PSEB Class 11 Mathematics September 2025 Exam Solutions – Step-by-Step Guide

The Punjab School Education Board (PSEB) Class 11 Mathematics September 2025 examination was an important assessment for students preparing for their board exams. The paper covered a balanced mix of topics, including permutations and combinations, sets, functions and relations, complex numbers, inequalities, and trigonometry.

To help students, we have created step-by-step solutions to all questions from this exam. Each problem is solved systematically, starting with the relevant formula and then moving through the logical steps until the final answer is reached. For example, questions from permutations and combinations are solved using factorial-based formulas and case-by-case reasoning. Set theory and functions/relations questions are explained with proper Venn diagrams and mappings. Problems from complex numbers include detailed use of conjugates, modulus, and arguments, while inequalities are solved using algebraic manipulations and number line representation. Trigonometric identities and equations are carefully simplified to help students understand common tricks and shortcuts.

Section A (MCQs with Solutions)

Q1(i).

Let \(A=\{1,3,4,5\}\). Which of the following statements is true?

1. \(5\notin A\)
2. \(\{3\}\in A\)
3. \(\{1,4\}\not\subset A\)
4. \(\{3\}\subset A\)

Solution:

A is false because 5 ∈ A.

B is ture because {3} is an element but a subset.

C is false because {1,4} ⊆ A.

D is false because {3} is element of A, not a subset of A.

Answer: B

Q1(ii).

If \(P = \{0,2,5\}\) and \(Q = \{2,0,6\}\), then \(Q - P\) is:

1. \(\{0,2\}\)
2. \(\{5\}\)
3. \(\{0,2,5,6\}\)
4. \(\{6\}\)

Solution:

\(Q - P = \{2,0,6\} - \{0,2,5\} \)

\(Q - P = \{6\} \)

Answer: D.

Q1(iii).

Interval form of \(\{x\in\mathbb{R}: x\le -2\}\) is:

1. \((0, 2]\)
2. \((-∞,-2)\)
3. \((-∞,-2]\)
4. \([-2,∞)\)

Solution:

All real numbers less than or equal to −2 are written as:

\((-\infty, -2]\)

Answer: C

Q1(iv).

\(\{-1\}\times\{1, 2\}=\)

1. \(\{1, 2\}\times\{-1\}\)
2. \(\{(-1, 1), (-1, 2)\}\)
3. \(\{(-1, 1), (-1, -2)\}\)
4. \(\{-1,2\}\)

Solution:

The Cartesian product of singletons gives:

\(\{-1\}\times\{1, 2\}=\{(-1,2)\}\)

=\(\{(-1, 1), (-1, 2)\}\)

Answer: B

Q1(v).

\(\{1,2\}\times\{-1\}=\)

1. \(\{(1,-1),(2,-1)\}\)
2. \(\{(-1,1),(-1,2)\}\)
3. \(\{(-1,-1),(-2,-1)\}\)
4. \(\{(1,2),(-1)\}\)

Solution:

Each element of {1,2} pairs with −1:

\(\{(1,-1),(2,-1)\}\)

Answer: A

Q1(vi).

Let \(R = \{(x^2 + 1,x)\); x is an odd prime number less than 7}. Domain of R is:

1. \(\{3,5,7\}\)
2. \(\{3,5\}\)
3. \(\{10,26\}\)
4. \(\{10,26,50\}\)

Solution:

Odd primes less than 7 are 3,5. So:

but here relation is \(x^2 + 1 \) so \[ 3^2 +1 =10\] and \[5^2+ 1 =26\]

\(R = \{(10,3), (26,5)\}\)

from above it is clear that domain is \(\{10, 26\}\)

\(\{10, 26\}\)

Answer: C

Q1(vii).

Domain of \(f(x)=\sqrt{x}\) is:

1. (0, ∞)
2. [0, ∞)
3. (-∞,0]
4. (-∞,0)

Solution:

we know that under the square root there can't be negative mumber so

\(x\geq 0 \). So :

\(\Rightarrow [0,\infty)\)

Answer: B

Q1(viii).

\(\sin120^\circ=\)

1. \(1/2\)
2. \(1/\sqrt{2}\)
3. \(\sqrt{3}/2\)
4. \(- \sqrt{3}/2\)

Solution:

120° = 180°−60°, in quadrant II where sine is positive. So:

\(\sin120^\circ=\sin(180^\circ - 60^\circ)\)

\(\sin(60^\circ)= \frac{\sqrt{3}}{2}\)

Answer: C

Q1(ix).

\(\sin13A\cos5A-\cos13A\sin5A=\)

1. \(\sin8A\)
2. \(\sin18A\)
3. \(\cos8A\)
4. \(\cos18A\)

Solution:

Formula: \[\sin C\cos D-\cos C\sin D=\sin(C-D)\]. So:

\(\sin(13A-5A)=\sin8A\)

Answer: A

Q1(x).

If \(\sin x = 0.04\) and \(\cos x = -0.04\), then \(x\) lies in quadrant:

1. I
2. II
3. III
4. IV

Solution:

Here \(\sin x\) is positive and \(\cos x\) is negative so the \(x\) should be in Quadrant II.

Answer: B

Q1(xi).

If \(\overline{x + iy} = 3 - 4i\), then \(x+y\) = ?

1. −1
2. −7
3. 1
4. 7

Solution:

From equations:

let us assume complex number is \(x + iy\), then its conjugate will be \[\overline{x + iy} = x - iy\] given that If \[\overline{x + iy} = 3 - 4i\] \[\Rightarrow x - iy = 3 - 4i \]

\(\Rightarrow x = 3, y = 4\)

So x+y= 7.

D

Q1(xii).

Imaginary part of \((1+i)^2\) is equal to:

1. 0
2. 2
3. 2i
4. −2

Solution:

\((1+i)^2=1+2i+i^2\)

\[1+2i-1=2i\]

The imaginary part is 2 (not 2i).

Answer: B

Q1(xii).

Value of \(i^{-101}\) is equal to:

1. 1
2. −1
3. i
4. −i

Solution (formula used):

Powers of \(i\) cycle every 4 terms:

\(i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1.\)

Steps:

\[i^{-101} = \frac{1 }{i^{101}}.\]

Now find \(i^{101}\):

101 ÷ 4 = 25 remainder 1 ⇒ \(i^{101} = i^1 = i.\)

So

\[i^{-101} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i \]

Answer: D.

Q1(xiii).

If S is solution set of \(x-1 > 3x-7\) for natural numbers x. Which is true?

1. 3 ∈ S
2. 4 ∈ S
3. −3 ∈ S
4. 2 ∈ S

Solution:

\[x - 1 > 3x - 7\] \[\Rightarrow -1 + 7 > 3x - x\] \[\Rightarrow 6 > 2x\] \[\Rightarrow x < 3\]

Since x ∈ N, possible values are 1 and 2. So S = {1,2}.

Checking options: only 2 ∈ S.

Answer: D

Q1(xiv).

Solution set of \(4 < 2x \le 6\):

1. \((4, 6]\)
2. \((2,3)\)
3. \((2, 3]\)
4. \([2,3]\)

Solution:

\[4 < 2x ≤ 6 \] \[⇒ 2 < x ≤ 3\] \[⇒ (2, 3]\]

Answer: C

Q1(xv).

The graph of a number line is shown. Which inequality does it represent?

1. x < 3
2. x ≤ 3
3. x ≥ 3
4. x > 3

Solution:

The line is shaded left of 3 and includes 3 (solid dot). That represents \(x ≤ 3\).

Answer: B

Q1(xvi).

Number of triangles that can be drawn through 10 points on a circle are:

1. 45
2. 90
3. 120
4. 240

Solution:

We know that for triangle we need 3 points, so Choose any 3 points out of 10 on a circle: \[{}^{10}C_{3} = \binom{10}{3} = \frac{10!}{3!7!} = 120 \].

Answer: C

Q1(xvii).

Total number of ways in which "EAGLE" can be arranged are:

1. 60
2. 120
3. 240
4. 480

Solution:

Word EAGLE has 5 letters with E repeated twice.

\(\frac{5!}{2!} = \frac{120}{2} = 60\)

Answer: A

Q1(xviii).

How many 4-digit numbers can be formed from digits {2,3,4,5} which are divisible by 5, no repetition is allowed?

1. 4
2. 8
3. 12
4. 24

Solution:

If number is divisible by 5 then, Last digit must be 5 (to be divisible by 5). Remaining 3 places can be filled with any of 4 digits without repetition.

then total possabilties are \(4 \times 3 \times 2 \times 1 = 24\)

Answer: D

Q1(xix).

Number of terms in expansion of \((1+x)^{59}\) is:

1. 3
2. 51
3. 60
4. 100

Solution:

In expansion of \((1+x)^n\), number of terms = n+1.

59 + 1 = 60

Answer: C

Q1(xx).

If (1+x)⁵ = 1 + 5x + 10x² + ax³ + 5x⁴ + x⁵, then a = ?

1. 5
2. 5
3. 10
4. 15

Solution:

\[ (1+x)^5 = \binom{5}{0}1^5x^0 + \binom{5}{1}1^4x^1\] \[+ \binom{5}{2}1^3x^2 + \binom{5}{3}1^2x^3 + \binom{5}{4}1x^4 \] \[+ \binom{5}{5}x^5 \]

Coefficient of x³ in \((1+x)^5\) is \(\binom{5}{3} = 10\).

But expansion shows coefficient a. So a = 10.

Answer C

Q2.

Let \(S\) = {vowels in the English alphabet that precede the letter \(k\)} and \(T\) = {letters in the word "EDUCATION" that precede \(k\)}. Find (i) \(T - S\) and (ii) \(S\cap T\).

Solution:

Step 1 — Determine \(S\). Vowels are \(a,e,i,o,u\). "Precede \(k\)" means alphabetically before \(k\)

\(S = \{A, E, I\}\).

Step 2 — Determine \(T\). The word "EDUCATION" has letters E, D, U, C, A, T, I, O, N.

\(T = \{E, D, U, C, A, T, I, O, N\}\).

(i) \(T - S\): remove from \(T\) the elements that are in \(S\).

\(T - S = \{D, U, C, T, O, N\}\)

(ii) \(S\cap T\): elements common to both sets.

\(S\cap T = \{A, E, I\}\).

Answer: (i) \(\{D, U, C, T, O, N\}\), (ii) \(\{A, E, I\}\).

Q3.

Draw the graph of the function \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=x^2\). Also write its domain and range.

Solution (formulas/ideas used):

- \(f(x)=x^2\) is a parabola with vertex at \((0,0)\). It opens upwards because coefficient of \(x^2\) is positive.

Graph description:

The curve has symmetry about the y-axis. As \(x\to\pm\infty\), \(f(x)\to\infty\). The minimum value is at \(x=0\) with \(f(0)=0\).

(You may draw a smooth curve through points \((−2,4), (−1,1), (0,0), (1,1), (2,4)\) forming the parabola.)

Domain: all real numbers, since any real input gives a real square.

Domain = \(\mathbb{R}\).

Range: squares are always ≥0, and every nonnegative real is attained (e.g. \(x=\sqrt{y}\)).

Range = \([0,\infty)\).

Answer: Graph — parabola with vertex (0,0), Domain \(\mathbb{R}\), Range \([0,\infty)\).

Q4.

A circular disc rotates 600 times in 10 minutes. Calculate the angle (in radians) turned by the disc in 15 seconds.

Solution (formulas used):

- One full rotation = \(2\pi\) radians./p>

Steps:

600 revolutions in 10 minutes ⇒ revolutions per minute = \(600/10 = 60\) rev/min.

Convert to rev/sec: \(60\ \text{rev/min} = 60/60 = 1\) rev/sec.

In 15 seconds the disc makes:

\( \text{rev/sec} \times 15\ \text{sec} = 15\ \text{revolutions}\)

Angle in radians = revolutions × \(2\pi\):

Angle = \(15 \times 2\pi = 30\pi\) radians.

Answer: \(30\pi\) radians.

Q5.

Solve the inequality: \(\;5 - 3(x+1) < 2(4 - x)\).

Solution (operations used):

Distribute and collect like terms; isolate \(x\).

\[5 - 3(x+1) < 2(4 - x)\]

\[ \Rightarrow 5 - 3x - 3 < 8 - 2x\]

\[\Rightarrow 2 - 3x < 8 - 2x\]

\[\Rightarrow -3x + 2x < 8 - 2\]

we knwo that ineqaulity changes when we multiply negative sign on both sides \(-x < 6 \quad\Rightarrow\quad x > -6 \)

Solution set: \(x > -6\) or \((-6, \infty )\).

Q6.

Find \(\operatorname{Re}\!\Big(\dfrac{-1}{1+i}\Big)\)).

Solution (formulas used):

First we will make it in "a + ib" form, To simplify a complex fraction multiply numerator and denominator by the conjugate of the denominator.

\[\dfrac{-1}{1+i} = \dfrac{-1}{1+i}\cdot\dfrac{1-i}{1-i}\] \[=\dfrac{-(1-i)}{(1+i)(1-i)}.\]

\[\Rightarrow (1+i)(1-i)=1^2 - i^2\] \[= 1-(-1)=2.\]

\(\dfrac{-(1-i)}{2} = \dfrac{-1 + i}{2} = -\dfrac{1}{2} + \dfrac{i}{2}.\)

so \[\operatorname{Re}\!\Big(\dfrac{-1}{1+i}\Big) = \operatorname{Re}\!\Big(-\dfrac{1}{2} + \dfrac{i}{2}. \]

The real part is : \(-\dfrac{1}{2}\).

Answer: \(\operatorname{Re}\!\Big(\dfrac{-1}{1+i}\Big) = -\dfrac{1}{2}.\)

Q7.

Find the number of 4-card combinations from a standard 52-card deck that contain exactly one king.

Solution (formulas used):

- Combination formula: \({}^{n}C_{r}=\dfrac{n!}{r!(n-r)!}\).

- Choose 1 king from the 4 kings, and choose the other 3 cards from the 48 non-king cards.

\[ \text{Number} = \binom{4}{1}\cdot\binom{48}{3}.\]

\[ \binom{48}{3}=\dfrac{48\cdot47\cdot46}{3\cdot2\cdot1}\] \[=\dfrac{103{,}776}{6}=17{,}296.\]

\[\binom{4}{1}=4\quad\Rightarrow\quad 4\times17{,}296 = 69{,}184.\]

Answer: \(\boxed{69{,}184}\) combinations.

Q8.

A committee of 5 members is to be formed from 7 boys and 9 girls. How many committees have at least 4 girls?

Solution (cases + formula used):

“At least 4 girls” means either 4 girls and 1 boy, or 5 girls.

\[\textbf{Case 1 (4 girls,1 boy)}: \binom{9}{4}\cdot\binom{7}{1}.\]

\[ \binom{9}{4}=\dfrac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2\cdot1}\] \[=126,\quad \binom{7}{1}=7.\]

\[\Rightarrow 126\cdot7 = 882.\]

\[\textbf{Case 2 (5 girls)}: \binom{9}{5} = \binom{9}{4} = 126.\]

\[ \textbf{Total} = 882 + 126 = 1008.\]

Answer: \(\boxed{1008}\) possible committees.

Q9.

Expand \(\displaystyle\Big(\tfrac{2}{x} - \tfrac{3x}{2}\Big)^{4}\).

Solution (formula used):

- Binomial theorem: \((a+b)^n=\sum_{k=0}^{n}\binom{n}{k} a^{\,n-k} b^{\,k}.\)

Set \(a=\tfrac{2}{x},\; b= -\tfrac{3x}{2},\; n=4.\)

\[\Big(\tfrac{2}{x} - \tfrac{3x}{2}\Big)^4\] \[= \sum_{k=0}^4 \binom{4}{k}\Big(\tfrac{2}{x}\Big)^{4-k}\Big(-\tfrac{3x}{2}\Big)^k.\]

Compute term by term:

\[k=0: \binom{4}{0}\Big(\tfrac{2}{x}\Big)^4 = \tfrac{16}{x^4}.\]

\[ k=1: \binom{4}{1}\Big(\tfrac{2}{x}\Big)^3\Big(-\tfrac{3x}{2}\Big)\] \[= 4\cdot \tfrac{8}{x^3}\cdot\big(-\tfrac{3x}{2}\big)\] \[= -\tfrac{48}{2}\cdot \tfrac{1}{x^2} = -\tfrac{24}{x^2}.\]

\[ k=2: \binom{4}{2}\Big(\tfrac{2}{x}\Big)^2\Big(\tfrac{9x^2}{4}\Big)\] \[= 6\cdot \tfrac{4}{x^2}\cdot \tfrac{9x^2}{4} = 6\cdot 9 = 54.\]

\[k=3: \binom{4}{3}\Big(\tfrac{2}{x}\Big)\Big(-\tfrac{27x^3}{8}\Big)\] \[= 4\cdot \tfrac{2}{x}\cdot \big(-\tfrac{27x^3}{8}\big) = -\tfrac{216}{8}x^2 = -27x^2.\]

\[k=4: \binom{4}{4}\Big(-\tfrac{3x}{2}\Big)^4 = \tfrac{81x^4}{16}.\]

Combine:

\[\Big(\tfrac{2}{x} - \tfrac{3x}{2}\Big)^4\] \[= \tfrac{16}{x^4} - \tfrac{24}{x^2} + 54 - 27x^2 + \tfrac{81}{16}x^4.\]

Answer: \(\dfrac{16}{x^4} - \dfrac{24}{x^2} + 54 - 27x^2 + \dfrac{81}{16}x^4.\)

Q10.

If \(cosec x = -\dfrac{25}{7}\), where \(x\) is in the 3rd quadrant, then find \(\sin\!\big(\tfrac{3\pi}{2}-x\big)\).

Solution (formulas used):

\(cosec x = \dfrac{1}{\sin x}\). \[ \sin(A-B)=\sin A \cos B - \cos A \sin B.\] \[\text{Here} A=\tfrac{3\pi}{2}, B=x.\] \(\text{Also}\) \[\sin(3\pi/2)=-1,\; \cos(3\pi/2)=0.\]

Step 1 — Find \(\sin x\)

\(cosec x = -25/7 \;\;\Rightarrow\;\; \sin x = -7/25.\)

Step 2 — Find \(\cos x\).

\[\cos^2 x = 1 - \sin^2 x \] \[= 1 - (7/25)^2 = 576/625.\]

\(\cos x = \pm 24/25.\)

In the 3rd quadrant both sine and cosine are negative, so \(\cos x = -24/25.\)

Step 3 — Use identity for \(\sin(3\pi/2 - x)\):

\[\sin(3\pi/2 - x) \] \[= \sin(3\pi/2)\cos x - \cos(3\pi/2)\sin x.\]

Now, \[\sin(3\pi/2)=-1,\; \cos(3\pi/2)=0.\]

\[\sin(3\pi/2 - x) =(-1)\cdot\cos x - 0\cdot\sin x \] \[= -\cos x.\]

Step 4 — Substitute \(\cos x\):

\(\sin(3\pi/2 - x) = -(-24/25) = 24/25.\)

Answer: \(\dfrac{24}{25}\).

Q11 (Or).

Prove that

\[-4\cos 300^\circ + \sqrt{3}\csc 1500^\circ \] \[- 2\sqrt{3}\sin(-780^\circ) = 3\]

Formulas / reduction facts used:

• \(\cos(300^\circ)=\tfrac{1}{2}.\)
• Reduce angles modulo \(360^\circ\): \(1500^\circ \equiv 1500-4\cdot360=60^\circ.\)
• \(\csc\theta=\dfrac{1}{\sin\theta},\quad \sin(-\theta)=-\sin\theta.\)
• \(\sin 60^\circ=\tfrac{\sqrt{3}}{2}.\)

Step 1 — first term:

\(-4\cos 300^\circ = -4\cdot \tfrac{1}{2} = -2.\)

Step 2 — second term (note 1500° → 60°):

\(\csc 1500^\circ = \csc 60^\circ = \dfrac{1}{\sin 60^\circ} = \dfrac{2}{\sqrt{3}}.\)

\(\sqrt{3}\csc 1500^\circ = \sqrt{3}\cdot\dfrac{2}{\sqrt{3}} = 2.\)

Step 3 — third term (reduce -780°):

\[\sin(-780^\circ) = -\sin(780^\circ),\quad 780^\circ\equiv 60^\circ,\; \sin 60^\circ=\tfrac{\sqrt{3}}{2},\]

\[ \sin(-780^\circ) = -\tfrac{\sqrt{3}}{2}.\]

\[-2\sqrt{3}\sin(-780^\circ) = -2\sqrt{3}\cdot\Big(-\tfrac{\sqrt{3}}{2}\Big)\] \[= \dfrac{-2\sqrt{3}\cdot(-\sqrt{3})}{2} = \dfrac{2\cdot 3}{2} = 3.\]

Combine all three terms:

(-2) + 2 + 3 = 3.

Therefore: \[-4\cos 300^\circ + \sqrt{3}\csc 1500^\circ\] \[- 2\sqrt{3}\sin(-780^\circ) = \boxed{3}.\]

Q12.

Find the real numbers \(x\) and \(y\) if \((x-iy)(3+5i)\) is the conjugate of \(-6-24i\).

Solution (formulas used):

Complex conjugate: \(\overline{a+bi}=a-bi\).

Steps:

The conjugate of \(-6-24i\) is \(-6+24i\).

\[(x-iy)(3+5i)\] \[=(3x+5x i) + (-3iy -5y i^2)\] \[= (3x+5y) + i(5x-3y).\]

Equate real and imaginary parts with \[-6+24i:\]

\[3x + 5y = -6,\qquad 5x - 3y = 24.\]

Solve the linear system. Multiply first eqn by 3: \(9x+15y=-18.\)

Multiply second by 5: \(25x-15y=120\). Add above two eqaution we get:

\[34x = 102\] \[\Rightarrow x = 3.\]

Then \[3(3) + 5y = -6 \Rightarrow 9 + 5y = -6\] \[\Rightarrow 5y = -15 \Rightarrow y = -3.\]

Answer: \(x=3,\ y=-3.\)

Q14.

Solve the inequalities: \[ 5(-x+4) - 3(2x+3) \ge 0,\] \[\quad 2x+19 \le 6x+47 \] and represent the solution graphically on a number line.

Solution:

First inequality:

\[ 5(-x+4) - 3(2x+3) ≥ 0 \] \[⇒ -5x+20 -6x-9 ≥ 0 \] \[⇒ -11x+11 ≥ 0 \] \[⇒ -11x ≥ -11 \] \[⇒ x ≤ 1\]

Second inequality:

\[2x+19 ≤ 6x+47\] \[⇒ 19-47 ≤ 4x\] \[⇒ -28 ≤ 4x\] \[⇒ x ≥ -7\]

Combined solution:

\[-7 ≤ x ≤ 1\]

Graphical representation: On a number line, shade the interval between -7 and 1 with closed dots at both ends.

Answer: \([-7,1]\).

Q15.

In how many ways can the letters of the word PERMUTATIONS be arranged? (a) If the word must start with R and end with S. (b) If there are always 7 letters between P and S (i.e. exactly 7 letters occur between them).

Solution — preliminaries and notation (formulas used):

The word PERMUTATIONS has 12 letters in total. List of letters and multiplicities:

P, E, R, M, U, T, A, T, I, O, N, S \(\quad\Rightarrow\quad T\) \(\text{ appears twice, all other letters once.}\)

- When arranging \(n\) objects with repeats, the number of distinct permutations is

\(\dfrac{n!}{n_1!\,n_2!\cdots}\)

- We will treat identical letters (the two T's) as indistinguishable, so divide by \(2!\) where appropriate.

---

(a) Start with R and end with S

Reasoning: Fix R in the first position and S in the last position. That leaves \(12-2=10\) middle places to fill with the remaining letters:

\(\{P, E, M, U, T, T, A, I, O, N\}\)

There are 10 letters to place, but two T's are identical. So the number of distinct arrangements of the middle 10 slots is

\(\dfrac{10!}{2!}.\)

Compute:

\[10! = 3\,628\,800 \quad\Rightarrow\quad \dfrac{10!}{2!}\] \[= \dfrac{3\,628\,800}{2} = 1\,814\,400.\]

Answer (a): \(\boxed{1\,814\,400}\) arrangements.

---

(b) Exactly 7 letters between P and S

Interpretation: “7 letters between P and S” means if one of them sits at position \(i\), the other sits at position \(i+8\). (So their positions differ by 8; there are exactly 7 slots in between.) We treat P and S as distinct letters, so two orders are possible: P before S or S before P.

Step 1 — possible starting positions for the earlier letter:

Positions in the 12-letter word are \(1,2,\dots,12\). If the earlier one is at position \(i\), we need \(i+8\le12\), so \(i\le4\). Thus \(i=1,2,3,4\) — there are 4 choices for the earlier index.

Step 2 — orders of the pair: For each choice of \(i\), the pair (P,S) can occur in either order: (P at \(i\), S at \(i+8\)) or (S at \(i\), P at \(i+8\)). So for each \(i\) there are \(2\) orderings. Hence number of placements for the P–S pair = \(4\times 2 = 8\).

Step 3 — arrange remaining letters: After placing P and S in their two specific positions, there remain \(12-2=10\) positions to fill with the remaining 10 letters:

\[ \{E, R, M, U, T, T, A, I, O, N\},\]

again with two identical T's. Number of arrangements of these 10 letters = \(10!/2!\).

Total count:

\(\text{Total} \) \[= (\text{number of P-S placements}) \times \dfrac{10!}{2!}\] \[= 8 \times \dfrac{10!}{2!}.\]

Numeric value:

\[\dfrac{10!}{2!} = 1\,814\,400 \quad\Rightarrow\quad\] \[\text{Total} = 8 \times 1\,814\,400 = 14\,515\,200.\]

Answer (b): \(\boxed{14\,515\,200}\) arrangements where P and S have exactly 7 letters between them.

---

Optional note (combined constraint): If the problem asked simultaneously that the word must start with R, end with S, and P and S have exactly 7 letters between them, then S is already fixed at position 12. "7 letters between P and S" forces P to be at position \(12-8=4\). So R at position 1, P at 4, S at 12 are fixed and we arrange the remaining 9 letters (with two T's) in the remaining 9 slots:

\[\text{Combined count} = \dfrac{9!}{2!}\] \[= \dfrac{362{,}880}{2} = 181{,}440.\]

So with all three constraints together you would get \(\boxed{181{,}440}\) arrangements.

Q16.

Find \(\sin\frac{x}{2},\ \cos\frac{x}{2}\) and \(\tan\frac{x}{2}\) when \(\sec x = -\frac{25}{7}\) and \(x\) lies in the 3rd quadrant.

Formulas / identities used:

\[\sec x=\dfrac{1}{\cos x}\].   \[\sin^2x+\cos^2x=1\].   Half-angle formulas:

\(\sin\frac{x}{2}=\pm\sqrt{\dfrac{1-\cos x}{2}},\quad\) \(\cos\frac{x}{2}=\pm\sqrt{\dfrac{1+\cos x}{2}},\quad\) \( \tan\frac{x}{2}=\dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}.\)

Step 1 — find \(\cos x\) and \(\sin x\).

Given \(\sec x=-\dfrac{25}{7}\), so

\(\cos x = -\dfrac{7}{25}.\)

Use \(\sin^2x = 1-\cos^2x\):

\[\sin^2 x = 1 - \Big(-\dfrac{7}{25}\Big)^2\] \[= 1 - \dfrac{49}{625} = \dfrac{625-49}{625} = \dfrac{576}{625}.\]

Thus \(\sin x = \pm \dfrac{24}{25}\). Since \(x\) is in the 3rd quadrant, both sine and cosine are negative, so

\(\sin x = -\dfrac{24}{25}.\)

Step 2 — determine the quadrant of \(x/2\).

If \(x\) is in quadrant III, then \(\pi < x < \tfrac{3\pi}{2}\). Dividing by 2 gives

\[\tfrac{\pi}{2} < \tfrac{x}{2} < \tfrac{3\pi}{4}\],

so \(x/2\) lies in quadrant II. Therefore:

\(\sin\frac{x}{2}>0,\quad \cos\frac{x}{2}<0,\) \(\quad \tan \frac{x}{2} <0\)

Step 3 — compute half-angle values (choose signs per quadrant):

Compute \(\sin\frac{x}{2}\):

\[\sin\frac{x}{2} = +\sqrt{\dfrac{1-\cos x}{2}}\] \[ = \sqrt{\dfrac{1 - (-7/25)}{2}}\] \[ = \sqrt{\dfrac{1+7/25}{2}} = \sqrt{\dfrac{(25+7)/25}{2}}\] \[ = \sqrt{\dfrac{32/25}{2}} = \sqrt{\dfrac{32}{50}} \] \[= \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}.\]

Compute \(\cos\frac{x}{2}\): (negative in QII)

\[\cos\frac{x}{2} = -\sqrt{\dfrac{1+\cos x}{2}}\] \[= -\sqrt{\dfrac{1 + (-7/25)}{2}}\] \[= -\sqrt{\dfrac{(25-7)/25}{2}}\] \[= -\sqrt{\dfrac{18/25}{2}} = -\sqrt{\dfrac{18}{50}}\] \[= -\sqrt{\dfrac{9}{25}} = -\dfrac{3}{5}.\]

Finally \(\tan\frac{x}{2} = \dfrac{\sin(x/2)}{\cos(x/2)}\):

\[ \tan\frac{x}{2} = \dfrac{\tfrac{4}{5}}{-\tfrac{3}{5}}\] \[= -\dfrac{4}{3}.\]

\(\sin\frac{x}{2}=\frac{4}{5}, \cos\frac{x}{2}=-\frac{3}{5}, \tan\frac{x}{2}=-\frac{4}{3}.\)

Q17.

How many numbers greater than 20,000,000 can be formed using the digits 2, 0, 3, 2, 4, 4, 2, 4?

Step 1 — Understand the problem.

We have 8 digits in total: {2, 0, 3, 2, 4, 4, 2, 4}. That is: three 2's, three 4's, one 0, and one 3. So the multiset of digits = {0, 2, 2, 2, 3, 4, 4, 4}.

We want to form 8-digit numbers greater than 20,000,000. Since the total number has 8 digits, the leading digit cannot be 0. So the first digit must be either 2, 3, or 4.

Step 2 — Total number of distinct arrangements (without restriction).

Total permutations of 8 objects with repeats (3 twos and 3 fours):

\(\dfrac{8!}{3!\,3!} = \dfrac{40320}{6\cdot6} = \dfrac{40320}{36} = 1120.\)

Step 3 — Restriction on first digit.

We cannot allow 0 as first digit. Let’s count separately:

Case A: First digit = 0.

Remaining digits: {2,2,2,3,4,4,4}. Total arrangements of these 7 letters:

\(\dfrac{7!}{3!\,3!} = \dfrac{5040}{6\cdot6} = \dfrac{5040}{36} = 140.\)

Case B: First digit ≠ 0 (2, 3, or 4).

Then all valid 8-digit numbers = Total arrangements − Invalid arrangements with leading 0.

1120 - 140 = 980.

Step 4 — Check the "greater than 20,000,000" condition.

Our valid numbers are 8-digit numbers starting with 2, 3, or 4.
- If first digit = 2: The number is ≥ 20,000,000 (since the next digit at least 0). ✔
- If first digit = 3 or 4: Also obviously ≥ 30,000,000 or 40,000,000. ✔

Hence all nonzero-first-digit arrangements satisfy "greater than 20,000,000."

Final Answer: The required number of 8-digit numbers = 980.

Q18.

Prove that

\[\cos^2 x + \cos^2\!\Big(x+\tfrac{\pi}{3}\Big) + \cos^2\!\Big(x-\tfrac{\pi}{3}\Big)\] \[= \dfrac{3}{2}.\]

Solution (identities used):

\[ 1) \cos^2\theta = \dfrac{1+\cos 2\theta}{2}.\] \[2) \cos(A+B)+\cos(A-B)=2\cos A\cos B.\]

Step 1 — expand each square:

\[\cos^2 x = \tfrac{1+\cos 2x}{2},\] \[\cos^2(x+\tfrac{\pi}{3}) = \tfrac{1+\cos(2x+\tfrac{2\pi}{3})}{2},\] \[\cos^2(x-\tfrac{\pi}{3}) = \tfrac{1+\cos(2x-\tfrac{2\pi}{3})}{2}.\]

Step 2 — add them up:

\[\cos^2 x + \cos^2(x+\tfrac{\pi}{3}) + \cos^2(x-\tfrac{\pi}{3})\] \[= \tfrac{1+1+1}{2} + \tfrac{\cos 2x + \cos(2x+\tfrac{2\pi}{3}) + \cos(2x-\tfrac{2\pi}{3})}{2}.\]

Step 3 — simplify cosine terms:

Use identity: \(\cos(A+B)+\cos(A-B)=2\cos A\cos B.\) here \(A=2x,\ B=\tfrac{2\pi}{3}\):

\[\cos(2x+\tfrac{2\pi}{3})+\cos(2x-\tfrac{2\pi}{3}) \] \[= 2\cos(2x)\cos(\tfrac{2\pi}{3}).\]

Since \(\cos(120^\circ)=\cos(\tfrac{2\pi}{3})=-\tfrac{1}{2}\), this becomes:

\[2\cos(2x)\cdot\Big(-\tfrac{1}{2}\Big) = -\cos(2x).\]

Step 4 — substitute back:

\[ = \tfrac{3}{2} + \tfrac{\cos 2x - \cos 2x}{2} = \tfrac{3}{2}.\]

Hence proved: \[\displaystyle \cos^2 x + \cos^2\!\Big(x+\tfrac{\pi}{3}\Big)\] \[+ \cos^2\!\Big(x-\tfrac{\pi}{3}\Big)=\dfrac{3}{2}.\]

Q18 (a).

Find the value of: \[ \cot\left(\tfrac{17\pi}{3}\right) \]

Step 1: Reduce angle into principal range \(0\) to \(2\pi\). \[\tfrac{17\pi}{3} = \tfrac{18\pi}{3} - \tfrac{\pi}{3} = 6\pi - \tfrac{\pi}{3}.\] Since cotangent has period \(\pi\): \[\cot\left(\tfrac{17\pi}{3}\right) = \cot\left(-\tfrac{\pi}{3}\right).\]

Step 2: Use odd function property: \[\cot(-\theta) = -\cot(\theta).\] So \[\cot(-\tfrac{\pi}{3}) = -\cot(\tfrac{\pi}{3}).\]

Step 3: \[\cot(\tfrac{\pi}{3}) = \tfrac{1}{\tan(\pi/3)} = \tfrac{1}{\sqrt{3}}.\]

Final Answer: \[ \cot\left(\tfrac{17\pi}{3}\right) = -\tfrac{1}{\sqrt{3}} \]

Q18 (b).

Prove that: \[ \cos\left(\tfrac{5\pi}{4} - x\right) - \cos\left(\tfrac{5\pi}{4} + x\right) = -\sqrt{2}\sin x \]

Formula used: \[\cos A - \cos B = -2 \sin\left(\tfrac{A+B}{2}\right)\sin\left(\tfrac{A-B}{2}\right).\]

Step 1: Let \(A = \tfrac{5\pi}{4} - x,\; B = \tfrac{5\pi}{4} + x.\)

Step 2: Compute: \[\frac{(A+B)}{2} = \tfrac{(5\pi/4 - x) + (5\pi/4 + x)}{2} = \tfrac{5\pi}{4}.\] \[\frac{(A-B)}{2} = \tfrac{(5\pi/4 - x) - (5\pi/4 + x)}{2} = -x.\]

Step 3: Apply identity: \(\cos A - \cos B = -2 \sin\left(\tfrac{5\pi}{4}\right)\sin(-x).\)

Step 4: Simplify values: \(\sin(\frac{5\pi}{4}) = \sin(\pi + \frac{\pi}{4}) = -\tfrac{1}{\sqrt{2}},\)\(\; \sin(-x) = -\sin x.\)
So RHS = \(-2 \cdot (-\tfrac{\sqrt{2}}{2}) \cdot (-\sin x).\)

= \(-2 \cdot -\tfrac{\sqrt{2}}{2} \cdot -\sin x = -\sqrt{2}\sin x.\)

Hence Proved ✅

Q. Sets Problem.

If \(A = \{t: t\in \mathbb{Z}, t^2 \leq 3\}\) \( B = \{y: y \text{ divides } 6\}\) \( C = \{x: x \in \mathbb{R}, x^2 - x = 0\},\) then find: (i) \((A \cap B)\cup C\) (ii) \(C \times (A-B)\)

Step 1: Find each set.

\(A = \{t \in \mathbb{Z}: t^2 \leq 3\} = \{-1,0,1\}.\) \(B = \{y: y \mid 6\}\) \[= \{-6,-3,-2,-1,1,2,3,6\}.\] \[C = \{x: x^2 - x = 0\}\] \[\Rightarrow x(x-1)=0\] \[\Rightarrow x=0,1.\; \therefore C=\{0,1\}.\]

Step 2: (i) Find (A ∩ B).
\(A \cap B\) \(= \{-1,0,1\} \cap \{-6,-3,-2,-1,1,2,3,6\}\) \(= \{-1,1\}.\)
So \((A \cap B)\cup C = \{-1,1\}\cup\{0,1\} \) \[= \{-1,0,1\}.\]

Step 3: (ii) Find A - B.
\(A - B =\) \[\{-1,0,1\} - \{-6,-3,-2,-1,1,2,3,6\}\] \(= \{0\}.\)
So \(C \times (A - B)\) \[= \{0,1\}\times \{0\}\] \[= \{(0,0),(1,0)\}.\]

Final Answer:
(i) \(\{-1,0,1\}\)
(ii) \(\{(0,0),(1,0)\}\)

Q. Complex Numbers.

If \(z_1 = 2 - i,\; z_2 = 1+i,\) find: \[ \left|\dfrac{z_1+z_2+1}{z_1-z_2+1}\right| \]

Step 1: Compute numerator \(z_1+z_2+1 = (2-i)+(1+i)+1 = 4.\)

Step 2: Compute denominator \[z_1-z_2+1 = (2-i)-(1+i)+1 \] \[ = (2-1+1)+(-i-i) = 2-2i.\]

thus we have \[ \left|\dfrac{z_1+z_2+1}{z_1-z_2+1}\right| = \left|\frac{4}{2 - 2i}\right| \]

Step 3: Fraction = \( \tfrac{4}{2-2i}.\)
Multiply numerator and denominator by conjugate \((2+2i)\): \[\left|\frac{4}{2 - 2i}\times \frac{(2+2i)}{(2+2i)} \right|=\left| \frac{8+8i}{8}\right| \] \[= \left |1+i\right|.\]

Step 4: Modulus = \(|1+i| = \sqrt{1^2+1^2} = \sqrt{2}.\)

Final Answer: \(\sqrt{2}\).

Q. Trigonometry (Half-angle).

Find \(\sin \tfrac{x}{2},\; \cos \tfrac{x}{2},\; \tan \tfrac{x}{2}\) when \(\sec x = -\tfrac{25}{7},\; x\) lies in 3rd quadrant.

Step 1: Since \(\sec x = -\tfrac{25}{7}\), we get \[\cos x = -\tfrac{7}{25}.\]

Step 2: Find \(\sin x\). \[\sin^2x = 1 - \cos^2x = 1 - \left(\tfrac{7}{25}\right)^2\] \[= 1 - \tfrac{49}{625} = \tfrac{576}{625}.\]
So \[\sin x = -\tfrac{24}{25}\] (negative in QIII).

Step 3: Use half-angle formulas: \[\sin \tfrac{x}{2} = \pm\sqrt{\tfrac{1-\cos x}{2}},\] \[\cos \tfrac{x}{2} = \pm\sqrt{\tfrac{1+\cos x}{2}},\] \[\tan \tfrac{x}{2} = \tfrac{\sin x}{1+\cos x}.\]

Step 4: Determine quadrant of \(\tfrac{x}{2}.\) If \(x\) is in QIII (\(\pi < x < 3\pi/2\)), then \(\tfrac{x}{2}\) lies in QII. In QII: \(\sin \tfrac{x}{2} > 0,\; \cos \tfrac{x}{2} < 0,\; \tan \tfrac{x}{2} < 0.\)

Step 5: Compute values. \[\sin \tfrac{x}{2} = +\sqrt{\tfrac{1-(-7/25)}{2}} = \sqrt{\tfrac{1+7/25}{2}}\] \[= \sqrt{\tfrac{32/25}{2}} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}.\]
\[\cos \tfrac{x}{2} = -\sqrt{\tfrac{1+(-7/25)}{2}}\] \[= -\sqrt{\tfrac{18/25}{2}} = -\sqrt{\tfrac{9}{25}} = -\tfrac{3}{5}.\]
\[\tan \tfrac{x}{2} = \tfrac{\sin(x)}{1+\cos(x)} \] \[= \tfrac{-24/25}{1-7/25} = \tfrac{-24/25}{18/25} = -\tfrac{24}{18} = -\tfrac{4}{3}.\]

Final Answer:
\(\sin \tfrac{x}{2} = \tfrac{4}{5},\; \cos \tfrac{x}{2} = -\tfrac{3}{5},\; \tan \tfrac{x}{2} = -\tfrac{4}{3}.\)