Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

$\begin{vmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{vmatrix} =$ _____________

Answer: c. 1

Solution: $\begin{vmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{vmatrix} = \sin \theta \cdot \sin \theta - (-\cos \theta) \cdot \cos \theta$ $= \sin^2 \theta + \cos^2 \theta = 1$

Q1.2 [1 mark]

If $f(x) = x^3 - 1$ then $f(-1) =$ _________

Answer: d. -2

Solution: $f(x) = x^3 - 1$ $f(-1) = (-1)^3 - 1 = -1 - 1 = -2$

Q1.3 [1 mark]

$\log 1 \times \log 2 \times \log 3 \times \log 4 =$ ______________

Answer: a. 0

Solution: Since $\log 1 = 0$ , we have: $\log 1 \times \log 2 \times \log 3 \times \log 4 = 0 \times \log 2 \times \log 3 \times \log 4 = 0$

Q1.4 [1 mark]

$\log x - \log y =$ _____________

Answer: b. $\log \frac{x}{y}$

Solution: Using logarithm property: $\log x - \log y = \log \frac{x}{y}$

Q1.5 [1 mark]

Principal Period of $\sin(2x + 7) =$ _________

Answer: c. $\pi$

Solution: For $\sin(ax + b)$ , the period is $\frac{2\pi}{|a|}$ Here, $a = 2$ , so period = $\frac{2\pi}{2} = \pi$

Q1.6 [1 mark]

$450° =$ __________ $radian$

Answer: c. $\frac{5\pi}{2}$

Solution: $450° = 450 \times \frac{\pi}{180} = \frac{450\pi}{180} = \frac{5\pi}{2}$ radians

Q1.7 [1 mark]

$\tan^{-1} x + \cot^{-1} x =$ _________

Answer: d. $\frac{\pi}{2}$

Solution: This is a standard identity: $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all $x > 0$

Q1.8 [1 mark]

$|2i - 3j + 4k| =$ _______

Answer: a. $\sqrt{29}$

Solution: $|2i - 3j + 4k| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$

Q1.9 [1 mark]

For vector $\vec{a} \times \vec{a} =$ _________

Answer: d. 0

Solution: The cross product of any vector with itself is always zero: $\vec{a} \times \vec{a} = 0$

Q1.10 [1 mark]

If two lines having slopes $m_1$ and $m_2$ are perpendicular to each other then _________

Answer: c. $m_1 \cdot m_2 = -1$

Solution: For perpendicular lines, the product of their slopes equals -1.

Q1.11 [1 mark]

If $x^2 + y^2 = 25$ then its radius ______

Answer: c. 5

Solution: Comparing with standard form $x^2 + y^2 = r^2$ : $r^2 = 25$ , so $r = 5$

Q1.12 [1 mark]

$\lim_{\theta \to 0} \frac{\sin 5\theta}{\tan 7\theta} =$ _________

Answer: b. $\frac{5}{7}$

Solution: $\lim_{\theta \to 0} \frac{\sin 5\theta}{\tan 7\theta} = \lim_{\theta \to 0} \frac{\sin 5\theta}{\frac{\sin 7\theta}{\cos 7\theta}} = \lim_{\theta \to 0} \frac{\sin 5\theta \cos 7\theta}{\sin 7\theta}$

$= \lim_{\theta \to 0} \frac{\sin 5\theta}{5\theta} \cdot \frac{7\theta}{\sin 7\theta} \cdot \frac{5\theta}{7\theta} \cdot \cos 7\theta$

$= 1 \times 1 \times \frac{5}{7} \times 1 = \frac{5}{7}$

Q1.13 [1 mark]

$\lim_{x \to 0} \frac{e^x - 1}{x} =$ ___________

Answer: c. 1

Solution: This is a standard limit: $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

Q1.14 [1 mark]

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} =$ _________

Answer: d. 2

Solution: $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x - 1} = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$

Q.2(A) [6 marks]

Attempt any two

Q2.1 [3 marks]

If $f(x) = \frac{1-x}{1+x}$ then prove that (1) $f(x) \cdot f(-x) = 1$ (2) $f(x) + f(\frac{1}{x}) = 0$

Answer:

Solution:

Part (1): Prove $f(x) \cdot f(-x) = 1$

$f(x) = \frac{1-x}{1+x}$

$f(-x) = \frac{1-(-x)}{1+(-x)} = \frac{1+x}{1-x}$

$f(x) \cdot f(-x) = \frac{1-x}{1+x} \cdot \frac{1+x}{1-x} = \frac{(1-x)(1+x)}{(1+x)(1-x)} = 1$

Part (2): Prove $f(x) + f(\frac{1}{x}) = 0$

$f(\frac{1}{x}) = \frac{1-\frac{1}{x}}{1+\frac{1}{x}} = \frac{\frac{x-1}{x}}{\frac{x+1}{x}} = \frac{x-1}{x+1}$

$f(x) + f(\frac{1}{x}) = \frac{1-x}{1+x} + \frac{x-1}{x+1} = \frac{1-x}{1+x} - \frac{1-x}{1+x} = 0$

Q2.2 [3 marks]

If $\begin{vmatrix} x & 2 & 3 \\ 5 & 0 & 7 \\ 3 & 1 & 2 \end{vmatrix} = 30$ then find the value of $x$

Answer:

Solution: Expanding along the second row (which has a zero): $\begin{vmatrix} x & 2 & 3 \\ 5 & 0 & 7 \\ 3 & 1 & 2 \end{vmatrix} = -5 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} + 0 - 7 \begin{vmatrix} x & 2 \\ 3 & 1 \end{vmatrix}$

$= -5(2 \times 2 - 3 \times 1) - 7(x \times 1 - 2 \times 3)$ $= -5(4 - 3) - 7(x - 6)$ $= -5(1) - 7x + 42$ $= -5 - 7x + 42$ $= 37 - 7x$

Given: $37 - 7x = 30$ $7x = 37 - 30 = 7$ $x = 1$

Q2.3 [3 marks]

Prove that $\tan 55° = \frac{\cos 10° + \sin 10°}{\cos 10° - \sin 10°}$

Answer:

Solution: We know that $55° = 45° + 10°$

Using the tangent addition formula: $\tan(45° + 10°) = \frac{\tan 45° + \tan 10°}{1 - \tan 45° \tan 10°}$

Since $\tan 45° = 1$ : $\tan 55° = \frac{1 + \tan 10°}{1 - \tan 10°}$

Now, $\tan 10° = \frac{\sin 10°}{\cos 10°}$

$\tan 55° = \frac{1 + \frac{\sin 10°}{\cos 10°}}{1 - \frac{\sin 10°}{\cos 10°}} = \frac{\frac{\cos 10° + \sin 10°}{\cos 10°}}{\frac{\cos 10° - \sin 10°}{\cos 10°}} = \frac{\cos 10° + \sin 10°}{\cos 10° - \sin 10°}$

Q.2(B) [8 marks]

Attempt any two

Q2.1 [4 marks]

Prove that $\frac{1}{\log_{xy} xyz} + \frac{1}{\log_{yz} xyz} + \frac{1}{\log_{zx} xyz} = 2$

Answer:

Solution: Using the change of base formula: $\frac{1}{\log_a b} = \log_b a$

$\frac{1}{\log_{xy} xyz} = \log_{xyz} (xy)$ $\frac{1}{\log_{yz} xyz} = \log_{xyz} (yz)$ $\frac{1}{\log_{zx} xyz} = \log_{xyz} (zx)$

LHS = $\log_{xyz} (xy) + \log_{xyz} (yz) + \log_{xyz} (zx)$ $= \log_{xyz} [(xy)(yz)(zx)]$ $= \log_{xyz} (x^2y^2z^2)$ $= \log_{xyz} (xyz)^2$ $= 2\log_{xyz} (xyz)$ $= 2 \times 1 = 2$ = RHS

Q2.2 [4 marks]

If $\log(\frac{a+b}{3}) = \frac{1}{2}(\log a + \log b)$ then prove that $a^2 + b^2 = 7ab$

Answer:

Solution: Given: $\log(\frac{a+b}{3}) = \frac{1}{2}(\log a + \log b)$

RHS: $\frac{1}{2}(\log a + \log b) = \frac{1}{2}\log(ab) = \log(ab)^{1/2} = \log\sqrt{ab}$

So: $\log(\frac{a+b}{3}) = \log\sqrt{ab}$

Taking antilog: $\frac{a+b}{3} = \sqrt{ab}$

Squaring both sides: $(\frac{a+b}{3})^2 = ab$

$\frac{(a+b)^2}{9} = ab$

$(a+b)^2 = 9ab$

$a^2 + 2ab + b^2 = 9ab$

$a^2 + b^2 = 9ab - 2ab = 7ab$

Q2.3 [4 marks]

If $\log x \times \frac{\log 16}{\log 32} = \log 256$ then find the value of $x$

Answer:

Solution: First, let's simplify the logarithmic terms: $\log 16 = \log 2^4 = 4\log 2$ $\log 32 = \log 2^5 = 5\log 2$ $\log 256 = \log 2^8 = 8\log 2$

$\frac{\log 16}{\log 32} = \frac{4\log 2}{5\log 2} = \frac{4}{5}$

Given equation becomes: $\log x \times \frac{4}{5} = 8\log 2$

$\log x = \frac{5 \times 8\log 2}{4} = 10\log 2$

$\log x = \log 2^{10} = \log 1024$

Therefore: $x = 1024$

Q.3(A) [6 marks]

Attempt any two

Q3.1 [3 marks]

Prove that $\frac{\sin(\frac{\pi}{2}+\theta)}{\cos(\pi-\theta)} + \frac{\cot(\frac{3\pi}{2}-\theta)}{\tan(\pi-\theta)} + \frac{\cosec(\frac{\pi}{2}-\theta)}{\sec(\pi+\theta)} = -3$

Answer:

Solution: Using trigonometric identities:

First term: $\sin(\frac{\pi}{2}+\theta) = \cos\theta$ $\cos(\pi-\theta) = -\cos\theta$ $\frac{\sin(\frac{\pi}{2}+\theta)}{\cos(\pi-\theta)} = \frac{\cos\theta}{-\cos\theta} = -1$

Second term: $\cot(\frac{3\pi}{2}-\theta) = \cot(2\pi - \frac{\pi}{2} - \theta) = \cot(-(\frac{\pi}{2} + \theta)) = -\cot(\frac{\pi}{2} + \theta) = -(-\tan\theta) = \tan\theta$ $\tan(\pi-\theta) = -\tan\theta$ $\frac{\cot(\frac{3\pi}{2}-\theta)}{\tan(\pi-\theta)} = \frac{\tan\theta}{-\tan\theta} = -1$

Third term: $\cosec(\frac{\pi}{2}-\theta) = \frac{1}{\sin(\frac{\pi}{2}-\theta)} = \frac{1}{\cos\theta}$ $\sec(\pi+\theta) = \frac{1}{\cos(\pi+\theta)} = \frac{1}{-\cos\theta}$ $\frac{\cosec(\frac{\pi}{2}-\theta)}{\sec(\pi+\theta)} = \frac{\frac{1}{\cos\theta}}{\frac{1}{-\cos\theta}} = \frac{-\cos\theta}{\cos\theta} = -1$

Therefore: LHS = $(-1) + (-1) + (-1) = -3$ = RHS

Q3.2 [3 marks]

Prove that $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \frac{\pi}{4}$

Answer:

Solution: Using the formula: $\tan^{-1}a + \tan^{-1}b = \tan^{-1}(\frac{a+b}{1-ab})$ when $ab < 1$

Let $a = \frac{1}{2}$ and $b = \frac{1}{3}$

$ab = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} < 1$ ✓

$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}})$

$= \tan^{-1}(\frac{\frac{3+2}{6}}{1 - \frac{1}{6}}) = \tan^{-1}(\frac{\frac{5}{6}}{\frac{5}{6}}) = \tan^{-1}(1) = \frac{\pi}{4}$

Q3.3 [3 marks]

Find the equation of the line passing through points $(1, 6)$ and $(-2, 5)$ . Also find the slope of the line.

Answer:

Solution: Step 1: Find the slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3}$

Step 2: Find the equation using point-slope form Using point $(1, 6)$ : $y - 6 = \frac{1}{3}(x - 1)$ $3(y - 6) = x - 1$ $3y - 18 = x - 1$ $x - 3y + 17 = 0$

Table: Line Properties

Property	Value
Slope	$\frac{1}{3}$
Equation	$x - 3y + 17 = 0$

Q.3(B) [8 marks]

Attempt any two

Q3.1 [4 marks]

Draw the graph of $y = \sin x$ ; $0 \leq x \leq \pi$

Answer:

Solution:

Table of Key Points:

$x$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\frac{2\pi}{3}$	$\frac{3\pi}{4}$	$\frac{5\pi}{6}$	$\pi$
$y = \sin x$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$

      y
      |
    1 +        *
      |      /   \
  √3/2+    /       \
      |   /         \
  √2/2+  /           \
      | /             \
  1/2 +                 *
      |                   \
    0 +---*---+---+---+---*---> x
      0  π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π

Properties:

Domain: $[0, \pi]$
Range: $[0, 1]$
Maximum: $1$ at $x = \frac{\pi}{2}$
Zeros: $x = 0$ and $x = \pi$

Q3.2 [4 marks]

Prove that $\frac{\sin \theta + \sin 2\theta + \sin 4\theta + \sin 5\theta}{\cos \theta + \cos 2\theta + \cos 4\theta + \cos 5\theta} = \tan 3\theta$

Answer:

Solution: We can group the terms strategically:

Numerator: $(\sin \theta + \sin 5\theta) + (\sin 2\theta + \sin 4\theta)$

Using sum-to-product formula: $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$

$\sin \theta + \sin 5\theta = 2\sin(\frac{\theta + 5\theta}{2})\cos(\frac{5\theta - \theta}{2}) = 2\sin(3\theta)\cos(2\theta)$

$\sin 2\theta + \sin 4\theta = 2\sin(\frac{2\theta + 4\theta}{2})\cos(\frac{4\theta - 2\theta}{2}) = 2\sin(3\theta)\cos(\theta)$

Numerator = $2\sin(3\theta)\cos(2\theta) + 2\sin(3\theta)\cos(\theta) = 2\sin(3\theta)[\cos(2\theta) + \cos(\theta)]$

Denominator: $(\cos \theta + \cos 5\theta) + (\cos 2\theta + \cos 4\theta)$

$\cos \theta + \cos 5\theta = 2\cos(\frac{\theta + 5\theta}{2})\cos(\frac{5\theta - \theta}{2}) = 2\cos(3\theta)\cos(2\theta)$

$\cos 2\theta + \cos 4\theta = 2\cos(\frac{2\theta + 4\theta}{2})\cos(\frac{4\theta - 2\theta}{2}) = 2\cos(3\theta)\cos(\theta)$

Denominator = $2\cos(3\theta)\cos(2\theta) + 2\cos(3\theta)\cos(\theta) = 2\cos(3\theta)[\cos(2\theta) + \cos(\theta)]$

Therefore: $\frac{\text{Numerator}}{\text{Denominator}} = \frac{2\sin(3\theta)[\cos(2\theta) + \cos(\theta)]}{2\cos(3\theta)[\cos(2\theta) + \cos(\theta)]} = \frac{\sin(3\theta)}{\cos(3\theta)} = \tan(3\theta)$

Q3.3 [4 marks]

The constant forces $i - j + k$ , $i + j - 3k$ and $4i + 5j - 6k$ act on a particle. Under the action of these forces, particle moves from point $3i - 2j + k$ to point $i + 3j - 4k$ . Find the total work done by the forces.

Answer:

Solution: Step 1: Find resultant force $\vec{F_{total}} = (i - j + k) + (i + j - 3k) + (4i + 5j - 6k)$ $= (1 + 1 + 4)i + (-1 + 1 + 5)j + (1 - 3 - 6)k$ $= 6i + 5j - 8k$

Step 2: Find displacement Initial position: $3i - 2j + k$ Final position: $i + 3j - 4k$ $\vec{d} = (i + 3j - 4k) - (3i - 2j + k) = -2i + 5j - 5k$

Step 3: Calculate work done $W = \vec{F_{total}} \cdot \vec{d} = (6i + 5j - 8k) \cdot (-2i + 5j - 5k)$ $W = 6(-2) + 5(5) + (-8)(-5) = -12 + 25 + 40 = 53$ units

Table: Work Calculation

Component	Force	Displacement	Work
x	6	-2	-12
y	5	5	25
z	-8	-5	40
Total			53

Q.4(A) [6 marks]

Attempt any two

Q4.1 [3 marks]

If $\vec{a} = 3i - j - 4k$ , $\vec{b} = 4j - 2i - 3k$ and $\vec{c} = 2j - k - i$ then find $|3\vec{a} - 2\vec{b} + 4\vec{c}|$

Answer:

Solution: First, let's rewrite the vectors in standard form: $\vec{a} = 3i - j - 4k$ $\vec{b} = -2i + 4j - 3k$ $\vec{c} = -i + 2j - k$

$3\vec{a} = 3(3i - j - 4k) = 9i - 3j - 12k$ $2\vec{b} = 2(-2i + 4j - 3k) = -4i + 8j - 6k$ $4\vec{c} = 4(-i + 2j - k) = -4i + 8j - 4k$

$3\vec{a} - 2\vec{b} + 4\vec{c} = (9i - 3j - 12k) - (-4i + 8j - 6k) + (-4i + 8j - 4k)$ $= 9i - 3j - 12k + 4i - 8j + 6k - 4i + 8j - 4k$ $= (9 + 4 - 4)i + (-3 - 8 + 8)j + (-12 + 6 - 4)k$ $= 9i - 3j - 10k$

$|3\vec{a} - 2\vec{b} + 4\vec{c}| = \sqrt{9^2 + (-3)^2 + (-10)^2} = \sqrt{81 + 9 + 100} = \sqrt{190}$

Q4.2 [3 marks]

For what value of $m$ , the vectors $2i - 3j + 5k$ and $mi - 6j - 8k$ are perpendicular to each other?

Answer:

Solution: For two vectors to be perpendicular, their dot product must be zero.

$\vec{A} = 2i - 3j + 5k$ $\vec{B} = mi - 6j - 8k$

$\vec{A} \cdot \vec{B} = 0$ $(2)(m) + (-3)(-6) + (5)(-8) = 0$ $2m + 18 - 40 = 0$ $2m - 22 = 0$ $m = 11$

Q4.3 [3 marks]

Find the equation of the circle having center $(4, 3)$ and passing through point $(7, -2)$

Answer:

Solution: Step 1: Find radius $r = \sqrt{(7-4)^2 + (-2-3)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$

Step 2: Write equation Using standard form: $(x - h)^2 + (y - k)^2 = r^2$ $(x - 4)^2 + (y - 3)^2 = 34$

Step 3: Expand $x^2 - 8x + 16 + y^2 - 6y + 9 = 34$ $x^2 + y^2 - 8x - 6y + 25 - 34 = 0$ $x^2 + y^2 - 8x - 6y - 9 = 0$

Table: Circle Properties

Property	Value
Center	$(4, 3)$
Radius	$\sqrt{34}$
Standard Form	$(x-4)^2 + (y-3)^2 = 34$
General Form	$x^2 + y^2 - 8x - 6y - 9 = 0$

Q.4(B) [8 marks]

Attempt any two

Q4.1 [4 marks]

Prove that the angle between vectors $i + 2j$ and $i + j + 3k$ is $\sin^{-1}\sqrt{\frac{46}{55}}$

Answer:

Solution: Let $\vec{A} = i + 2j$ and $\vec{B} = i + j + 3k$

Step 1: Calculate dot product $\vec{A} \cdot \vec{B} = (1)(1) + (2)(1) + (0)(3) = 1 + 2 + 0 = 3$

Step 2: Calculate magnitudes $|\vec{A}| = \sqrt{1^2 + 2^2 + 0^2} = \sqrt{5}$ $|\vec{B}| = \sqrt{1^2 + 1^2 + 3^2} = \sqrt{11}$

Step 3: Find cosine of angle $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{3}{\sqrt{5} \times \sqrt{11}} = \frac{3}{\sqrt{55}}$

Step 4: Find sine of angle $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{9}{55} = \frac{55 - 9}{55} = \frac{46}{55}$

$\sin \theta = \sqrt{\frac{46}{55}}$

Therefore: $\theta = \sin^{-1}\sqrt{\frac{46}{55}}$

Q4.2 [4 marks]

If $\vec{x} = -2k + 3i$ and $\vec{y} = 5i + 2j - 4k$ then find the value of $|(\vec{x} + \vec{y}) \times (\vec{x} - \vec{y})|$

Answer:

Solution: First, let's rewrite in standard form: $\vec{x} = 3i + 0j - 2k$ $\vec{y} = 5i + 2j - 4k$

$\vec{x} + \vec{y} = (3 + 5)i + (0 + 2)j + (-2 - 4)k = 8i + 2j - 6k$ $\vec{x} - \vec{y} = (3 - 5)i + (0 - 2)j + (-2 + 4)k = -2i - 2j + 2k$

$(\vec{x} + \vec{y}) \times (\vec{x} - \vec{y}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & 2 & -6 \\ -2 & -2 & 2 \end{vmatrix}$

$= \hat{i}(2 \times 2 - (-6) \times (-2)) - \hat{j}(8 \times 2 - (-6) \times (-2)) + \hat{k}(8 \times (-2) - 2 \times (-2))$ $= \hat{i}(4 - 12) - \hat{j}(16 - 12) + \hat{k}(-16 + 4)$ $= -8\hat{i} - 4\hat{j} - 12\hat{k}$

$|(\vec{x} + \vec{y}) \times (\vec{x} - \vec{y})| = \sqrt{(-8)^2 + (-4)^2 + (-12)^2}$ $= \sqrt{64 + 16 + 144} = \sqrt{224} = 4\sqrt{14}$

Q4.3 [4 marks]

Evaluate: $\lim_{n \to \infty} (\sqrt{n^2 + n + 1} - n)$

Answer:

Solution: We have the indeterminate form $\infty - \infty$ . Let's rationalize:

$\lim_{n \to \infty} (\sqrt{n^2 + n + 1} - n)$

Multiply and divide by the conjugate: $= \lim_{n \to \infty} \frac{(\sqrt{n^2 + n + 1} - n)(\sqrt{n^2 + n + 1} + n)}{\sqrt{n^2 + n + 1} + n}$

$= \lim_{n \to \infty} \frac{(n^2 + n + 1) - n^2}{\sqrt{n^2 + n + 1} + n}$

$= \lim_{n \to \infty} \frac{n + 1}{\sqrt{n^2 + n + 1} + n}$

Divide numerator and denominator by $n$ : $= \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{\sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} + 1}$

$= \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Q.5(A) [6 marks]

Attempt any two

Q5.1 [3 marks]

Evaluate: $\lim_{x \to -2} \frac{x^3 + 2x^2 + x + 2}{x^2 + x - 2}$

Answer:

Solution: Direct substitution at $x = -2$ : Numerator: $(-2)^3 + 2(-2)^2 + (-2) + 2 = -8 + 8 - 2 + 2 = 0$ Denominator: $(-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0$

We get $\frac{0}{0}$ form, so we need to factor.

Factoring numerator: $x^3 + 2x^2 + x + 2$ $= x^2(x + 2) + 1(x + 2) = (x + 2)(x^2 + 1)$

Factoring denominator: $x^2 + x - 2$ $= (x + 2)(x - 1)$

$\lim_{x \to -2} \frac{x^3 + 2x^2 + x + 2}{x^2 + x - 2} = \lim_{x \to -2} \frac{(x + 2)(x^2 + 1)}{(x + 2)(x - 1)}$

$= \lim_{x \to -2} \frac{x^2 + 1}{x - 1} = \frac{(-2)^2 + 1}{-2 - 1} = \frac{4 + 1}{-3} = \frac{5}{-3} = -\frac{5}{3}$

Q5.2 [3 marks]

Evaluate: $\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos^2 x}$

Answer:

Solution: Direct substitution at $x = \frac{\pi}{2}$ : Numerator: $1 - \sin \frac{\pi}{2} = 1 - 1 = 0$ Denominator: $\cos^2 \frac{\pi}{2} = 0^2 = 0$

We get $\frac{0}{0}$ form.

Using the identity: $\cos^2 x = 1 - \sin^2 x$

$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos^2 x} = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{1 - \sin^2 x}$

$= \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{(1 - \sin x)(1 + \sin x)}$

$= \lim_{x \to \frac{\pi}{2}} \frac{1}{1 + \sin x}$

Substituting $x = \frac{\pi}{2}$ : $= \frac{1}{1 + 1} = \frac{1}{2}$

Q5.3 [3 marks]

Evaluate: $\lim_{x \to \infty} (1 + \frac{5}{x})^{2x}$

Answer:

Solution: Let $y = (1 + \frac{5}{x})^{2x}$

Taking natural logarithm: $\ln y = 2x \ln(1 + \frac{5}{x})$

$\lim_{x \to \infty} \ln y = \lim_{x \to \infty} 2x \ln(1 + \frac{5}{x})$

Let $t = \frac{5}{x}$ , then as $x \to \infty$ , $t \to 0$ and $x = \frac{5}{t}$

$= \lim_{t \to 0} 2 \cdot \frac{5}{t} \ln(1 + t) = \lim_{t \to 0} 10 \cdot \frac{\ln(1 + t)}{t}$

Using the standard limit $\lim_{t \to 0} \frac{\ln(1 + t)}{t} = 1$ :

$= 10 \times 1 = 10$

Therefore: $\lim_{x \to \infty} y = e^{10}$

Q.5(B) [8 marks]

Attempt any two

Q5.1 [4 marks]

Find the equation of the line passing through point $(2, 4)$ and perpendicular to line $5x - 7y + 11 = 0$

Answer:

Solution: Step 1: Find slope of given line $5x - 7y + 11 = 0$ $7y = 5x + 11$ $y = \frac{5}{7}x + \frac{11}{7}$ Slope of given line = $\frac{5}{7}$

Step 2: Find slope of perpendicular line For perpendicular lines: $m_1 \times m_2 = -1$ $\frac{5}{7} \times m_2 = -1$ $m_2 = -\frac{7}{5}$

Step 3: Use point-slope form $y - y_1 = m(x - x_1)$ $y - 4 = -\frac{7}{5}(x - 2)$ $y - 4 = -\frac{7}{5}x + \frac{14}{5}$ $y = -\frac{7}{5}x + \frac{14}{5} + 4$ $y = -\frac{7}{5}x + \frac{14 + 20}{5}$ $y = -\frac{7}{5}x + \frac{34}{5}$

Multiplying by 5: $5y = -7x + 34$ $7x + 5y - 34 = 0$

Q5.2 [4 marks]

If the equation of circle is $2x^2 + 2y^2 + 4x - 8y - 6 = 0$ then find its center and radius

Answer:

Solution: Step 1: Simplify by dividing by 2 $x^2 + y^2 + 2x - 4y - 3 = 0$

Step 2: Complete the square $(x^2 + 2x) + (y^2 - 4y) = 3$ $(x^2 + 2x + 1) + (y^2 - 4y + 4) = 3 + 1 + 4$ $(x + 1)^2 + (y - 2)^2 = 8$

Table: Circle Properties

Property	Value
Center	$(-1, 2)$
Radius	$\sqrt{8} = 2\sqrt{2}$

Q5.3 [4 marks]

Find the equation of tangent and normal of circle $x^2 + y^2 - 2x + 4y - 20 = 0$ at point $(-2, 2)$

Answer:

Solution: Step 1: Find center of circle $x^2 + y^2 - 2x + 4y - 20 = 0$ Completing the square: $(x^2 - 2x + 1) + (y^2 + 4y + 4) = 20 + 1 + 4$ $(x - 1)^2 + (y + 2)^2 = 25$

Center: $(1, -2)$ , Radius: $5$

Step 2: Find slope of radius to point $(-2, 2)$ $m_{radius} = \frac{2 - (-2)}{-2 - 1} = \frac{4}{-3} = -\frac{4}{3}$

Step 3: Find slope of tangent Tangent is perpendicular to radius: $m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}$

Step 4: Equation of tangent Using point-slope form at $(-2, 2)$ : $y - 2 = \frac{3}{4}(x - (-2))$ $y - 2 = \frac{3}{4}(x + 2)$ $4(y - 2) = 3(x + 2)$ $4y - 8 = 3x + 6$ $3x - 4y + 14 = 0$

Step 5: Equation of normal Normal has slope $m_{radius} = -\frac{4}{3}$ : $y - 2 = -\frac{4}{3}(x + 2)$ $3(y - 2) = -4(x + 2)$ $3y - 6 = -4x - 8$ $4x + 3y + 2 = 0$

Table: Line Equations

Line	Equation
Tangent	$3x - 4y + 14 = 0$
Normal	$4x + 3y + 2 = 0$

Mathematics Formula Cheat Sheet for Winter Exams

Determinants

2×2 Matrix: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
3×3 Matrix: Expand along row/column with most zeros
Properties: $|A| = 0$ if any row/column is zero

Functions

Composition: $(f \circ g)(x) = f(g(x))$
Even function: $f(-x) = f(x)$
Odd function: $f(-x) = -f(x)$

Logarithms

Basic properties:
- $\log_a a = 1$
- $\log 1 = 0$
- $\log x - \log y = \log \frac{x}{y}$
- $\log x + \log y = \log(xy)$
Change of base: $\frac{1}{\log_a b} = \log_b a$

Trigonometry

Periods

$\sin(ax + b)$ has period $\frac{2\pi}{|a|}$
$\cos(ax + b)$ has period $\frac{2\pi}{|a|}$
$\tan(ax + b)$ has period $\frac{\pi}{|a|}$

Angle Conversions

Degrees to radians: $\text{radians} = \text{degrees} \times \frac{\pi}{180}$

Inverse Trigonometric Identities

$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$
$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$
$\tan^{-1} a + \tan^{-1} b = \tan^{-1}(\frac{a+b}{1-ab})$ when $ab < 1$

Allied Angles

$\sin(\frac{\pi}{2} + \theta) = \cos \theta$
$\cos(\pi - \theta) = -\cos \theta$
$\tan(\pi - \theta) = -\tan \theta$
$\cot(\frac{3\pi}{2} - \theta) = \tan \theta$

Sum-to-Product Formulas

$\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$
$\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$

Vectors

Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
Dot Product: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$
Cross Product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Properties:
- $\vec{a} \times \vec{a} = 0$
- $\vec{a} \perp \vec{b}$ iff $\vec{a} \cdot \vec{b} = 0$
Work done: $W = \vec{F} \cdot \vec{d}$

Coordinate Geometry

Lines

Slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$
Two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
Perpendicular lines: $m_1 \times m_2 = -1$
Point-slope form: $y - y_1 = m(x - x_1)$

Circles

Standard form: $(x - h)^2 + (y - k)^2 = r^2$
General form: $x^2 + y^2 + 2gx + 2fy + c = 0$
Center: $(-g, -f)$ , Radius: $\sqrt{g^2 + f^2 - c}$
Tangent at point $(x_1, y_1)$ : $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$

Limits

Standard limits:
- $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
- $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
- $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$
- $\lim_{x \to \infty} (1 + \frac{a}{x})^x = e^a$
Rationalization: For expressions like $\sqrt{A} - \sqrt{B}$ , multiply by $\frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} + \sqrt{B}}$

Problem-Solving Strategies

For Function Problems

Check domain restrictions
Use algebraic manipulation for compositions
Verify results by substitution

For Logarithmic Proofs

Use change of base formula strategically
Convert complex expressions to simpler forms
Apply logarithm properties systematically

For Trigonometric Identities

Look for sum-to-product opportunities
Use allied angle formulas
Factor expressions when possible

For Vector Problems

Write vectors in component form
Use properties of dot and cross products
Check perpendicularity using dot product

For Limit Problems

Try direct substitution first
Factor and cancel for $\frac{0}{0}$ forms
Use rationalization for radical expressions
Apply standard limit formulas

For Circle Problems

Complete the square to find center and radius
Use slope relationships for tangent and normal
Remember tangent is perpendicular to radius

Common Mistakes to Avoid

Sign errors in determinant calculations
Forgetting domain restrictions in logarithmic functions
Angle measure confusion (degrees vs radians)
Not simplifying trigonometric expressions fully
Calculation errors in vector operations
Incomplete factorization in limit problems

Exam Success Tips

Show all working steps clearly
Verify answers when possible
Use proper mathematical notation
Draw diagrams for geometry problems
Manage time effectively across all questions

Best of luck with your Winter 2023 Mathematics exam! 🎯