & 23x-1 \end{vmatrix}

Expand along

C_1

A = (x+1)\big[(-3x+1)(23x-1) + 5x(14x+1)\big] - 1 \cdot \big[2x(23x-1) - 3x(14x+1)\big] + 0

Simplify each bracket:

First bracket:

(-3x+1)(23x-1) = -69x^2 + 3x + 23x - 1 = -69x^2 + 26x - 1

Add

5x(14x+1) = 70x^2 + 5x

gives

x^2 + 31x - 1

Second bracket:

2x(23x-1) - 3x(14x+1) = 46x^2 - 2x - 42x^2 - 3x = 4x^2 - 5x

Thus

A = (x+1)(x^2 + 31x - 1) - (4x^2 - 5x)

Expand:

(x+1)(x^2+31x-1) = x^3 + 31x^2 - x + x^2 + 31x - 1 = x^3 + 32x^2 + 30x - 1

Subtract

(4x^2 - 5x)

A = x^3 + 32x^2 + 30x - 1 - 4x^2 + 5x = x^3 + 28x^2 + 35x - 1

Given

A = ax^3 + bx^2 + cx + d

, comparing coefficients gives

c = 35

Answer:

c = 35

(Option C).

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