Correct Answer: A. 2y - x -5 = 0

Explanation

We are given the equation \(x^{2} + y^{2} - 4x - 2y = 0\)

\(y = x^{2} + y^{2} - 4x - 2y \)

Using the method of implicit differentiation,

\(\frac{\mathrm d y}{\mathrm d x} = 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x}\)

For the tangent, \(\frac{\mathrm d y}{\mathrm d x} = 0\),

\(\therefore 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x} = 0\)

\((2y - 2)\frac{\mathrm d y}{\mathrm d x} = 4 - 2x \implies \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2x}{2y - 2}\)

At (1, 3), \(\frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}\)

Equation: \(\frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1\)

= \(2y - x - 6 + 1 = 2y - x - 5 = 0\)

Correct Answer: C. -3

Explanation

\(y = 4x^{3} + kx^{2} - 6x + 4\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)

At P(1, m)

\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)

\(6 + 2k = 0 \implies k = -3\)

Correct Answer: B. 240

Explanation

\((1 - 2x)^{6} = ^{6}C_{0}(1^{6}) + ^{6}C_{1}(1^{5})(-2x^{1}) + ^{6}C_{2}(1^{4})(-2x^{2}) + ^{6}C_{3}(1^{3})(-2x^{3}) + ^{6}C_{4}(1^{2})(-2x^{4}) + ...\)

\(\text{The coefficient of} x^{4} = 15 \times 16 = 240\)

Correct Answer: C. 4

Explanation

Given the equation of the circle \(x^{2} + y^{2} - 8x - 2y + 1 = 0\).

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2} \equiv x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Comparing the RHS of the equation above with the equation rewritten as \(x^{2} + y^{2} - 8x - 2y = -1\), we have

\(-2a = -8; -2b = -2 \implies a = 4, b = 1\)

\(\therefore r^{2} - 4^{2} - 1^{2} = -1 \implies r^{2} = -1 + 16 + 1 = 16\)

\(r = \sqrt{16} = 4\)

Explanation

(a)(i) The matrix of rotation of point anticlockwise through angle 60° about the origin is given by:

\(M = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} = \begin{pmatrix} \cos 60° & -\sin 60° \\ \sin 60° & \cos 60° \end{pmatrix}\)

= \(\begin{pmatrix} \frac{1}{2} & \frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\)

(ii) Using the transformation,

\(T :P \to MP\) or \(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

\(T : \begin{pmatrix} 3 \\ 5 \end{pmatrix} \to \begin{pmatrix} \frac{1}{2} & \frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 3 \\ 5 \end{pmatrix}\)

Image of P, P' = \(\begin{pmatrix} \frac{3 + 5\sqrt{3}}{2} \\ \frac{3\sqrt{3} - 5}{2} \end{pmatrix}\)

(b) \(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 2x + 3y \\ 3x - y \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

(i) \(N = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix}\)

(ii) \(N^{2} + aN + bI = 0\)

\(\begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} ^{2} + a \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + b \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + a \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + b \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 4 + 9 & 6 - 3 \\ 6 - 3 & 9 + 1 \end{pmatrix} + \begin{pmatrix} 2a & 3a \\ 3a & -a \end{pmatrix} + \begin{pmatrix} b & 0 \\ 0 & b \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

Adding corresponding elements of each matrix,

\(13 + 2a + b = 0 \implies 2a + b = -13 ... (1)\)

\(3 + 3a = 0 \implies 3a = -3 ; a = -1\)

\(2(-1) + b = -13 \implies -2 + b = -13\)

\(b = -13 + 2 = -11\)

\(\therefore a = -1 ; b = -11\)

Correct Answer: A. 7920

Explanation

= \(\frac{12!}{2! 3! 7!} \)

\(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{2 \times 3 \times 2 \times 7!}\)

= \(7,920\)

Correct Answer: D. \(- \cos x\)

Correct Answer: D. \(- 4\sqrt{3} i + 4j\)

Explanation

\(F = F\cos \theta i + F \sin \theta j\)

\((8 N, 150°) = 8 \cos 150 i + 8 \sin 150 j\)

= \(- 8 \cos 30 i + 8 \sin 30 j\)

= \(-8(\frac{\sqrt{3}}{2}) i + 8(\frac{1}{2} j\)

= \(-4\sqrt{3} i + 4 j\)

Correct Answer: B. -1

Explanation

\(\begin{pmatrix} 2 & x \\ 3 & 5 \end{pmatrix} = 13\)

\(\begin{vmatrix} 2 & x \\ 3 & 5 \end{vmatrix} = (2 \times 5) - (3 \times x) = 13\)

\(10 - 3x = 13 \implies -3x = 3; x = -1\)

Correct Answer: B. Birds fly if and only if the sky is blue?