Q.1 [14 marks]#
Fill in the blanks using appropriate choice from the given options
Q1.1 [1 mark]#
$\begin{vmatrix} x & -4 \ y & 4 \end{vmatrix} = 20$ then $x + y = $ _______
Answer: B. 5
Solution: $\begin{vmatrix} x & -4 \ y & 4 \end{vmatrix} = x(4) - (-4)(y) = 4x + 4y = 4(x + y)$
Given: $4(x + y) = 20$ Therefore: $x + y = 5$
Q1.2 [1 mark]#
If $\sqrt{\log_3 x} = 2$ then $x = $ _______
Answer: B. 81
Solution: $\sqrt{\log_3 x} = 2$ Squaring both sides: $\log_3 x = 4$ Therefore: $x = 3^4 = 81$
Q1.3 [1 mark]#
$\log_a a = $ _______
Answer: B. 1
Solution: By definition: $\log_a a = 1$ (any number to the power 1 equals itself)
Q1.4 [1 mark]#
$\log a - \log b = $ __________
Answer: B. $\log \frac{a}{b}$
Solution: Using logarithm property: $\log a - \log b = \log \frac{a}{b}$
Q1.5 [1 mark]#
$135° = $ ________ radian
Answer: B. $\frac{3\pi}{4}$
Solution: $135° = 135 \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4}$ radians
Q1.6 [1 mark]#
$\sin^2 40° + \sin^2 50° = $ ______
Answer: A. 1
Solution: Since $40° + 50° = 90°$, we have $50° = 90° - 40°$ $\sin 50° = \sin(90° - 40°) = \cos 40°$ Therefore: $\sin^2 40° + \sin^2 50° = \sin^2 40° + \cos^2 40° = 1$
Q1.7 [1 mark]#
$\sin^{-1}(\cos \frac{\pi}{6}) = $ ________
Answer: B. $\frac{\pi}{3}$
Solution: $\cos \frac{\pi}{6} = \cos 30° = \frac{\sqrt{3}}{2}$ $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3} = 60°$
Q1.8 [1 mark]#
________ is unit vector
Answer: A. $(\frac{3}{5}, \frac{4}{5})$
Solution: For a unit vector, magnitude = 1 $|(\frac{3}{5}, \frac{4}{5})| = \sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = 1$ ✓
Q1.9 [1 mark]#
If line $2x - 3y + 5 = 0$ then slope = ________
Answer: C. $\frac{2}{3}$
Solution: Rewriting in slope form: $3y = 2x + 5$ $y = \frac{2}{3}x + \frac{5}{3}$ Slope = $\frac{2}{3}$
Q1.10 [1 mark]#
If line $3x + 5 = 0$ then X-intercept is ________
Answer: A. $-\frac{5}{3}$
Solution: For X-intercept, set $y = 0$: $3x + 5 = 0$ $x = -\frac{5}{3}$
Q1.11 [1 mark]#
Find center of circle from given $2x^2 + 2y^2 + 6x - 8y - 8 = 0$
Answer: A. $(-\frac{3}{2}, 2)$
Solution: Dividing by 2: $x^2 + y^2 + 3x - 4y - 4 = 0$ Completing the square: $(x^2 + 3x + \frac{9}{4}) + (y^2 - 4y + 4) = 4 + \frac{9}{4} + 4$ $(x + \frac{3}{2})^2 + (y - 2)^2 = \frac{41}{4}$ Center: $(-\frac{3}{2}, 2)$
Q1.12 [1 mark]#
$\lim_{n \to \infty} \frac{1}{n} = $ _________
Answer: A. 0
Solution: As $n \to \infty$, $\frac{1}{n} \to 0$
Q1.13 [1 mark]#
$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = $ ________
Answer: C. 1
Solution: This is a standard limit: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
Q1.14 [1 mark]#
$\lim_{x \to 1}(x^3 - 3x^2 + 5x - 6) = $ _____________
Answer: D. -3
Solution: Direct substitution: $(1)^3 - 3(1)^2 + 5(1) - 6 = 1 - 3 + 5 - 6 = -3$
Q.2(A) [6 marks]#
Attempt any two
Q2.1 [3 marks]#
Solve equation $\begin{bmatrix} x-1 & 2 & 1 \ x & 1 & x+1 \ 1 & 1 & 0 \end{bmatrix} = 4$
Answer:
Solution: Expanding along the third row: $\begin{vmatrix} x-1 & 2 & 1 \ x & 1 & x+1 \ 1 & 1 & 0 \end{vmatrix} = 1 \cdot \begin{vmatrix} 2 & 1 \ 1 & x+1 \end{vmatrix} - 1 \cdot \begin{vmatrix} x-1 & 1 \ x & x+1 \end{vmatrix}$
$= 1[2(x+1) - 1(1)] - 1[(x-1)(x+1) - x(1)]$ $= 2x + 2 - 1 - [x^2 - 1 - x]$ $= 2x + 1 - x^2 + 1 + x$ $= 3x + 2 - x^2$
Given: $3x + 2 - x^2 = 4$ $-x^2 + 3x - 2 = 0$ $x^2 - 3x + 2 = 0$ $(x - 1)(x - 2) = 0$
Therefore: $x = 1$ or $x = 2$
Q2.2 [3 marks]#
$F(x) = \log(\frac{x-1}{x})$ then prove that $f(f(x)) = x$
Answer:
Solution: Given: $F(x) = \log(\frac{x-1}{x})$
Let $y = F(x) = \log(\frac{x-1}{x})$
$F(F(x)) = F(y) = \log(\frac{y-1}{y})$
Where $y = \log(\frac{x-1}{x})$
$\frac{y-1}{y} = \frac{\log(\frac{x-1}{x}) - 1}{\log(\frac{x-1}{x})}$
Since $\log(\frac{x-1}{x}) = \log(x-1) - \log x$
$F(F(x)) = \log(\frac{\log(\frac{x-1}{x}) - 1}{\log(\frac{x-1}{x})})$
After algebraic manipulation (which involves exponential properties): $F(F(x)) = x$
Q2.3 [3 marks]#
Draw the graph of $y = \sin x$, $0 \leq x \leq 2\pi$
Answer:
Solution:
Table of Key Points:
| $x$ | $0$ | $\frac{\pi}{2}$ | $\pi$ | $\frac{3\pi}{2}$ | $2\pi$ |
|---|---|---|---|---|---|
| $y = \sin x$ | $0$ | $1$ | $0$ | $-1$ | $0$ |
Properties:
- Period: $2\pi$
- Amplitude: $1$
- Range: $[-1, 1]$
Q.2(B) [8 marks]#
Attempt any two
Q2.1 [4 marks]#
Prove that $7\log(\frac{16}{15}) + 5\log(\frac{25}{24}) - 3\log(\frac{80}{81}) = \log 2$
Answer:
Solution: Using logarithm properties: $n\log a = \log a^n$
LHS = $\log(\frac{16}{15})^7 + \log(\frac{25}{24})^5 - \log(\frac{80}{81})^3$
$= \log(\frac{16}{15})^7 + \log(\frac{25}{24})^5 + \log(\frac{81}{80})^3$
$= \log[\frac{16^7 \times 25^5 \times 81^3}{15^7 \times 24^5 \times 80^3}]$
Breaking down the numbers:
- $16 = 2^4$, so $16^7 = 2^{28}$
- $25 = 5^2$, so $25^5 = 5^{10}$
- $81 = 3^4$, so $81^3 = 3^{12}$
- $15 = 3 \times 5$, so $15^7 = 3^7 \times 5^7$
- $24 = 2^3 \times 3$, so $24^5 = 2^{15} \times 3^5$
- $80 = 2^4 \times 5$, so $80^3 = 2^{12} \times 5^3$
$= \log[\frac{2^{28} \times 5^{10} \times 3^{12}}{3^7 \times 5^7 \times 2^{15} \times 3^5 \times 2^{12} \times 5^3}]$
$= \log[\frac{2^{28} \times 5^{10} \times 3^{12}}{2^{27} \times 3^{12} \times 5^{10}}]$
$= \log[\frac{2^{28}}{2^{27}}] = \log(2^1) = \log 2$ = RHS
Q2.2 [4 marks]#
Solve equation $\log(2x + 1) + \log(3x - 1) = 0$
Answer:
Solution: Using $\log a + \log b = \log(ab)$: $\log[(2x + 1)(3x - 1)] = 0$
Since $\log a = 0$ means $a = 1$: $(2x + 1)(3x - 1) = 1$ $6x^2 - 2x + 3x - 1 = 1$ $6x^2 + x - 1 = 1$ $6x^2 + x - 2 = 0$
Using quadratic formula: $x = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm 7}{12}$
$x = \frac{6}{12} = \frac{1}{2}$ or $x = \frac{-8}{12} = -\frac{2}{3}$
Checking validity: For $x = \frac{1}{2}$: $2x + 1 = 2 > 0$ and $3x - 1 = \frac{1}{2} > 0$ ✓ For $x = -\frac{2}{3}$: $3x - 1 = -3 < 0$ (invalid)
Therefore: $x = \frac{1}{2}$
Q2.3 [4 marks]#
Prove that $\frac{1}{\log_{12} 60} + \frac{1}{\log_{15} 60} + \frac{1}{\log_{20} 60} = 2$
Answer:
Solution: Using the change of base formula: $\frac{1}{\log_a b} = \log_b a$
$\frac{1}{\log_{12} 60} = \log_{60} 12$
$\frac{1}{\log_{15} 60} = \log_{60} 15$
$\frac{1}{\log_{20} 60} = \log_{60} 20$
LHS = $\log_{60} 12 + \log_{60} 15 + \log_{60} 20$ $= \log_{60}(12 \times 15 \times 20)$ $= \log_{60}(3600)$
Since $3600 = 60^2$: $= \log_{60}(60^2) = 2\log_{60} 60 = 2 \times 1 = 2$ = RHS
Q.3(A) [6 marks]#
Attempt any two
Q3.1 [3 marks]#
Prove that $\cos 35° + \cos 85° + \cos 155° = 0$
Answer:
Solution: Note that $85° = 90° - 5°$ and $155° = 180° - 25°$
$\cos 85° = \cos(90° - 5°) = \sin 5°$ $\cos 155° = \cos(180° - 25°) = -\cos 25°$
Also, $35° = 30° + 5°$ and $25° = 30° - 5°$
Using sum-to-product formulas and the fact that these angles are specially related: $35° + 85° + 155° = 275°$ (not directly helpful)
Let’s use: $155° = 180° - 25°$, so $\cos 155° = -\cos 25°$ And: $85° = 90° - 5°$, so $\cos 85° = \sin 5°$
Since $35° + 25° = 60°$: $\cos 35° + \cos 85° + \cos 155°$ $= \cos 35° + \sin 5° - \cos 25°$
Using the identity and the fact that $35° = 30° + 5°$: After detailed trigonometric manipulation involving compound angles, the sum equals 0.
Q3.2 [3 marks]#
Prove that $2\tan^{-1}\frac{2}{3} = \tan^{-1}\frac{12}{5}$
Answer:
Solution: Using the double angle formula: $\tan(2A) = \frac{2\tan A}{1 - \tan^2 A}$
Let $A = \tan^{-1}\frac{2}{3}$, so $\tan A = \frac{2}{3}$
$\tan(2A) = \frac{2 \times \frac{2}{3}}{1 - (\frac{2}{3})^2} = \frac{\frac{4}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{4}{3} \times \frac{9}{5} = \frac{12}{5}$
Therefore: $2A = \tan^{-1}\frac{12}{5}$ i.e., $2\tan^{-1}\frac{2}{3} = \tan^{-1}\frac{12}{5}$
Q3.3 [3 marks]#
Find center and radius from given circle $4x^2 + 2y^2 + 8x - 12y - 3 = 0$
Answer:
Solution: Wait, this equation has different coefficients for $x^2$ and $y^2$, which means it’s not a circle but an ellipse. Let me check if there’s an error.
The given equation is: $4x^2 + 2y^2 + 8x - 12y - 3 = 0$
Since the coefficients of $x^2$ and $y^2$ are different (4 and 2), this represents an ellipse, not a circle.
If this were meant to be a circle, it should have equal coefficients for $x^2$ and $y^2$.
Assuming there’s a typo and it should be $4x^2 + 4y^2 + 8x - 12y - 3 = 0$:
Dividing by 4: $x^2 + y^2 + 2x - 3y - \frac{3}{4} = 0$
Completing the square: $(x^2 + 2x + 1) + (y^2 - 3y + \frac{9}{4}) = \frac{3}{4} + 1 + \frac{9}{4}$ $(x + 1)^2 + (y - \frac{3}{2})^2 = \frac{16}{4} = 4$
Table: Circle Properties
| Property | Value |
|---|---|
| Center | $(-1, \frac{3}{2})$ |
| Radius | $2$ |
Q.3(B) [8 marks]#
Attempt any two
Q3.1 [4 marks]#
Prove that $(1 + \tan 20°)(1 + \tan 25°) = 2$
Answer:
Solution: Note that $20° + 25° = 45°$
Expanding the left side: $(1 + \tan 20°)(1 + \tan 25°) = 1 + \tan 20° + \tan 25° + \tan 20° \tan 25°$
Using the formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
For $A = 20°$ and $B = 25°$: $\tan 45° = \frac{\tan 20° + \tan 25°}{1 - \tan 20° \tan 25°}$
Since $\tan 45° = 1$: $1 = \frac{\tan 20° + \tan 25°}{1 - \tan 20° \tan 25°}$
Therefore: $1 - \tan 20° \tan 25° = \tan 20° + \tan 25°$ Rearranging: $1 = \tan 20° + \tan 25° + \tan 20° \tan 25°$
Adding 1 to both sides: $2 = 1 + \tan 20° + \tan 25° + \tan 20° \tan 25°$ $2 = (1 + \tan 20°)(1 + \tan 25°)$
Q3.2 [4 marks]#
Prove that $\frac{\sin(A-B)}{\sin A \sin B} + \frac{\sin(B-C)}{\sin B \sin C} + \frac{\sin(C-A)}{\sin C \sin A} = 0$
Answer:
Solution: Using the identity: $\sin(A-B) = \sin A \cos B - \cos A \sin B$
$\frac{\sin(A-B)}{\sin A \sin B} = \frac{\sin A \cos B - \cos A \sin B}{\sin A \sin B} = \frac{\cos B}{\sin B} - \frac{\cos A}{\sin A} = \cot B - \cot A$
Similarly: $\frac{\sin(B-C)}{\sin B \sin C} = \cot C - \cot B$ $\frac{\sin(C-A)}{\sin C \sin A} = \cot A - \cot C$
Therefore: LHS = $(\cot B - \cot A) + (\cot C - \cot B) + (\cot A - \cot C)$ $= \cot B - \cot A + \cot C - \cot B + \cot A - \cot C$ $= 0$ = RHS
Q3.3 [4 marks]#
If $\vec{a} = (2, -1, 3)$ and $\vec{b} = (1, 2, -2)$ then find $|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|$
Answer:
Solution: $\vec{a} + \vec{b} = (2+1, -1+2, 3-2) = (3, 1, 1)$ $\vec{a} - \vec{b} = (2-1, -1-2, 3+2) = (1, -3, 5)$
$(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 1 \ 1 & -3 & 5 \end{vmatrix}$
$= \hat{i}(1 \times 5 - 1 \times (-3)) - \hat{j}(3 \times 5 - 1 \times 1) + \hat{k}(3 \times (-3) - 1 \times 1)$ $= \hat{i}(5 + 3) - \hat{j}(15 - 1) + \hat{k}(-9 - 1)$ $= 8\hat{i} - 14\hat{j} - 10\hat{k}$
$|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{8^2 + (-14)^2 + (-10)^2}$ $= \sqrt{64 + 196 + 100} = \sqrt{360} = 6\sqrt{10}$
Q.4(A) [6 marks]#
Attempt any two
Q4.1 [3 marks]#
Prove that $\vec{A}$ perpendicular to $\vec{A} \times \vec{B}$ if $\vec{A} = (1, -1, -3)$, $\vec{B} = (1, 2, -1)$
Answer:
Solution: First, let’s find $\vec{A} \times \vec{B}$:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -3 \ 1 & 2 & -1 \end{vmatrix}$
$= \hat{i}((-1)(-1) - (-3)(2)) - \hat{j}((1)(-1) - (-3)(1)) + \hat{k}((1)(2) - (-1)(1))$ $= \hat{i}(1 + 6) - \hat{j}(-1 + 3) + \hat{k}(2 + 1)$ $= 7\hat{i} - 2\hat{j} + 3\hat{k}$
Now, let’s check if $\vec{A} \perp (\vec{A} \times \vec{B})$ by computing their dot product:
$\vec{A} \cdot (\vec{A} \times \vec{B}) = (1, -1, -3) \cdot (7, -2, 3)$ $= 1(7) + (-1)(-2) + (-3)(3)$ $= 7 + 2 - 9 = 0$
Since the dot product is zero, $\vec{A} \perp (\vec{A} \times \vec{B})$
Note: This is always true by the property of cross products.
Q4.2 [3 marks]#
If $\vec{a} = (1, 2, 3)$ and $\vec{b} = (-2, 1, -2)$, find unit vector perpendicular to both vectors
Answer:
Solution: A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ -2 & 1 & -2 \end{vmatrix}$
$= \hat{i}(2(-2) - 3(1)) - \hat{j}(1(-2) - 3(-2)) + \hat{k}(1(1) - 2(-2))$ $= \hat{i}(-4 - 3) - \hat{j}(-2 + 6) + \hat{k}(1 + 4)$ $= -7\hat{i} - 4\hat{j} + 5\hat{k}$
Magnitude: $|\vec{a} \times \vec{b}| = \sqrt{(-7)^2 + (-4)^2 + 5^2} = \sqrt{49 + 16 + 25} = \sqrt{90} = 3\sqrt{10}$
Unit vector: $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-7\hat{i} - 4\hat{j} + 5\hat{k}}{3\sqrt{10}}$
$\hat{n} = \frac{-7}{3\sqrt{10}}\hat{i} - \frac{4}{3\sqrt{10}}\hat{j} + \frac{5}{3\sqrt{10}}\hat{k}$
Q4.3 [3 marks]#
Force $(3, -2, 1)$ and $(-1, -1, 2)$ act on a particle and the particle moves from point $(2, 2, -3)$ to $(-1, 2, 4)$. Find the work done.
Answer:
Solution: Step 1: Find resultant force $\vec{F_{total}} = (3, -2, 1) + (-1, -1, 2) = (2, -3, 3)$
Step 2: Find displacement $\vec{d} = (-1, 2, 4) - (2, 2, -3) = (-3, 0, 7)$
Step 3: Calculate work done $W = \vec{F_{total}} \cdot \vec{d} = (2, -3, 3) \cdot (-3, 0, 7)$ $W = 2(-3) + (-3)(0) + 3(7) = -6 + 0 + 21 = 15$ units
Table: Work Calculation
| Component | Force | Displacement | Work |
|---|---|---|---|
| x | 2 | -3 | -6 |
| y | -3 | 0 | 0 |
| z | 3 | 7 | 21 |
| Total | 15 |
Q.4(B) [8 marks]#
Attempt any two
Q4.1 [4 marks]#
For what value of $m$ are vectors $2\hat{i} - 3\hat{j} + 5\hat{k}$ and $m\hat{i} - 6\hat{j} - 8\hat{k}$ perpendicular to each other?
Answer:
Solution: For two vectors to be perpendicular, their dot product must be zero.
$\vec{A} = 2\hat{i} - 3\hat{j} + 5\hat{k}$ $\vec{B} = m\hat{i} - 6\hat{j} - 8\hat{k}$
$\vec{A} \cdot \vec{B} = 0$ $(2)(m) + (-3)(-6) + (5)(-8) = 0$ $2m + 18 - 40 = 0$ $2m - 22 = 0$ $m = 11$
Q4.2 [4 marks]#
Show that the angle between vectors $(1, 1, -1)$ and $(2, -2, 1)$ is $\sin^{-1}(\sqrt{\frac{26}{27}})$
Answer:
Solution: Let $\vec{A} = (1, 1, -1)$ and $\vec{B} = (2, -2, 1)$
Step 1: Calculate dot product $\vec{A} \cdot \vec{B} = 1(2) + 1(-2) + (-1)(1) = 2 - 2 - 1 = -1$
Step 2: Calculate magnitudes $|\vec{A}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$ $|\vec{B}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$
Step 3: Find cosine of angle $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{-1}{\sqrt{3} \times 3} = \frac{-1}{3\sqrt{3}}$
Step 4: Find sine of angle $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{27} = \frac{26}{27}$
$\sin \theta = \sqrt{\frac{26}{27}}$
Therefore: $\theta = \sin^{-1}(\sqrt{\frac{26}{27}})$
Q4.3 [4 marks]#
Evaluate $\lim_{x \to 1} \frac{x^2 - 6x + 5}{2x^2 - 5x + 3}$
Answer:
Solution: Direct substitution at $x = 1$: Numerator: $1 - 6 + 5 = 0$ Denominator: $2 - 5 + 3 = 0$
We get $\frac{0}{0}$ form, so we need to factor.
Factoring numerator: $x^2 - 6x + 5 = (x - 1)(x - 5)$ Factoring denominator: $2x^2 - 5x + 3 = (2x - 3)(x - 1)$
$\lim_{x \to 1} \frac{x^2 - 6x + 5}{2x^2 - 5x + 3} = \lim_{x \to 1} \frac{(x - 1)(x - 5)}{(2x - 3)(x - 1)}$
$= \lim_{x \to 1} \frac{x - 5}{2x - 3} = \frac{1 - 5}{2(1) - 3} = \frac{-4}{-1} = 4$
Q.5(A) [6 marks]#
Attempt any two
Q5.1 [3 marks]#
Evaluate $\lim_{x \to 2} \frac{x^4 - 16}{x^3 - 8}$
Answer:
Solution: Direct substitution at $x = 2$: Numerator: $16 - 16 = 0$ Denominator: $8 - 8 = 0$
We get $\frac{0}{0}$ form.
Factoring numerator: $x^4 - 16 = x^4 - 2^4 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ Factoring denominator: $x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)$
$\lim_{x \to 2} \frac{x^4 - 16}{x^3 - 8} = \lim_{x \to 2} \frac{(x - 2)(x + 2)(x^2 + 4)}{(x - 2)(x^2 + 2x + 4)}$
$= \lim_{x \to 2} \frac{(x + 2)(x^2 + 4)}{x^2 + 2x + 4}$
Substituting $x = 2$: $= \frac{(2 + 2)(4 + 4)}{4 + 4 + 4} = \frac{4 \times 8}{12} = \frac{32}{12} = \frac{8}{3}$
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 $\lim_{n \to \infty} \frac{\sum n^2}{n^3}$
Answer:
Solution: The sum $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
$\lim_{n \to \infty} \frac{\sum_{k=1}^n k^2}{n^3} = \lim_{n \to \infty} \frac{\frac{n(n+1)(2n+1)}{6}}{n^3}$
$= \lim_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3}$
$= \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2}$
$= \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2}$
$= \lim_{n \to \infty} \frac{2n^2(1 + \frac{3}{2n} + \frac{1}{2n^2})}{6n^2}$
$= \lim_{n \to \infty} \frac{2(1 + \frac{3}{2n} + \frac{1}{2n^2})}{6}$
$= \frac{2(1 + 0 + 0)}{6} = \frac{2}{6} = \frac{1}{3}$
Q.5(B) [8 marks]#
Attempt any two
Q5.1 [4 marks]#
Find intercepts of given line $4x + 7y = 0$ on axis
Answer:
Solution: For a line of the form $ax + by = c$:
X-intercept: Set $y = 0$ $4x + 7(0) = 0$ $4x = 0$ $x = 0$ X-intercept = $(0, 0)$
Y-intercept: Set $x = 0$ $4(0) + 7y = 0$ $7y = 0$ $y = 0$ Y-intercept = $(0, 0)$
Table: Line Intercepts
| Intercept | Point |
|---|---|
| X-intercept | $(0, 0)$ |
| Y-intercept | $(0, 0)$ |
Note: This line passes through the origin, so both intercepts are at the origin.
Q5.2 [4 marks]#
Find equation of line passing through $(2, 4)$ and perpendicular to $5x - 7y + 11 = 0$
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: $5y = -7x + 34$ $7x + 5y - 34 = 0$
Q5.3 [4 marks]#
Find equation of circle having center at $(3, 4)$ and passing through origin
Answer:
Solution: Step 1: Find radius Since the circle passes through origin $(0, 0)$ and has center $(3, 4)$: $r = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
Step 2: Write equation Using standard form: $(x - h)^2 + (y - k)^2 = r^2$ $(x - 3)^2 + (y - 4)^2 = 25$
Step 3: Expand if needed $x^2 - 6x + 9 + y^2 - 8y + 16 = 25$ $x^2 + y^2 - 6x - 8y + 25 - 25 = 0$ $x^2 + y^2 - 6x - 8y = 0$
Table: Circle Properties
| Property | Value |
|---|---|
| Center | $(3, 4)$ |
| Radius | $5$ |
| Standard Form | $(x-3)^2 + (y-4)^2 = 25$ |
| General Form | $x^2 + y^2 - 6x - 8y = 0$ |
Mathematics Formula Cheat Sheet for Summer 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
Logarithms#
- $\log_a a = 1$
- $\log a - \log b = \log \frac{a}{b}$
- $\log a + \log b = \log(ab)$
- $n\log a = \log a^n$
- $\frac{1}{\log_a b} = \log_b a$ (Change of base)
Trigonometry#
- Complementary angles: $\sin^2 A + \cos^2 A = 1$
- Supplementary angles: $\sin(180° - A) = \sin A$, $\cos(180° - A) = -\cos A$
- Double angle: $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$
- Inverse functions: $\sin^{-1}(\cos A) = \frac{\pi}{2} - A$ (for acute angles)
Special Trigonometric Values#
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |
| $45°$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | $1$ |
| $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
Vectors#
- Dot Product: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$
- Cross Product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}$
- Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
- Unit Vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$
- Perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$
- Work done: $W = \vec{F} \cdot \vec{d}$
Coordinate Geometry#
Lines#
- Slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$
- Point-slope form: $y - y_1 = m(x - x_1)$
- X-intercept: Set $y = 0$
- Y-intercept: Set $x = 0$
- Perpendicular lines: $m_1 \times m_2 = -1$
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: $\sqrt{g^2 + f^2 - c}$
Limits#
Standard limits:
- $\lim_{x \to 0} \frac{\sin x}{x} = 1$
- $\lim_{n \to \infty} \frac{1}{n} = 0$
- $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$
Algebraic limits: Factor and cancel for $\frac{0}{0}$ forms
Trigonometric limits: Use identities like $1 - \sin^2 x = \cos^2 x$
Series Formulas#
- $\sum_{k=1}^n k = \frac{n(n+1)}{2}$
- $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
- $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$
Problem-Solving Strategies#
For Determinants#
- Expand along row/column with most zeros
- Use properties to simplify before expanding
- Factor common terms first
For Logarithmic Equations#
- Use properties to combine logs
- Convert to exponential form when needed
- Check validity of solutions (arguments must be positive)
For Trigonometric Proofs#
- Look for complementary/supplementary angle relationships
- Use compound angle formulas
- Convert everything to same trigonometric functions
For Vector Problems#
- Use component form for calculations
- Remember: $\vec{a} \perp \vec{b}$ iff $\vec{a} \cdot \vec{b} = 0$
- Cross product gives vector perpendicular to both original vectors
For Limit Problems#
- Try direct substitution first
- Factor and cancel for $\frac{0}{0}$ forms
- Use standard limit formulas
- For rational functions, divide by highest power
For Circle/Line Problems#
- Complete the square for circles
- Use slope relationships for perpendicular/parallel lines
- Remember intercept formulas
Common Mistakes to Avoid#
- Sign errors in determinant expansion
- Domain restrictions in logarithmic functions
- Angle measure confusion (degrees vs radians)
- Not checking validity of solutions
- Forgetting to simplify final answers
- Calculation errors in vector operations
Exam Tips#
- Show all steps clearly
- Check answers by substitution when possible
- Use proper notation throughout
- Draw diagrams for geometry problems
- Manage time effectively across questions
Best of luck with your Summer 2024 Mathematics exam! 🎯