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SECTION A

Question 1

Given that  $$\sec \theta = \frac{25}{24} $$, compute the value of  $$\cot \theta + \cos \theta $$.

Solution: First, let’s find  $$\cos \theta $$ using the relationship  $$\sec \theta = \frac{1}{\cos \theta} $$:

 $$\sec \theta = \frac{25}{24} $$

 $$\cos \theta = \frac{1}{\sec \theta} = \frac{24}{25} $$

Now, to find  $$\cot \theta $$, we need  $$\sin \theta $$. Using the Pythagorean identity:  $$\sin^2 \theta + \cos^2 \theta = 1 $$

 $$\sin^2 \theta = 1 – \cos^2 \theta = 1 – \left(\frac{24}{25}\right)^2 = 1 – \frac{576}{625} = \frac{625-576}{625} = \frac{49}{625} $$

 $$\sin \theta = \frac{7}{25} $$ (taking the positive value since  $$\sec \theta $$ is positive, suggesting we’re in the first quadrant)

Now we can calculate  $$\cot \theta $$:  $$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{24/25}{7/25} = \frac{24}{7} $$

Finally, we compute  $$\cot \theta + \cos \theta $$:  $$\cot \theta + \cos \theta = \frac{24}{7} + \frac{24}{25} = \frac{24 \cdot 25 + 24 \cdot 7}{7 \cdot 25} = \frac{24(25+7)}{175} = \frac{24 \cdot 32}{175} = \frac{768}{175} $$

Therefore,  $$\cot \theta + \cos \theta = \frac{768}{175} $$

Question 2

If  $$\alpha $$ and  $$\beta $$ are the roots of the equation  $$x^2 – 5x + 6 = 0 $$, find  $$\alpha – \beta $$.

Solution: From Vieta’s formulas, for a quadratic equation  $$ax^2 + bx + c = 0 $$, the sum and product of roots are given by:

•  $$\alpha + \beta = -\frac{b}{a} $$
•  $$\alpha \cdot \beta = \frac{c}{a} $$

For  $$x^2 – 5x + 6 = 0 $$, we have  $$a = 1 $$,  $$b = -5 $$, and  $$c = 6 $$.

Therefore:

•  $$\alpha + \beta = -\frac{-5}{1} = 5 $$
•  $$\alpha \cdot \beta = \frac{6}{1} = 6 $$

To find  $$\alpha – \beta $$, we use the relationship:  $$(\alpha – \beta)^2 = (\alpha + \beta)^2 – 4\alpha\beta $$

 $$(\alpha – \beta)^2 = 5^2 – 4 \cdot 6 = 25 – 24 = 1 $$

 $$\alpha – \beta = \pm 1 $$

Since we’re just asked for  $$\alpha – \beta $$ without specifying which is which, either value is valid. Conventionally, if we consider  $$\alpha > \beta $$, then  $$\alpha – \beta = 1 $$.

Question 3

Express  $$\left(\frac{2z+1}{z-1}\right)^2 $$ in the form  $$a + ib $$.

Solution: Let’s denote the complex expression as  $$\left(\frac{2z+1}{z-1}\right)^2 $$.

We need to simplify this expression. First, let’s work with  $$\frac{2z+1}{z-1} $$:

 $$\frac{2z+1}{z-1} = \frac{2z-2+2+1}{z-1} = \frac{2(z-1)+3}{z-1} = 2 + \frac{3}{z-1} $$

Now we need to square this result:  $$\left(2 + \frac{3}{z-1}\right)^2 = 4 + \frac{12}{z-1} + \frac{9}{(z-1)^2} $$

This is a general form. To express in the form  $$a + ib $$, we would need specific values of  $$z $$. However, since we’re not given specific values, we can leave it in this algebraic form.

Note: The question likely intended to substitute a specific complex value for  $$z $$, but it appears to be missing from the image. Without a specific value of  $$z $$, we cannot express the final answer in the exact form  $$a + ib $$.

Question 4

If the 16th term of an AP is 3 times the 4th term, prove that the 23rd term is 5 times the 3rd term.

Solution: Let’s denote the first term of the AP by  $$a $$ and the common difference by  $$d $$.

The  $$n $$-th term of an AP is given by  $$T_n = a + (n-1)d $$.

Given information:  $$T_{16} = 3T_4 $$

Let’s substitute:  $$a + (16-1)d = 3[a + (4-1)d] $$  $$a + 15d = 3(a + 3d) $$  $$a + 15d = 3a + 9d $$  $$a – 3a = 9d – 15d $$  $$-2a = -6d $$  $$a = 3d $$

Now, we need to prove that  $$T_{23} = 5T_3 $$:

 $$T_{23} = a + (23-1)d = a + 22d = 3d + 22d = 25d $$  $$T_3 = a + (3-1)d = a + 2d = 3d + 2d = 5d $$

Therefore:  $$T_{23} = 25d = 5 \cdot 5d = 5T_3 $$

Thus, the 23rd term is 5 times the 3rd term.

Question 5

Solve the equation  $$1^x = 3 $$.

Solution: We need to solve  $$1^x = 3 $$.

Note that  $$1 $$ raised to any power is always equal to  $$1 $$:  $$1^x = 1 $$ for any value of  $$x $$.

However, our equation states  $$1^x = 3 $$, which is impossible since  $$1 $$ raised to any power cannot equal  $$3 $$.

Therefore, the equation  $$1^x = 3 $$ has no solution.

SECTION B: ALGEBRA

Question 6(a)

Given that  $$x – 1 $$ is a factor of the polynomial  $$px^2 + qx + r $$, and when the polynomial is divided by  $$x + 1 $$ and  $$x – 2 $$ leaves remainders of  $$2 $$ and  $$8 $$, respectively. Calculate the values of  $$p $$,  $$q $$ and  $$r $$.

Solution: If  $$x – 1 $$ is a factor of  $$px^2 + qx + r $$, then  $$p(1)^2 + q(1) + r = 0 $$, which simplifies to:  $$p + q + r = 0 $$ … (1)

When  $$px^2 + qx + r $$ is divided by  $$x + 1 $$, the remainder is  $$2 $$. By the remainder theorem, if  $$x = -1 $$, then  $$p(-1)^2 + q(-1) + r = 2 $$:  $$p – q + r = 2 $$ … (2)

When  $$px^2 + qx + r $$ is divided by  $$x – 2 $$, the remainder is  $$8 $$. By the remainder theorem, if  $$x = 2 $$, then  $$p(2)^2 + q(2) + r = 8 $$:  $$4p + 2q + r = 8 $$ … (3)

From equations (1) and (2):  $$p + q + r = 0 $$ … (1)  $$p – q + r = 2 $$ … (2)

Subtracting (2) from (1):  $$2q = -2 $$  $$q = -1 $$

Substituting  $$q = -1 $$ into equation (1):  $$p + (-1) + r = 0 $$  $$p + r = 1 $$ … (4)

Now from equations (3) and (4):  $$4p + 2(-1) + r = 8 $$  $$4p – 2 + r = 8 $$  $$4p + r = 10 $$ … (5)

Subtracting (4) from (5):  $$3p = 9 $$  $$p = 3 $$

Substituting  $$p = 3 $$ into equation (4):  $$3 + r = 1 $$  $$r = -2 $$

Therefore:  $$p = 3 $$,  $$q = -1 $$, and  $$r = -2 $$

Question 6(b)

Obtain the first four terms of the binomial expansion of  $$(x – y)^{10} $$ and use it to estimate the value of  $$(0.99)^{10} $$, correct to 3 decimal points.

Solution: Using the binomial theorem:

 $$(x – y)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^{10-k} (-y)^k $$

The first four terms are:  $$(x – y)^{10} = \binom{10}{0} x^{10} (-y)^0 + \binom{10}{1} x^9 (-y)^1 + \binom{10}{2} x^8 (-y)^2 + \binom{10}{3} x^7 (-y)^3 + … $$

 $$= x^{10} – 10x^9y + 45x^8y^2 – 120x^7y^3 + … $$

To estimate  $$(0.99)^{10} $$, we can rewrite it as  $$(1 – 0.01)^{10} $$.

Setting  $$x = 1 $$ and  $$y = 0.01 $$ in our expansion:

 $$(0.99)^{10} = (1 – 0.01)^{10} = 1^{10} – 10 \cdot 1^9 \cdot 0.01 + 45 \cdot 1^8 \cdot (0.01)^2 – 120 \cdot 1^7 \cdot (0.01)^3 + … $$

 $$= 1 – 0.1 + 0.0045 – 0.00012 + … $$

 $$= 0.90438… \approx 0.904 $$ (to 3 decimal places)

Question 7(a)

Prove that  $$2^{4n} – 1 $$ is always divisible by 5.

Solution: We can prove this by showing that  $$2^{4n} \equiv 1 \pmod{5} $$ for all positive integers  $$n $$.

Let’s examine the powers of 2 modulo 5:  $$2^1 \equiv 2 \pmod{5} $$  $$2^2 \equiv 4 \pmod{5} $$  $$2^3 \equiv 8 \equiv 3 \pmod{5} $$  $$2^4 \equiv 16 \equiv 1 \pmod{5} $$

We observe that  $$2^4 \equiv 1 \pmod{5} $$, which means the pattern repeats every 4 powers.

For any positive integer  $$n $$:  $$2^{4n} = (2^4)^n \equiv 1^n \equiv 1 \pmod{5} $$

Therefore:  $$2^{4n} – 1 \equiv 1 – 1 \equiv 0 \pmod{5} $$

This proves that  $$2^{4n} – 1 $$ is always divisible by 5 for all positive integers  $$n $$.

Question 7(b)

Solve the following simultaneous linear equations using Cramer’s rule:  $$x + 2y – 2z = 1 $$  $$x – 3y + 8z = 6 $$  $$2x – 3y – z = -2 $$

Solution: Cramer’s rule states that for a system of equations  $$AX = B $$, each variable  $$x_i $$ can be calculated as  $$x_i = \frac{|A_i|}{|A|} $$ where  $$|A| $$ is the determinant of coefficient matrix, and  $$|A_i| $$ is the determinant of matrix  $$A $$ with its  $$i $$-th column replaced by the constant column  $$B $$.

For our system, we have:  $$A = \begin{pmatrix} 1 & 2 & -2 \ 1 & -3 & 8 \ 2 & -3 & -1 \end{pmatrix} $$

 $$B = \begin{pmatrix} 1 \ 6 \ -2 \end{pmatrix} $$

First, calculate  $$|A| $$:  $$|A| = 1 \cdot \begin{vmatrix} -3 & 8 \ -3 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 1 & 8 \ 2 & -1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & -3 \ 2 & -3 \end{vmatrix} $$

 $$= 1 \cdot ((-3)(-1) – 8(-3)) – 2 \cdot (1(-1) – 8(2)) + (-2) \cdot (1(-3) – (-3)(2)) $$

 $$= 1 \cdot (3 + 24) – 2 \cdot (-1 – 16) + (-2) \cdot (-3 – (-6)) $$

 $$= 1 \cdot 27 – 2 \cdot (-17) + (-2) \cdot 3 $$

 $$= 27 + 34 – 6 = 55 $$

Next, calculate  $$|A_x| $$ (replace first column with constants):  $$|A_x| = \begin{vmatrix} 1 & 2 & -2 \ 6 & -3 & 8 \ -2 & -3 & -1 \end{vmatrix} $$

 $$= 1 \cdot \begin{vmatrix} -3 & 8 \ -3 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 6 & 8 \ -2 & -1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 6 & -3 \ -2 & -3 \end{vmatrix} $$

 $$= 1 \cdot (3 + 24) – 2 \cdot (-6 – (-16)) + (-2) \cdot (-18 – 6) $$

 $$= 27 – 2 \cdot 10 + (-2) \cdot (-24) $$

 $$= 27 – 20 + 48 = 55 $$

Calculate  $$|A_y| $$ (replace second column with constants):  $$|A_y| = \begin{vmatrix} 1 & 1 & -2 \ 1 & 6 & 8 \ 2 & -2 & -1 \end{vmatrix} $$

 $$= 1 \cdot \begin{vmatrix} 6 & 8 \ -2 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 8 \ 2 & -1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & 6 \ 2 & -2 \end{vmatrix} $$

 $$= 1 \cdot (-6 – (-16)) – 1 \cdot (-1 – 16) + (-2) \cdot (-2 – 12) $$

 $$= 1 \cdot 10 – 1 \cdot (-17) + (-2) \cdot (-14) $$

 $$= 10 + 17 + 28 = 55 $$

Calculate  $$|A_z| $$ (replace third column with constants):  $$|A_z| = \begin{vmatrix} 1 & 2 & 1 \ 1 & -3 & 6 \ 2 & -3 & -2 \end{vmatrix} $$

 $$= 1 \cdot \begin{vmatrix} -3 & 6 \ -3 & -2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 1 & 6 \ 2 & -2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -3 \ 2 & -3 \end{vmatrix} $$

 $$= 1 \cdot (6 – (-18)) – 2 \cdot (-2 – 12) + 1 \cdot (-3 – (-6)) $$

 $$= 1 \cdot 24 – 2 \cdot (-14) + 1 \cdot 3 $$

 $$= 24 + 28 + 3 = 55 $$

Using Cramer’s rule:  $$x = \frac{|A_x|}{|A|} = \frac{55}{55} = 1 $$  $$y = \frac{|A_y|}{|A|} = \frac{55}{55} = 1 $$  $$z = \frac{|A_z|}{|A|} = \frac{55}{55} = 1 $$

Therefore, the solution is  $$x = 1 $$,  $$y = 1 $$,  $$z = 1 $$.

Question 8(a)

Solve the equation  $$\log_5 (x^2 + 9) – \log_5 (x + 1) = 1 $$

Solution: Starting with the equation  $$\log_5 (x^2 + 9) – \log_5 (x + 1) = 1 $$

Using the logarithm property  $$\log_a(m) – \log_a(n) = \log_a(\frac{m}{n}) $$:

 $$\log_5 \left(\frac{x^2 + 9}{x + 1}\right) = 1 $$

Since  $$\log_5(5) = 1 $$, we can rewrite:

 $$\frac{x^2 + 9}{x + 1} = 5 $$

Multiply both sides by  $$(x + 1) $$:  $$x^2 + 9 = 5(x + 1) $$  $$x^2 + 9 = 5x + 5 $$  $$x^2 – 5x + 4 = 0 $$

Using the quadratic formula:  $$x = \frac{5 \pm \sqrt{25 – 16}}{2} = \frac{5 \pm 3}{2} $$

 $$x = 4 $$ or  $$x = 1 $$

We need to check if these solutions satisfy the original equation by ensuring the expressions inside the logarithms are positive:

For  $$x = 4 $$:  $$x^2 + 9 = 16 + 9 = 25 > 0 $$ ✓  $$x + 1 = 4 + 1 = 5 > 0 $$ ✓

For  $$x = 1 $$:  $$x^2 + 9 = 1 + 9 = 10 > 0 $$ ✓  $$x + 1 = 1 + 1 = 2 > 0 $$ ✓

Both solutions are valid:  $$x = 4 $$ or  $$x = 1 $$

Question 8(b)

Solve the equation  $$49^x – 42^x = 36^x $$.

Solution: Starting with  $$49^x – 42^x = 36^x $$

We can rewrite  $$49 = 7^2 $$,  $$42 = 6 \times 7 $$, and  $$36 = 6^2 $$:  $$(7^2)^x – (6 \times 7)^x = (6^2)^x $$  $$7^{2x} – (6^x \times 7^x) = 6^{2x} $$

Factoring out  $$7^x $$:  $$7^{2x} – 7^x \cdot 6^x = 6^{2x} $$  $$7^x(7^x – 6^x) = 6^{2x} $$

Dividing both sides by  $$6^x $$:  $$\frac{7^x(7^x – 6^x)}{6^x} = 6^x $$  $$\frac{7^x \cdot 7^x}{6^x} – \frac{7^x \cdot 6^x}{6^x} = 6^x $$  $$\left(\frac{7}{6}\right)^x \cdot 7^x – 7^x = 6^x $$

Let  $$t = \left(\frac{7}{6}\right)^x $$, then:  $$t \cdot 7^x – 7^x = 6^x $$  $$7^x(t – 1) = 6^x $$

Taking the equation  $$7^x(t – 1) = 6^x $$:  $$7^x(t – 1) = 6^x $$  $$\frac{7^x}{6^x}(t – 1) = 1 $$  $$\left(\frac{7}{6}\right)^x(t – 1) = 1 $$  $$t(t – 1) = 1 $$  $$t^2 – t – 1 = 0 $$

Using the quadratic formula:  $$t = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} $$

Since  $$t = \left(\frac{7}{6}\right)^x $$ must be positive, both solutions  $$t = \frac{1 + \sqrt{5}}{2} $$ and  $$t = \frac{1 – \sqrt{5}}{2} $$ are possible if they’re positive.

 $$\frac{1 + \sqrt{5}}{2} \approx 1.618 $$ (positive) ✓  $$\frac{1 – \sqrt{5}}{2} \approx -0.618 $$ (negative) ✗

So  $$t = \frac{1 + \sqrt{5}}{2} $$.

Now we solve for  $$x $$:  $$\left(\frac{7}{6}\right)^x = \frac{1 + \sqrt{5}}{2} $$

Taking the natural logarithm of both sides:  $$x \ln\left(\frac{7}{6}\right) = \ln\left(\frac{1 + \sqrt{5}}{2}\right) $$  $$x = \frac{\ln\left(\frac{1 + \sqrt{5}}{2}\right)}{\ln\left(\frac{7}{6}\right)} $$

This is the value of  $$x $$ that solves the original equation.

SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

Question 9(a)

Given that  $$z_1 = 1 – 2i $$,  $$z_2 = 3 + 2i $$ and  $$z_3 = 1 + i $$, simplify the following: (i)  $$z_1z_2z_3 $$ (ii)  $$z_1z_2 – z_2z_3 $$ (iii)  $$\frac{z_1z_3}{z_1} $$

Solution:

(i)  $$z_1z_2z_3 $$

First, let’s calculate  $$z_1z_2 $$:  $$z_1z_2 = (1 – 2i)(3 + 2i) $$  $$= 3 + 2i – 6i – 4i^2 $$  $$= 3 – 4i – 4(-1) $$ (since  $$i^2 = -1 $$)  $$= 3 – 4i + 4 $$  $$= 7 – 4i $$

Now, let’s calculate  $$z_1z_2z_3 $$:  $$z_1z_2z_3 = (7 – 4i)(1 + i) $$  $$= 7 + 7i – 4i – 4i^2 $$  $$= 7 + 3i – 4(-1) $$  $$= 7 + 3i + 4 $$  $$= 11 + 3i $$

(ii)  $$z_1z_2 – z_2z_3 $$

We already calculated  $$z_1z_2 = 7 – 4i $$. Now let’s calculate  $$z_2z_3 $$:  $$z_2z_3 = (3 + 2i)(1 + i) $$  $$= 3 + 3i + 2i + 2i^2 $$  $$= 3 + 5i + 2(-1) $$  $$= 3 + 5i – 2 $$  $$= 1 + 5i $$

Now we calculate  $$z_1z_2 – z_2z_3 $$:  $$z_1z_2 – z_2z_3 = (7 – 4i) – (1 + 5i) $$  $$= 7 – 4i – 1 – 5i $$  $$= 6 – 9i $$

(iii)  $$\frac{z_1z_3}{z_1} $$

Simplifying this expression:  $$\frac{z_1z_3}{z_1} = z_3 = 1 + i $$

Question 9(b)

Show that if  $$\omega $$ is one complex cube root of unity then  $$\omega^2 $$ is also another complex cube root of unity and  $$1 + \omega + \omega^2 = 0 $$.

Solution: A complex cube root of unity is a complex number  $$z $$ such that  $$z^3 = 1 $$.

If  $$\omega $$ is a cube root of unity, then  $$\omega^3 = 1 $$.

We need to prove that  $$\omega^2 $$ is also a cube root of unity:  $$(\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1 $$

Therefore,  $$\omega^2 $$ is also a cube root of unity.

Now, let’s prove that  $$1 + \omega + \omega^2 = 0 $$:

We know that  $$\omega^3 = 1 $$, which means  $$\omega^3 – 1 = 0 $$. We can factorize  $$\omega^3 – 1 $$ as follows:  $$\omega^3 – 1 = (\omega – 1)(\omega^2 + \omega + 1) $$

Since  $$\omega^3 – 1 = 0 $$ and  $$\omega \neq 1 $$ (as  $$\omega $$ is a non-trivial cube root of unity), we must have:  $$\omega^2 + \omega + 1 = 0 $$

Rearranging:  $$1 + \omega + \omega^2 = 0 $$

Which is what we needed to prove.

Question 9(c)

Given that  $$z = x + iy $$, show that  $$|z + i| = |1 + i| $$ is the circle  $$x^2 + y^2 + 2y – 1 = 0 $$.

Solution: We know that  $$z = x + iy $$, so  $$z + i = x + i(y+1) $$.

The condition  $$|z + i| = |1 + i| $$ means:  $$|x + i(y+1)| = |1 + i| $$

First, let’s calculate  $$|1 + i| $$:  $$|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} $$

Now, the condition becomes:  $$|x + i(y+1)| = \sqrt{2} $$

Squaring both sides:  $$|x + i(y+1)|^2 = 2 $$  $$x^2 + (y+1)^2 = 2 $$  $$x^2 + y^2 + 2y + 1 = 2 $$  $$x^2 + y^2 + 2y – 1 = 0 $$

Therefore, the equation of the circle is  $$x^2 + y^2 + 2y – 1 = 0 $$.

Question 10(a)

Given that  $$\tan A = \frac{1}{7} $$,  $$\tan B = \frac{1}{3} $$ and  $$A + B + C = 45° $$, find the value of  $$\tan C $$.

Solution: We know that  $$A + B + C = 45° $$, so  $$C = 45° – (A + B) $$.

Using the tangent addition formula:  $$\tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} $$

 $$\tan(A + B) = \frac{\frac{1}{7} + \frac{1}{3}}{1 – \frac{1}{7} \cdot \frac{1}{3}} = \frac{\frac{3 + 7}{21}}{1 – \frac{1}{21}} = \frac{\frac{10}{21}}{\frac{20}{21}} = \frac{10}{20} = \frac{1}{2} $$

Now, using the formula for  $$\tan(C) $$ when  $$C = 45° – (A + B) $$:  $$\tan(45° – (A + B)) = \frac{1 – \tan(A + B)}{1 + \tan(A + B)} $$

 $$\tan C = \frac{1 – \frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} $$

Therefore,  $$\tan C = \frac{1}{3} $$

Question 10(b)

Find the square roots of  $$-i $$ and  $$3 + 4i $$.

Solution:

1. Finding the square roots of  $$-i $$:

We need to find a complex number  $$z = a + bi $$ such that  $$z^2 = -i $$.

Let’s write  $$-i $$ in polar form:  $$-i = 1e^{-i\pi/2} $$

The square roots are:  $$\sqrt{-i} = \sqrt{1}e^{-i\pi/4} = e^{-i\pi/4} = \cos(-\pi/4) + i\sin(-\pi/4) = \frac{\sqrt{2}}{2} – i\frac{\sqrt{2}}{2} $$

and  $$\sqrt{-i} = \sqrt{1}e^{-i\pi/4 + i\pi} = e^{i(3\pi/4)} = \cos(3\pi/4) + i\sin(3\pi/4) = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $$

Therefore, the square roots of  $$-i $$ are  $$\frac{\sqrt{2}}{2} – i\frac{\sqrt{2}}{2} $$ and  $$-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $$

PAPER 2 SECTION A (20%)

1. Evaluate

∫π4π22x(tan⁡2x−sec⁡2x)dx\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2x \left( \tan^2 x – \sec^2 x \right) dx

Solution:

Recall:

tan⁡2x=sec⁡2x−1⇒tan⁡2x−sec⁡2x=−1\tan^2 x = \sec^2 x – 1 \Rightarrow \tan^2 x – \sec^2 x = -1

So the integral becomes:

∫π4π22x(−1) dx=−2∫π4π2x dx=−2[x22]π4π2=−[x2]π4π2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2x(-1)\,dx = -2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x\,dx = -2 \left[ \frac{x^2}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\left[ x^2 \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} =−((π2)2−(π4)2)=−(π24−π216)=−(4π2−π216)=−3π216= -\left( \left( \frac{\pi}{2} \right)^2 – \left( \frac{\pi}{4} \right)^2 \right) = -\left( \frac{\pi^2}{4} – \frac{\pi^2}{16} \right) = -\left( \frac{4\pi^2 – \pi^2}{16} \right) = -\frac{3\pi^2}{16}

2. Find the sine of the angle between the vectors

a⃗=3i−2j+k,b⃗=i+2j−3k\vec{a} = 3\mathbf{i} – 2\mathbf{j} + \mathbf{k}, \quad \vec{b} = \mathbf{i} + 2\mathbf{j} – 3\mathbf{k}

Solution:

Use the formula:

sin⁡θ=∣a⃗×b⃗∣∣a⃗∣∣b⃗∣\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}

Compute a⃗×b⃗\vec{a} \times \vec{b}:

a⃗×b⃗=∣ijk3−2112−3∣=i((−2)(−3)−(1)(2))−j(3(−3)−(1)(1))+k(3(2)−(−2)(1))\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 1 \\ 1 & 2 & -3 \\ \end{vmatrix} = \mathbf{i}((-2)(-3) – (1)(2)) – \mathbf{j}(3(-3) – (1)(1)) + \mathbf{k}(3(2) – (-2)(1)) =i(6−2)−j(−9−1)+k(6+2)=4i+10j+8k= \mathbf{i}(6 – 2) – \mathbf{j}(-9 – 1) + \mathbf{k}(6 + 2) = 4\mathbf{i} + 10\mathbf{j} + 8\mathbf{k} ∣a⃗×b⃗∣=42+102+82=16+100+64=180|\vec{a} \times \vec{b}| = \sqrt{4^2 + 10^2 + 8^2} = \sqrt{16 + 100 + 64} = \sqrt{180} ∣a⃗∣=32+(−2)2+12=9+4+1=14|\vec{a}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} ∣b⃗∣=12+22+(−3)2=1+4+9=14|\vec{b}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} sin⁡θ=18014⋅14=18014=6514=357\sin \theta = \frac{\sqrt{180}}{\sqrt{14} \cdot \sqrt{14}} = \frac{\sqrt{180}}{14} = \frac{6\sqrt{5}}{14} = \frac{3\sqrt{5}}{7}

3. Solve the equation

dydx=e2x+y\frac{dy}{dx} = e^{2x + y}

Solution:

Separate variables:

e−ydy=e2xdx⇒∫e−ydy=∫e2xdxe^{-y} dy = e^{2x} dx \Rightarrow \int e^{-y} dy = \int e^{2x} dx −e−y=12e2x+C⇒e−y=−12e2x−C⇒y=−ln⁡(−12e2x−C)- e^{-y} = \frac{1}{2} e^{2x} + C \Rightarrow e^{-y} = -\frac{1}{2} e^{2x} – C \Rightarrow y = -\ln \left( -\frac{1}{2} e^{2x} – C \right)

4. Differentiate esin⁡xcos⁡2xe^{\sin x} \cos 2x

Solution:

Let y=esin⁡xcos⁡2xy = e^{\sin x} \cos 2x

Use product rule:

dydx=esin⁡x⋅cos⁡2x⋅cos⁡x−2esin⁡x⋅sin⁡2x=esin⁡x(cos⁡x⋅cos⁡2x−2sin⁡2x)\frac{dy}{dx} = e^{\sin x} \cdot \cos 2x \cdot \cos x – 2e^{\sin x} \cdot \sin 2x = e^{\sin x} \left( \cos x \cdot \cos 2x – 2 \sin 2x \right)

5. Find the value of xx for which the vectors

xi−28j+k,xi−25kx\mathbf{i} – 28\mathbf{j} + \mathbf{k}, \quad x\mathbf{i} – 25\mathbf{k}

are perpendicular.

Solution:

Vectors are perpendicular if dot product = 0.

Let:

u⃗=xi−28j+k,v⃗=xi−25k\vec{u} = x\mathbf{i} – 28\mathbf{j} + \mathbf{k}, \quad \vec{v} = x\mathbf{i} – 25\mathbf{k} u⃗⋅v⃗=x⋅x+0⋅(−28)+1⋅(−25)=x2−25\vec{u} \cdot \vec{v} = x \cdot x + 0 \cdot (-28) + 1 \cdot (-25) = x^2 – 25

Set equal to 0:

x2−25=0⇒x=±5x^2 – 25 = 0 \Rightarrow x = \pm 5

Let’s continue with SECTION B: CALCULUS of the IJMB 2024 A-Level Mathematics Paper II.

SECTION B (40%)

6(a)

If

f(x)={x2+3x−2,x<−1×3+3×2−x+1,x≥−1f(x) = \begin{cases} x^2 + 3x – 2, & x < -1 \\ x^3 + 3x^2 – x + 1, & x \geq -1 \end{cases}

Determine whether f(x)f(x) is continuous at x=−1x = -1.

Solution:

A function is continuous at x=−1x = -1 if:

lim⁡x→−1−f(x)=lim⁡x→−1+f(x)=f(−1)\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1)

Left-hand limit:
Use the first part (since x<−1x < -1):

lim⁡x→−1−f(x)=(−1)2+3(−1)−2=1−3−2=−4\lim_{x \to -1^-} f(x) = (-1)^2 + 3(-1) – 2 = 1 – 3 – 2 = -4

Right-hand limit and f(−1)f(-1):
Use second part (since x≥−1x \geq -1):

lim⁡x→−1+f(x)=(−1)3+3(−1)2−(−1)+1=−1+3+1+1=4\lim_{x \to -1^+} f(x) = (-1)^3 + 3(-1)^2 – (-1) + 1 = -1 + 3 + 1 + 1 = 4

Since −4≠4-4 \neq 4,

lim⁡x→−1−f(x)≠lim⁡x→−1+f(x)\lim_{x \to -1^-} f(x) \neq \lim_{x \to -1^+} f(x)

Therefore, f(x)f(x) is not continuous at x=−1x = -1.

6(b)

Find the coordinates of the turning point(s) of

y=2×3−15×2+36x−1y = 2x^3 – 15x^2 + 36x – 1

Solution:

Find dydx\frac{dy}{dx}:

dydx=6×2−30x+36\frac{dy}{dx} = 6x^2 – 30x + 36

Set derivative to 0:

6×2−30x+36=0⇒x2−5x+6=0⇒(x−2)(x−3)=0⇒x=2, 36x^2 – 30x + 36 = 0 \Rightarrow x^2 – 5x + 6 = 0 \Rightarrow (x – 2)(x – 3) = 0 \Rightarrow x = 2, \, 3

Find y-values:

For x=2x = 2:

y=2(8)−15(4)+36(2)−1=16−60+72−1=27y = 2(8) – 15(4) + 36(2) – 1 = 16 – 60 + 72 – 1 = 27

For x=3x = 3:

y=2(27)−15(9)+36(3)−1=54−135+108−1=26y = 2(27) – 15(9) + 36(3) – 1 = 54 – 135 + 108 – 1 = 26

Coordinates of turning points:

• (2,27)(2, 27)
  
• (3,26)(3, 26)
  

Determine nature of turning points:

Second derivative:

d2ydx2=12x−30\frac{d^2y}{dx^2} = 12x – 30

At x=2x = 2: 12(2)−30=−612(2) – 30 = -6 → Maximum
At x=3x = 3: 12(3)−30=612(3) – 30 = 6 → Minimum

7(a)

Find the volume generated when the region bounded by y=4x−x2y = 4x – x^2, y=0y = 0, and the y-axis is rotated about the x-axis.

Solution:

The region is from x=0x = 0 to where y=0y = 0:

4x−x2=0⇒x(4−x)=0⇒x=0, 44x – x^2 = 0 \Rightarrow x(4 – x) = 0 \Rightarrow x = 0, \, 4

Volume of revolution about x-axis:

V=π∫04(4x−x2)2dxV = \pi \int_0^4 (4x – x^2)^2 dx

First expand:

(4x−x2)2=16×2−8×3+x4(4x – x^2)^2 = 16x^2 – 8x^3 + x^4 V=π∫04(x4−8×3+16×2)dx=π[x55−2×4+16×33]04V = \pi \int_0^4 (x^4 – 8x^3 + 16x^2) dx = \pi \left[ \frac{x^5}{5} – 2x^4 + \frac{16x^3}{3} \right]_0^4

At x=4x = 4:

455=10245,2(44)=2(256)=512,16(64)3=10243\frac{4^5}{5} = \frac{1024}{5}, \quad 2(4^4) = 2(256) = 512, \quad \frac{16(64)}{3} = \frac{1024}{3} V=π(10245−512+10243)V = \pi \left( \frac{1024}{5} – 512 + \frac{1024}{3} \right)

Find LCM of 5, 1, 3 = 15:

=π(307215−768015+512015)=π⋅51215= \pi \left( \frac{3072}{15} – \frac{7680}{15} + \frac{5120}{15} \right) = \pi \cdot \frac{512}{15}

So,

V=512π15 units3V = \frac{512\pi}{15} \text{ units}^3

7(b)

Find the equation of the tangent to

y=ln⁡(x2+3x−2)at x=1y = \ln (x^2 + 3x – 2) \quad \text{at } x = 1

Solution:

Let’s find the derivative:

dydx=1×2+3x−2⋅(2x+3)\frac{dy}{dx} = \frac{1}{x^2 + 3x – 2} \cdot (2x + 3)

At x=1x = 1:

x2+3x−2=1+3−2=2,2x+3=5⇒dydx=52x^2 + 3x – 2 = 1 + 3 – 2 = 2,\quad 2x + 3 = 5 \Rightarrow \frac{dy}{dx} = \frac{5}{2}

Point on the curve:

y=ln⁡(2)⇒Point is (1,ln⁡2)y = \ln(2) \Rightarrow \text{Point is } (1, \ln 2)

Tangent:

y−ln⁡2=52(x−1)⇒y=52x−52+ln⁡2y – \ln 2 = \frac{5}{2}(x – 1) \Rightarrow y = \frac{5}{2}x – \frac{5}{2} + \ln 2

SECTION C (20%)

8(a)

Solve the differential equation:

dydx=y+xy−x\frac{dy}{dx} = \frac{y + x}{y – x}

Solution:

Let’s use the substitution method.

Let

v=yx⇒y=vx⇒dydx=v+xdvdxv = \frac{y}{x} \Rightarrow y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}

Substitute into the equation:

v+xdvdx=vx+xvx−x=x(v+1)x(v−1)=v+1v−1v + x\frac{dv}{dx} = \frac{vx + x}{vx – x} = \frac{x(v + 1)}{x(v – 1)} = \frac{v + 1}{v – 1}

Subtract vv from both sides:

xdvdx=v+1v−1−v=v+1−v(v−1)v−1=v+1−(v2−v)v−1=v+1−v2+vv−1=−v2+2v+1v−1x\frac{dv}{dx} = \frac{v + 1}{v – 1} – v = \frac{v + 1 – v(v – 1)}{v – 1} = \frac{v + 1 – (v^2 – v)}{v – 1} = \frac{v + 1 – v^2 + v}{v – 1} = \frac{-v^2 + 2v + 1}{v – 1}

So,

xdvdx=−v2+2v+1v−1x\frac{dv}{dx} = \frac{-v^2 + 2v + 1}{v – 1}

Let’s integrate:

v−1−v2+2v+1dv=1xdx\frac{v – 1}{-v^2 + 2v + 1} dv = \frac{1}{x} dx

Factor denominator:

−v2+2v+1=−(v2−2v−1)-v^2 + 2v + 1 = -(v^2 – 2v – 1)

This integral is a bit messy, so alternatively, use substitution earlier:

Try:

dydx=y+xy−x⇒Homogeneous equation, try v=yx\frac{dy}{dx} = \frac{y + x}{y – x} \Rightarrow \text{Homogeneous equation, try } v = \frac{y}{x}

So y=vx⇒dydx=v+xdvdxy = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}

Substitute into original:

v+xdvdx=vx+xvx−x=x(v+1)x(v−1)=v+1v−1⇒xdvdx=v+1v−1−v=−v2+2v+1v−1⇒v−1−v2+2v+1dv=1xdxv + x\frac{dv}{dx} = \frac{vx + x}{vx – x} = \frac{x(v + 1)}{x(v – 1)} = \frac{v + 1}{v – 1} \Rightarrow x\frac{dv}{dx} = \frac{v + 1}{v – 1} – v = \frac{-v^2 + 2v + 1}{v – 1} \Rightarrow \frac{v – 1}{-v^2 + 2v + 1} dv = \frac{1}{x} dx

Now integrate both sides (LHS is rational, partial fractions can be used):

8(b)

Find the magnitude and direction of the vector

v⃗=3i−4j\vec{v} = 3\mathbf{i} – 4\mathbf{j}

Solution:

Magnitude:

∣v⃗∣=32+(−4)2=9+16=25=5|\vec{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Direction (angle with positive x-axis):

θ=tan⁡−1(−43)=−53.13∘\theta = \tan^{-1} \left( \frac{-4}{3} \right) = -53.13^\circ

So the direction is 53.13° below the x-axis (or 306.87∘306.87^\circ from the positive x-axis in standard position).

9(a)

If the acceleration of a particle is given by

d2xdt2=6t, with initial velocity 5 and initial displacement 2,\frac{d^2x}{dt^2} = 6t, \text{ with initial velocity } 5 \text{ and initial displacement } 2,

find the displacement as a function of time.

Solution:

Integrate acceleration to get velocity:

dvdt=6t⇒v=∫6t dt=3t2+C\frac{dv}{dt} = 6t \Rightarrow v = \int 6t \, dt = 3t^2 + C

Given: initial velocity v(0)=5v(0) = 5

v=3t2+C⇒5=0+C⇒C=5⇒v=3t2+5v = 3t^2 + C \Rightarrow 5 = 0 + C \Rightarrow C = 5 \Rightarrow v = 3t^2 + 5

Now integrate velocity to get displacement:

x=∫(3t2+5)dt=t3+5t+C′x = \int (3t^2 + 5) dt = t^3 + 5t + C’

Given: x(0)=2⇒2=0+0+C′⇒C′=2x(0) = 2 \Rightarrow 2 = 0 + 0 + C’ \Rightarrow C’ = 2

Final answer:

x(t)=t3+5t+2x(t) = t^3 + 5t + 2

9(b)

Two forces: F⃗1=2i+3j\vec{F}_1 = 2\mathbf{i} + 3\mathbf{j},
F⃗2=−i+4j\vec{F}_2 = -\mathbf{i} + 4\mathbf{j}.
Find the resultant force and its magnitude.

Solution:

Resultant:

R⃗=F⃗1+F⃗2=(2−1)i+(3+4)j=i+7j\vec{R} = \vec{F}_1 + \vec{F}_2 = (2 – 1)\mathbf{i} + (3 + 4)\mathbf{j} = \mathbf{i} + 7\mathbf{j}

Magnitude:

∣R⃗∣=12+72=1+49=50=52|\vec{R}| = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}

Direction:

θ=tan⁡−1(71)=tan⁡−1(7)≈81.87∘\theta = \tan^{-1}\left( \frac{7}{1} \right) = \tan^{-1}(7) \approx 81.87^\circ

Continuing with the last question of the IJMB 2024 A-Level Mathematics Paper II – SECTION C:

10(a)

Solve the differential equation:

dydx=y2sin⁡x\frac{dy}{dx} = y^2 \sin x

Solution:

This is a separable differential equation.

dydx=y2sin⁡x⇒1y2dy=sin⁡x dx\frac{dy}{dx} = y^2 \sin x \Rightarrow \frac{1}{y^2} dy = \sin x \, dx

Integrate both sides:

∫y−2dy=∫sin⁡x dx⇒−1y=−cos⁡x+C\int y^{-2} dy = \int \sin x \, dx \Rightarrow -\frac{1}{y} = -\cos x + C

Multiply both sides by –1:

1y=cos⁡x−C⇒y=1cos⁡x−C\frac{1}{y} = \cos x – C \Rightarrow y = \frac{1}{\cos x – C}

Final Answer:

y(x)=1cos⁡x−Cy(x) = \frac{1}{\cos x – C}

10(b)

Given vectors a⃗=3i+2j\vec{a} = 3\mathbf{i} + 2\mathbf{j}, b⃗=i+4j\vec{b} = \mathbf{i} + 4\mathbf{j},
find a unit vector in the direction of a⃗+b⃗\vec{a} + \vec{b}.

Solution:

First, find a⃗+b⃗\vec{a} + \vec{b}:

a⃗+b⃗=(3+1)i+(2+4)j=4i+6j\vec{a} + \vec{b} = (3 + 1)\mathbf{i} + (2 + 4)\mathbf{j} = 4\mathbf{i} + 6\mathbf{j}

Magnitude of a⃗+b⃗\vec{a} + \vec{b}:

∣a⃗+b⃗∣=42+62=16+36=52=213|\vec{a} + \vec{b}| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}

Now divide by the magnitude to get the unit vector:

u^=1213(4i+6j)=4213i+6213j=213i+313j\hat{u} = \frac{1}{2\sqrt{13}}(4\mathbf{i} + 6\mathbf{j}) = \frac{4}{2\sqrt{13}}\mathbf{i} + \frac{6}{2\sqrt{13}}\mathbf{j} = \frac{2}{\sqrt{13}}\mathbf{i} + \frac{3}{\sqrt{13}}\mathbf{j}

Final Answer:

u^=213i+313j\boxed{\hat{u} = \frac{2}{\sqrt{13}}\mathbf{i} + \frac{3}{\sqrt{13}}\mathbf{j}}

SECTION A

Question 1: Calculate the gradient of the line joining the points (7, 50) and (-7, 8).

To find the gradient of a line joining two points  $$(x_1, y_1) $$ and  $$(x_2, y_2) $$, we use the formula:

 $$ $$m = \frac{y_2 – y_1}{x_2 – x_1} $$ $$

Given:

– Point 1:  $$(7, 50) $$

– Point 2:  $$(-7, 8) $$

Substituting these values:

 $$ $$m = \frac{8 – 50}{-7 – 7} = \frac{-42}{-14} = \frac{42}{14} = 3 $$ $$

Therefore, the gradient of the line joining the points  $$(7, 50) $$ and  $$(-7, 8) $$ is  $$3 $$.

Question 2: Calculate the mean of the distribution f(x)=e−3xf(x) = e^{-3x} f(x)=e−3x where 0≤x<∞0 \leq x < \infty 0≤x<∞.

For a continuous probability distribution  $$f(x) $$, the mean (expected value) is calculated using:

 $$ $$\mu = \int_{-\infty}^{\infty} x \cdot f(x) \, dx $$ $$

For our distribution  $$f(x) = e^{-3x} $$ where  $$0 \leq x < \infty $$, we first need to verify if it’s a valid probability density function by checking if  $$\int_{0}^{\infty} e^{-3x} \, dx = 1 $$.

 $$ $$\int_{0}^{\infty} e^{-3x} \, dx = \left[ -\frac{1}{3} e^{-3x} \right]_{0}^{\infty} = -\frac{1}{3} \left( 0 – 1 \right) = \frac{1}{3} $$ $$

Since this doesn’t equal 1, we need to normalize by multiplying by 3:

 $$ $$f(x) = 3e^{-3x} $$ $$

Now, we can calculate the mean:

 $$ $$\mu = \int_{0}^{\infty} x \cdot 3e^{-3x} \, dx = 3 \int_{0}^{\infty} x \cdot e^{-3x} \, dx $$ $$

Using integration by parts with  $$u = x $$ and  $$dv = e^{-3x} \, dx $$:

 $$ $$\mu = 3 \left[ -\frac{x}{3} e^{-3x} \right]_{0}^{\infty} + 3 \int_{0}^{\infty} \frac{1}{3} e^{-3x} \, dx $$ $$

 $$ $$\mu = 3 \left( 0 + 0 \right) + 3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{3} $$ $$

Therefore, the mean of the distribution is  $$\frac{1}{3} $$.

Question 3: Find the equation of the line through the mid-point of (2,3), (-5,1) and perpendicular to the line 2x+y=7002x + y = 700 2x+y=700.

Step 1: Find the mid-point of  $$(2,3) $$ and  $$(-5,1) $$.

The mid-point formula is:  $$(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) $$

Mid-point  $$= (\frac{2 + (-5)}{2}, \frac{3 + 1}{2}) = (\frac{-3}{2}, \frac{4}{2}) = (-\frac{3}{2}, 2) $$

Step 2: Find the slope of the line  $$2x + y = 700 $$.

Rewriting in slope-intercept form:  $$y = -2x + 700 $$

So, the slope of this line is  $$-2 $$.

Step 3: The slope of a perpendicular line is the negative reciprocal of the original slope.

Since the slope of  $$2x + y = 700 $$ is  $$-2 $$, the slope of the perpendicular line is  $$\frac{1}{2} $$.

Step 4: Use the point-slope form of a line equation.

The point-slope form is:  $$y – y_1 = m(x – x_1) $$

where  $$(x_1, y_1) $$ is the point  $$(-\frac{3}{2}, 2) $$ and  $$m = \frac{1}{2} $$

 $$ $$y – 2 = \frac{1}{2}(x – (-\frac{3}{2})) $$ $$

 $$ $$y – 2 = \frac{1}{2}(x + \frac{3}{2}) $$ $$

 $$ $$y – 2 = \frac{1}{2}x + \frac{3}{4} $$ $$

 $$ $$y = \frac{1}{2}x + \frac{3}{4} + 2 $$ $$

 $$ $$y = \frac{1}{2}x + \frac{3}{4} + \frac{8}{4} $$ $$

 $$ $$y = \frac{1}{2}x + \frac{11}{4} $$ $$

Therefore, the equation of the required line is  $$y = \frac{1}{2}x + \frac{11}{4} $$ or  $$2y = x + \frac{11}{2} $$ or  $$2y – x = \frac{11}{2} $$.

Question 4: Find the equation of the circle whose end points of diameter are (7, -3) and (-1,2).

To find the equation of a circle given the endpoints of a diameter:

Step 1: Find the center of the circle, which is the midpoint of the diameter.

Center  $$= (\frac{7 + (-1)}{2}, \frac{-3 + 2}{2}) = (\frac{6}{2}, \frac{-1}{2}) = (3, -\frac{1}{2}) $$

Step 2: Calculate the radius, which is half the distance between the endpoints.

 $$ $$r = \frac{1}{2} \sqrt{(7-(-1))^2 + (-3-2)^2} = \frac{1}{2} \sqrt{8^2 + (-5)^2} = \frac{1}{2} \sqrt{64 + 25} = \frac{1}{2} \sqrt{89} $$ $$

Step 3: Write the standard equation of a circle.

The standard form is:  $$(x – h)^2 + (y – k)^2 = r^2 $$, where  $$(h, k) $$ is the center.

 $$ $$(x – 3)^2 + (y – (-\frac{1}{2}))^2 = (\frac{\sqrt{89}}{2})^2 $$ $$

 $$ $$(x – 3)^2 + (y + \frac{1}{2})^2 = \frac{89}{4} $$ $$

Therefore, the equation of the circle is  $$(x – 3)^2 + (y + \frac{1}{2})^2 = \frac{89}{4} $$.

Question 5: Find the coordinates of the vertex and focus of the parabola x2=16−4yx^2 = 16 – 4y x2=16−4y.

Let’s rewrite the equation to match the standard form for a parabola.

 $$ $$x^2 = 16 – 4y $$ $$

 $$ $$x^2 = -4(y – 4) $$ $$

This is in the form  $$x^2 = -4a(y – k) $$ where the axis of symmetry is parallel to the y-axis.

Comparing with the standard form, we have  $$a = 1 $$ and  $$k = 4 $$.

For a parabola with equation  $$x^2 = -4a(y – k) $$:

– The vertex is at  $$(0, k) $$

– The focus is at  $$(0, k-a) $$

– The directrix is the line  $$y = k+a $$

Therefore:

– The vertex is at  $$(0, 4) $$

– The focus is at  $$(0, 4-1) = (0, 3) $$

– The directrix is the line  $$y = 4+1 = 5 $$

I’ll continue with more questions if you’d like. Would you prefer I solve more questions from Section A, or should I proceed to questions from Sections B and C?

Retry

B

continue

Edit

I’ll continue with the solutions for the remaining questions, moving on to Section B.

SECTION B: COORDINATE GEOMETRY

Question 6(a): Obtain the equation of the conic whose parametric equations are given by x=1+2tan⁡θx = 1 + 2 \tan \theta x=1+2tanθ and y=−1+3sec⁡θy = -1 + 3 \sec \theta y=−1+3secθ, and describe it completely.

We need to eliminate the parameter  $$\theta $$ to find the Cartesian equation of the curve.

Step 1: From the first equation, solve for  $$\tan \theta $$.

 $$ $$x = 1 + 2\tan \theta $$ $$

 $$ $$\tan \theta = \frac{x-1}{2} $$ $$

Step 2: Use the identity  $$\sec^2 \theta = 1 + \tan^2 \theta $$

 $$ $$\sec^2 \theta = 1 + \left(\frac{x-1}{2}\right)^2 = 1 + \frac{(x-1)^2}{4} = \frac{4 + (x-1)^2}{4} $$ $$

Step 3: From the second equation, solve for  $$\sec \theta $$.

 $$ $$y = -1 + 3\sec \theta $$ $$

 $$ $$\sec \theta = \frac{y+1}{3} $$ $$

Step 4: Square both sides.

 $$ $$\sec^2 \theta = \left(\frac{y+1}{3}\right)^2 = \frac{(y+1)^2}{9} $$ $$

Step 5: Equate the expressions for  $$\sec^2 \theta $$ from Steps 2 and 4.

 $$ $$\frac{4 + (x-1)^2}{4} = \frac{(y+1)^2}{9} $$ $$

Step 6: Cross multiply.

 $$ $$9(4 + (x-1)^2) = 4(y+1)^2 $$ $$

 $$ $$36 + 9(x-1)^2 = 4(y+1)^2 $$ $$

 $$ $$36 + 9(x^2 – 2x + 1) = 4(y^2 + 2y + 1) $$ $$

 $$ $$36 + 9x^2 – 18x + 9 = 4y^2 + 8y + 4 $$ $$

 $$ $$9x^2 – 18x – 4y^2 – 8y + 41 = 0 $$ $$

This is the equation of a hyperbola. To analyze it completely:

Let’s convert it to standard form by completing the squares:

 $$ $$9(x^2 – 2x + 1) – 4(y^2 + 2y + 1) = -41 + 9 – 4 $$ $$

 $$ $$9(x – 1)^2 – 4(y + 1)^2 = -36 $$ $$

 $$ $$\frac{9(x – 1)^2}{36} – \frac{4(y + 1)^2}{36} = -1 $$ $$

 $$ $$\frac{(x – 1)^2}{4} – \frac{(y + 1)^2}{9} = -1 $$ $$

Taking the negative of both sides:

 $$ $$\frac{(y + 1)^2}{9} – \frac{(x – 1)^2}{4} = 1 $$ $$

This is a hyperbola with:

– Center at  $$(1, -1) $$

– Transverse axis parallel to the y-axis

– Semi-major axis  $$a = 3 $$

– Semi-minor axis  $$b = 2 $$

– Eccentricity  $$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} $$

– Foci at  $$(1, -1 \pm c) $$ where  $$c = ae = 3 \cdot \frac{\sqrt{13}}{3} = \sqrt{13} $$

Question 6(b): Sketch the curve r=(1+sin⁡θ)r = (1 + \sin \theta) r=(1+sinθ) in polar form for 0≤θ≤360°0 \leq \theta \leq 360° 0≤θ≤360°.

The equation  $$r = 1 + \sin \theta $$ represents a cardioid (heart-shaped curve).

Key points to plot:

– At  $$\theta = 0° $$:  $$r = 1 + \sin(0) = 1 + 0 = 1 $$

– At  $$\theta = 90° $$:  $$r = 1 + \sin(90°) = 1 + 1 = 2 $$ (maximum distance)

– At  $$\theta = 180° $$:  $$r = 1 + \sin(180°) = 1 + 0 = 1 $$

– At  $$\theta = 270° $$:  $$r = 1 + \sin(270°) = 1 + (-1) = 0 $$ (curve passes through origin)

– At  $$\theta = 360° $$:  $$r = 1 + \sin(360°) = 1 + 0 = 1 $$

The curve looks like a heart shape with a cusp at the origin. It has:

– Maximum distance from origin: 2 units at  $$\theta = 90° $$

– Minimum distance from origin: 0 units at  $$\theta = 270° $$

– The curve passes through the point  $$(1,0) $$ when  $$\theta = 0° $$ or  $$\theta = 180° $$

Question 6(c): Show that r=11+cos⁡θr = \frac{1}{1+\cos \theta} r=1+cosθ1​ is the polar form of a parabola.

To show that  $$r = \frac{1}{1+\cos \theta} $$ represents a parabola, we need to convert it to Cartesian form and verify it matches the standard form of a parabola.

Step 1: Recall the relationships between polar and Cartesian coordinates:

–  $$x = r\cos \theta $$

–  $$y = r\sin \theta $$

–  $$r^2 = x^2 + y^2 $$

Step 2: From the given equation  $$r = \frac{1}{1+\cos \theta} $$, we have:

 $$ $$r(1+\cos \theta) = 1 $$ $$

 $$ $$r + r\cos \theta = 1 $$ $$

Step 3: Substitute  $$r\cos \theta = x $$:

 $$ $$r + x = 1 $$ $$

 $$ $$r = 1 – x $$ $$

Step 4: Now use  $$r^2 = x^2 + y^2 $$:

 $$ $$(1-x)^2 = x^2 + y^2 $$ $$

 $$ $$1 – 2x + x^2 = x^2 + y^2 $$ $$

 $$ $$1 – 2x = y^2 $$ $$

 $$ $$y^2 = -2x + 1 $$ $$

This is the standard form of a parabola with:

– Vertex at  $$(\frac{1}{2}, 0) $$

– Focus at  $$(0,0) $$ (the origin in the polar coordinate system)

– Directrix is the vertical line  $$x = 1 $$

– The parabola opens to the left (negative x-direction)

Therefore,  $$r = \frac{1}{1+\cos \theta} $$ is indeed the polar form of a parabola.

Question 7(a): Show that the equation 9×2+25y2−54x−100y−44=09x^2 + 25y^2 – 54x – 100y – 44 = 0 9×2+25y2−54x−100y−44=0 represents an Ellipse. Find the coordinates of its center.

First, let’s rearrange to group the terms with  $$x $$ and  $$y $$:

 $$ $$9x^2 – 54x + 25y^2 – 100y – 44 = 0 $$ $$

Now complete the squares for both  $$x $$ and  $$y $$ terms:

For the  $$x $$ terms:

 $$ $$9x^2 – 54x = 9(x^2 – 6x) = 9(x^2 – 6x + 9 – 9) = 9(x – 3)^2 – 81 $$ $$

For the  $$y $$ terms:

 $$ $$25y^2 – 100y = 25(y^2 – 4y) = 25(y^2 – 4y + 4 – 4) = 25(y – 2)^2 – 100 $$ $$

Substituting back:

 $$ $$9(x – 3)^2 – 81 + 25(y – 2)^2 – 100 – 44 = 0 $$ $$

 $$ $$9(x – 3)^2 + 25(y – 2)^2 = 81 + 100 + 44 = 225 $$ $$

 $$ $$\frac{9(x – 3)^2}{225} + \frac{25(y – 2)^2}{225} = 1 $$ $$

 $$ $$\frac{(x – 3)^2}{25} + \frac{(y – 2)^2}{9} = 1 $$ $$

This is the standard form of an ellipse  $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 $$ with:

– Center at  $$(h,k) = (3,2) $$

– Semi-major axis  $$a = 5 $$ (along the x-axis)

– Semi-minor axis  $$b = 3 $$ (along the y-axis)

Therefore, the equation represents an ellipse with center at  $$(3,2) $$.

Question 7(b): Find the equation of a straight line parallel to the line x−3y=5x – 3y = 5 x−3y=5 which passes through the mid-point of the line joining the points (3,−2)(3,-2) (3,−2) and (7,8)(7,8) (7,8).

Step 1: Find the mid-point of the line joining  $$(3,-2) $$ and  $$(7,8) $$.

Mid-point  $$= \left(\frac{3+7}{2}, \frac{-2+8}{2}\right) = \left(\frac{10}{2}, \frac{6}{2}\right) = (5, 3) $$

Step 2: The slope of the line  $$x – 3y = 5 $$ is determined by rearranging to slope-intercept form:

 $$ $$x – 3y = 5 $$ $$

 $$ $$-3y = -x + 5 $$ $$

 $$ $$y = \frac{1}{3}x – \frac{5}{3} $$ $$

So the slope of the given line is  $$\frac{1}{3} $$.

Step 3: Since we want a line parallel to  $$x – 3y = 5 $$, our new line will have the same slope:  $$\frac{1}{3} $$.

Step 4: Use the point-slope form to find the equation of the new line:

 $$ $$y – y_1 = m(x – x_1) $$ $$

 $$ $$y – 3 = \frac{1}{3}(x – 5) $$ $$

 $$ $$y – 3 = \frac{1}{3}x – \frac{5}{3} $$ $$

 $$ $$y = \frac{1}{3}x – \frac{5}{3} + 3 $$ $$

 $$ $$y = \frac{1}{3}x + \frac{4}{3} $$ $$

Converting back to general form:

 $$ $$3y = x + 4 $$ $$

 $$ $$x – 3y = -4 $$ $$

Therefore, the equation of the required line is  $$x – 3y = -4 $$.

Question 7(c): Find the value of 2x+y2x + y 2x+y, if (2,3)(2,3) (2,3) is the mid-point of (x,3)(x,3) (x,3) and (−3,y)(-3,y) (−3,y).

If  $$(2,3) $$ is the mid-point of  $$(x,3) $$ and  $$(-3,y) $$, then:

 $$ $$2 = \frac{x + (-3)}{2} $$ $$

 $$ $$3 = \frac{3 + y}{2} $$ $$

Solving for  $$x $$ from the first equation:

 $$ $$2 = \frac{x – 3}{2} $$ $$

 $$ $$4 = x – 3 $$ $$

 $$ $$x = 7 $$ $$

Solving for  $$y $$ from the second equation:

 $$ $$3 = \frac{3 + y}{2} $$ $$

 $$ $$6 = 3 + y $$ $$

 $$ $$y = 3 $$ $$

Now we can find the value of  $$2x + y $$:

 $$ $$2x + y = 2(7) + 3 = 14 + 3 = 17 $$ $$

Therefore, the value of  $$2x + y $$ is  $$17 $$.

Question 8(a): Determine the condition for the line y=mxy = mx y=mx to be tangent to the parabola y2=4a(x−5)y^2 = 4a(x – 5) y2=4a(x−5). Write down the general equation of the tangent to the parabola.

A line is tangent to a parabola if it intersects the parabola at exactly one point.

Step 1: Find the intersection of the line  $$y = mx $$ and the parabola  $$y^2 = 4a(x – 5) $$.

Substitute  $$y = mx $$ into the parabola equation:

 $$ $$m^2x^2 = 4a(x – 5) $$ $$

 $$ $$m^2x^2 = 4ax – 20a $$ $$

 $$ $$m^2x^2 – 4ax + 20a = 0 $$ $$

Step 2: For the line to be tangent, this quadratic equation must have exactly one solution (a repeated root). This occurs when the discriminant is zero:

 $$ $$b^2 – 4ac = 0 $$ $$

 $$ $$(-4a)^2 – 4(m^2)(20a) = 0 $$ $$

 $$ $$16a^2 – 80am^2 = 0 $$ $$

 $$ $$16a^2 = 80am^2 $$ $$

 $$ $$a = 5m^2 $$ $$

Therefore, the condition for the line  $$y = mx $$ to be tangent to the parabola is  $$a = 5m^2 $$.

Step 3: For the general equation of a tangent to the parabola  $$y^2 = 4a(x – 5) $$, we can use the formula:

For a point  $$(x_0, y_0) $$ on the parabola, the equation of the tangent at that point is:

 $$ $$y y_0 = 2a(x + x_0 – 10) $$ $$

Given that  $$y_0^2 = 4a(x_0 – 5) $$, we can rewrite as:

 $$ $$y y_0 = 2a(x + x_0 – 10) $$ $$

This is the general form of the tangent to the parabola  $$y^2 = 4a(x – 5) $$.

I’ll continue with more questions in my next response. Let me know if you’d like me to focus on any specific section or question next.

Retry

B

continue

Edit

I’ll continue with the solutions for the remaining questions.

Question 8(b): Let A, B and C be the midpoints of the sides ST, RT and SR of the triangle △RST respectively, whose coordinates are R(2, -7), S(4, -5), T(6, 3). Find the coordinates of the point of intersection P of the lines RA and SB. Show that TC passes through the point P.

Step 1: Find the coordinates of the midpoints A, B, and C.

– Midpoint A of side ST:

   $$ $$A = \left(\frac{S_x + T_x}{2}, \frac{S_y + T_y}{2}\right) = \left(\frac{4 + 6}{2}, \frac{-5 + 3}{2}\right) = (5, -1) $$ $$

– Midpoint B of side RT:

   $$ $$B = \left(\frac{R_x + T_x}{2}, \frac{R_y + T_y}{2}\right) = \left(\frac{2 + 6}{2}, \frac{-7 + 3}{2}\right) = (4, -2) $$ $$

– Midpoint C of side SR:

   $$ $$C = \left(\frac{S_x + R_x}{2}, \frac{S_y + R_y}{2}\right) = \left(\frac{4 + 2}{2}, \frac{-5 + (-7)}{2}\right) = (3, -6) $$ $$

Step 2: Find the equation of line RA.

Line RA passes through R(2, -7) and A(5, -1).

The slope of RA is:

 $$ $$m_{RA} = \frac{-1 – (-7)}{5 – 2} = \frac{6}{3} = 2 $$ $$

Equation of line RA using point-slope form:

 $$ $$y – (-7) = 2(x – 2) $$ $$

 $$ $$y + 7 = 2x – 4 $$ $$

 $$ $$y = 2x – 11 $$ $$

Step 3: Find the equation of line SB.

Line SB passes through S(4, -5) and B(4, -2).

The slope of SB is:

 $$ $$m_{SB} = \frac{-2 – (-5)}{4 – 4} = \frac{3}{0} $$ $$

Since the slope is undefined, the equation of line SB is:

 $$ $$x = 4 $$ $$

Step 4: Find the point of intersection P of lines RA and SB.

From the equation of SB, we know that the x-coordinate of P is 4.

Substituting x = 4 into the equation of RA:

 $$ $$y = 2(4) – 11 = 8 – 11 = -3 $$ $$

Therefore, the point of intersection P is (4, -3).

Step 5: Check if TC passes through point P.

Line TC passes through T(6, 3) and C(3, -6).

The slope of TC is:

 $$ $$m_{TC} = \frac{-6 – 3}{3 – 6} = \frac{-9}{-3} = 3 $$ $$

Equation of line TC using point-slope form:

 $$ $$y – 3 = 3(x – 6) $$ $$

 $$ $$y – 3 = 3x – 18 $$ $$

 $$ $$y = 3x – 15 $$ $$

To check if P(4, -3) lies on this line, substitute:

 $$ $$-3 \stackrel{?}{=} 3(4) – 15 = 12 – 15 = -3 $$ $$

Since the equation is satisfied, P(4, -3) lies on line TC, which means TC passes through point P.

Question 8(c): Calculate the slope of the line parallel to the line joining the points (3,7) and (-2, -5).

To find the slope of a line joining two points  $$(x_1, y_1) $$ and  $$(x_2, y_2) $$, we use the formula:

 $$ $$m = \frac{y_2 – y_1}{x_2 – x_1} $$ $$

For the line joining the points  $$(3,7) $$ and  $$(-2, -5) $$:

 $$ $$m = \frac{-5 – 7}{-2 – 3} = \frac{-12}{-5} = \frac{12}{5} = 2.4 $$ $$

Since a parallel line has the same slope, the slope of the line parallel to the given line is also  $$\frac{12}{5} $$ or  $$2.4 $$.

SECTION C: STATISTICS

Question 9(a): A biased coin has the probability for a head 23\frac{2}{3} 32​. The coin is tossed six times. Calculate the probability that:

Given:

– The probability of getting a head is  $$p = \frac{2}{3} $$

– The probability of getting a tail is  $$q = 1 – p = 1 – \frac{2}{3} = \frac{1}{3} $$

– The coin is tossed 6 times

We’ll use the binomial probability formula:

 $$ $$P(X = k) = \binom{n}{k} p^k q^{n-k} $$ $$

(i) All six trials result in a tail:

This means getting 0 heads and 6 tails.

 $$ $$P(X = 0) = \binom{6}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^6 = 1 \times 1 \times \left(\frac{1}{3}\right)^6 = \frac{1}{729} $$ $$

(ii) None of the six results in a tail:

This means getting 6 heads and 0 tails.

 $$ $$P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^0 = 1 \times \left(\frac{2}{3}\right)^6 \times 1 = \frac{64}{729} $$ $$

(iii) At least three heads appear in the six trials:

This means  $$X \geq 3 $$, or  $$P(X=3) + P(X=4) + P(X=5) + P(X=6) $$

 $$ $$P(X \geq 3) = \sum_{k=3}^{6} \binom{6}{k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{6-k} $$ $$

 $$ $$P(X = 3) = \binom{6}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^3 = 20 \times \frac{8}{27} \times \frac{1}{27} = \frac{160}{729} $$ $$

 $$ $$P(X = 4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{240}{729} $$ $$

 $$ $$P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{192}{729} $$ $$

 $$ $$P(X = 6) = \frac{64}{729} $$ $$ (from part ii)

 $$ $$P(X \geq 3) = \frac{160}{729} + \frac{240}{729} + \frac{192}{729} + \frac{64}{729} = \frac{656}{729} $$ $$

(iv) Not more than three tails appear:

This means at least 3 heads (same as part iii).

 $$ $$P(\text{tails} \leq 3) = P(\text{heads} \geq 3) = \frac{656}{729} $$ $$

(v) The first two results in a head, and the remaining four in a tail:

This is a specific sequence: HH(4 tails)

 $$ $$P(HHTTTT) = \left(\frac{2}{3}\right)^2 \times \left(\frac{1}{3}\right)^4 = \frac{4}{9} \times \frac{1}{81} = \frac{4}{729} $$ $$

(vi) The heads and tails appear alternatively:

There are two possible alternating sequences: HTHTHTHT… or THTHTHTH…

First sequence: HTHTHT

Second sequence: THTHTH

 $$ $$P(\text{alternating}) = P(HTHTHT) + P(THTHTH) $$ $$

 $$ $$P(HTHTHT) = \left(\frac{2}{3}\right)^3 \times \left(\frac{1}{3}\right)^3 = \frac{8}{27} \times \frac{1}{27} = \frac{8}{729} $$ $$

 $$ $$P(THTHTH) = \left(\frac{1}{3}\right)^3 \times \left(\frac{2}{3}\right)^3 = \frac{1}{27} \times \frac{8}{27} = \frac{8}{729} $$ $$

 $$ $$P(\text{alternating}) = \frac{8}{729} + \frac{8}{729} = \frac{16}{729} $$ $$

Question 9(b): Find the Harmonic and Geometric means of 3, 5 and 7.

Given numbers: 3, 5, and 7

(1) Harmonic Mean (H):

 $$ $$H = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + … + \frac{1}{x_n}} $$ $$

For our values:

 $$ $$H = \frac{3}{\frac{1}{3} + \frac{1}{5} + \frac{1}{7}} $$ $$

 $$ $$H = \frac{3}{\frac{35 + 21 + 15}{105}} $$ $$

 $$ $$H = \frac{3}{\frac{71}{105}} $$ $$

 $$ $$H = 3 \times \frac{105}{71} $$ $$

 $$ $$H = \frac{315}{71} \approx 4.44 $$ $$

(2) Geometric Mean (G):

 $$ $$G = \sqrt[n]{x_1 \times x_2 \times … \times x_n} $$ $$

For our values:

 $$ $$G = \sqrt[3]{3 \times 5 \times 7} $$ $$

 $$ $$G = \sqrt[3]{105} $$ $$

 $$ $$G \approx 4.73 $$ $$

Therefore:

– Harmonic Mean =  $$\frac{315}{71} \approx 4.44 $$

– Geometric Mean =  $$\sqrt[3]{105} \approx 4.73 $$

Question 10(a): Two dice are tossed once, what is the probability that the:

When two dice are tossed, there are 36 possible outcomes (6×6 = 36), each with equal probability.

(i) Sum is less than 7:

Favorable outcomes: (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1)

Total favorable outcomes = 15

 $$ $$P(\text{sum} < 7) = \frac{15}{36} = \frac{5}{12} $$ $$

(ii) Same number appeared:

Favorable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

Total favorable outcomes = 6

 $$ $$P(\text{same number}) = \frac{6}{36} = \frac{1}{6} $$ $$

(iii) Difference is 2:

Favorable outcomes: (1,3), (2,4), (3,5), (4,6), (3,1), (4,2), (5,3), (6,4)

Total favorable outcomes = 8

 $$ $$P(\text{difference} = 2) = \frac{8}{36} = \frac{2}{9} $$ $$

(iv) Product is greater than 8:

Favorable outcomes: (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)

Total favorable outcomes = 20

 $$ $$P(\text{product} > 8) = \frac{20}{36} = \frac{5}{9} $$ $$

(v) First number is even:

Favorable outcomes: all outcomes with first die showing 2, 4, or 6

Total favorable outcomes = 18

 $$ $$P(\text{first number even}) = \frac{18}{36} = \frac{1}{2} $$ $$

Question 10(b): If the diameters of ball bearings are normally distributed with mean 60.5cm and standard deviation 2.5cm, determine the probability of the ball bearing with diameter between 58.5cm and 63.5cm.

Given:

– Mean (μ) = 60.5 cm

– Standard deviation (σ) = 2.5 cm

– We want to find P(58.5 < X < 63.5)

Step 1: Standardize to Z-scores.

For X = 58.5:

 $$ $$Z_1 = \frac{58.5 – 60.5}{2.5} = \frac{-2}{2.5} = -0.8 $$ $$

For X = 63.5:

 $$ $$Z_2 = \frac{63.5 – 60.5}{2.5} = \frac{3}{2.5} = 1.2 $$ $$

Step 2: Find the probability using the standard normal distribution table or formula.

 $$ $$P(58.5 < X < 63.5) = P(-0.8 < Z < 1.2) $$ $$

 $$ $$P(58.5 < X < 63.5) = P(Z < 1.2) – P(Z < -0.8) $$ $$

 $$ $$P(58.5 < X < 63.5) = \Phi(1.2) – \Phi(-0.8) $$ $$

 $$ $$P(58.5 < X < 63.5) = \Phi(1.2) – (1 – \Phi(0.8)) $$ $$

 $$ $$P(58.5 < X < 63.5) = \Phi(1.2) + \Phi(0.8) – 1 $$ $$

From standard normal tables:

–  $$\Phi(1.2) \approx 0.8849 $$

–  $$\Phi(0.8) \approx 0.7881 $$

 $$ $$P(58.5 < X < 63.5) = 0.8849 + 0.7881 – 1 = 0.673 $$ $$

Therefore, the probability of a ball bearing having a diameter between 58.5 cm and 63.5 cm is approximately 0.673 or 67.3%.

Question 11(a): If f(x)=mx2f(x) = mx^2 f(x)=mx2 for 0≤x≤20 \leq x \leq 2 0≤x≤2 is a probability density function for a continuous random variable x, calculate:

For a function to be a valid PDF:

1.  $$f(x) \geq 0 $$ for all  $$x $$ in the domain

2.  $$\int_{-\infty}^{\infty} f(x) dx = 1 $$

(i) The value of m:

Using condition 2 and our domain  $$0 \leq x \leq 2 $$:

 $$ $$\int_{0}^{2} mx^2 dx = 1 $$ $$

 $$ $$m \left[ \frac{x^3}{3} \right]_{0}^{2} = 1 $$ $$

 $$ $$m \left( \frac{8}{3} – 0 \right) = 1 $$ $$

 $$ $$m = \frac{3}{8} $$ $$

(ii) P(0≤x≤1):

 $$ $$P(0 \leq x \leq 1) = \int_{0}^{1} \frac{3}{8}x^2 dx $$ $$

 $$ $$P(0 \leq x \leq 1) = \frac{3}{8} \left[ \frac{x^3}{3} \right]_{0}^{1} $$ $$

 $$ $$P(0 \leq x \leq 1) = \frac{3}{8} \left( \frac{1}{3} – 0 \right) $$ $$

 $$ $$P(0 \leq x \leq 1) = \frac{3}{8} \times \frac{1}{3} = \frac{1}{8} $$ $$

(iii) The mean:

 $$ $$\mu = \int_{0}^{2} x \cdot f(x) dx = \int_{0}^{2} x \cdot \frac{3}{8}x^2 dx = \frac{3}{8} \int_{0}^{2} x^3 dx $$ $$

 $$ $$\mu = \frac{3}{8} \left[ \frac{x^4}{4} \right]_{0}^{2} $$ $$

 $$ $$\mu = \frac{3}{8} \left( \frac{16}{4} – 0 \right) $$ $$

 $$ $$\mu = \frac{3}{8} \times 4 = \frac{12}{8} = \frac{3}{2} $$ $$

(iv) The variance:

 $$ $$\sigma^2 = \int_{0}^{2} (x-\mu)^2 \cdot f(x) dx $$ $$

 $$ $$\sigma^2 = \int_{0}^{2} \left(x-\frac{3}{2}\right)^2 \cdot \frac{3}{8}x^2 dx $$ $$

An alternative approach is to use the formula:

 $$ $$\sigma^2 = E(X^2) – \mu^2 $$ $$

First, find  $$E(X^2) $$:

 $$ $$E(X^2) = \int_{0}^{2} x^2 \cdot f(x) dx = \int_{0}^{2} x^2 \cdot \frac{3}{8}x^2 dx = \frac{3}{8} \int_{0}^{2} x^4 dx $$ $$

 $$ $$E(X^2) = \frac{3}{8} \left[ \frac{x^5}{5} \right]_{0}^{2} $$ $$

 $$ $$E(X^2) = \frac{3}{8} \left( \frac{32}{5} – 0 \right) $$ $$

 $$ $$E(X^2) = \frac{3}{8} \times \frac{32}{5} = \frac{96}{40} = \frac{12}{5} $$ $$

Now calculate the variance:

 $$ $$\sigma^2 = E(X^2) – \mu^2 = \frac{12}{5} – \left(\frac{3}{2}\right)^2 = \frac{12}{5} – \frac{9}{4} $$ $$

 $$ $$\sigma^2 = \frac{48 – 45}{20} = \frac{3}{20} = 0.15 $$ $$

Question 11(b): On an average, two accidents occur every day on a busy highway and it is known that the number of accidents per day follows poison law. Find the probability that on a certain day:

Given:

– Mean number of accidents per day (λ) = 2

– The number of accidents follows a Poisson distribution

The Poisson probability mass function is:

 $$ $$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} $$ $$

(i) There are 2 accidents:

 $$ $$P(X = 2) = \frac{e^{-2} \times 2^2}{2!} = \frac{e^{-2} \times 4}{2} = 2e^{-2} = 2 \times 0.1353 = 0.2706 $$ $$

(ii) There is no accident:

 $$ $$P(X = 0) = \frac{e^{-2} \times 2^0}{0!} = e^{-2} = 0.1353 $$ $$

(iii) Find the least value, n such that the probability for n accidents on a particular day is <0.1.

We need to find the smallest n such that P(X = n) < 0.1:

 $$ $$P(X = n) = \frac{e^{-2} \times 2^n}{n!} < 0.1 $$ $$

Let’s check for different values of n:

For n = 5:

 $$ $$P(X = 5) = \frac{e^{-2} \times 2^5}{5!} = \frac{e^{-2} \times 32}{120} = \frac{32 \times 0.1353}{120} = \frac{4.3296}{120} = 0.0361 $$ $$

For n = 4:

 $$ $$P(X = 4) = \frac{e^{-2} \times 2^4}{4!} = \frac{e^{-2} \times 16}{24} = \frac{16 \times 0.1353}{24} = \frac{2.1648}{24} = 0.0902 $$ $$

Since P(X = 4) < 0.1 and P(X = 3) ≈ 0.1804 > 0.1, the least value of n is 4.

[Approximately take e^{-2} = 1/7]