CBSE Class 11 Mathematics Revision Assignment Set 04

Question 1. Prove by the principle of Mathematical induction for all natural numbers: \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\)
Answer: Let the given statement be \(P(n)\):
\(P(n) : \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\)
Step 1: For \(n = 1\),
LHS = \(\frac{1}{1 \cdot 2} = \frac{1}{2}\)
RHS = \(\frac{1}{1 + 1} = \frac{1}{2}\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)} = \frac{k}{k + 1}\)
Step 3: We need to prove that \(P(k + 1)\) is true, i.e.,
\(\frac{1}{1 \cdot 2} + \dots + \frac{1}{k(k + 1)} + \frac{1}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}\)
Using the induction hypothesis, LHS becomes:
\(= \frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)}\)
\(= \frac{1}{k + 1} \left( k + \frac{1}{k + 2} \right)\)
\(= \frac{1}{k + 1} \left( \frac{k(k + 2) + 1}{k + 2} \right)\)
\(= \frac{1}{k + 1} \left( \frac{k^2 + 2k + 1}{k + 2} \right)\)
\(= \frac{1}{k + 1} \left( \frac{(k + 1)^2}{k + 2} \right) = \frac{k + 1}{k + 2}\)
Thus, LHS = RHS for \(n = k + 1\). Hence, by PMI, the statement is true for all natural numbers \(n\).
In simple words: We check the formula for 1, assume it holds for k, and prove it works for k + 1.

Exam Tip: Always write the three formal steps clearly: base case for \(n = 1\), induction hypothesis for \(n = k\), and inductive step for \(n = k + 1\).

Question 2. Prove by the principle of Mathematical induction for all natural numbers: \(\sin \theta + \sin 2\theta + \sin 3\theta + \dots + \sin n\theta = \frac{\sin\left(\frac{n\theta}{2}\right)\sin\left(\frac{(n + 1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Answer: Let \(P(n) : \sin \theta + \sin 2\theta + \dots + \sin n\theta = \frac{\sin\left(\frac{n\theta}{2}\right)\sin\left(\frac{(n + 1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Step 1: For \(n = 1\),
LHS = \(\sin \theta\)
RHS = \(\frac{\sin\left(\frac{\theta}{2}\right)\sin(\theta)}{\sin\left(\frac{\theta}{2}\right)} = \sin \theta\)
Thus, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(\sin \theta + \sin 2\theta + \dots + \sin k\theta = \frac{\sin\left(\frac{k\theta}{2}\right)\sin\left(\frac{(k + 1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Step 3: For \(n = k + 1\), add \(\sin(k + 1)\theta\) to both sides:
\(\text{LHS} = \frac{\sin\left(\frac{k\theta}{2}\right)\sin\left(\frac{(k + 1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} + \sin(k + 1)\theta\)
\(= \frac{1}{\sin\left(\frac{\theta}{2}\right)} \left[ \sin\left(\frac{k\theta}{2}\right)\sin\left(\frac{k\theta + \theta}{2}\right) + \sin\left(\frac{\theta}{2}\right)\sin(k + 1)\theta \right]\)
Using the product-to-sum identity \(2\sin A \sin B = \cos(A - B) - \cos(A + B)\), simplify and combine terms to get:
\(= \frac{\sin\left(\frac{(k + 1)\theta}{2}\right)\sin\left(\frac{(k + 2)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Thus, \(P(k + 1)\) is true. Hence proved by induction.
In simple words: Substitute k + 1 into the expression and use trigonometric formulas to simplify the sum.

Exam Tip: Memorize standard trigonometric transformation formulas as they are essential for simplifying induction steps involving series of sines and cosines.

Question 3. Prove by the principle of Mathematical induction for all natural numbers: \((11^{n + 2} + 12^{2n + 1})\) is divisible by 133.
Answer: Let \(P(n) : 11^{n + 2} + 12^{2n + 1}\) is divisible by 133.
Step 1: For \(n = 1\),
\(11^{1 + 2} + 12^{2(1) + 1} = 11^3 + 12^3 = 1331 + 1728 = 3059\)
Since \(\frac{3059}{133} = 23\), \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true, i.e., \(11^{k + 2} + 12^{2k + 1} = 133m\) for some integer \(m\).
Step 3: For \(n = k + 1\),
\(11^{k + 1 + 2} + 12^{2(k + 1) + 1} = 11^{k + 3} + 12^{2k + 3}\)
\(= 11 \cdot 11^{k + 2} + 12^2 \cdot 12^{2k + 1}\)
\(= 11 \cdot 11^{k + 2} + 144 \cdot 12^{2k + 1}\)
\(= 11(11^{k + 2} + 12^{2k + 1}) + (144 - 11)12^{2k + 1}\)
\(= 11(133m) + 133(12^{2k + 1})\)
\(= 133(11m + 12^{2k + 1})\)
This is clearly divisible by 133. Thus, \(P(k + 1)\) is true.
In simple words: Express the expression for k + 1 in terms of the assumption for k to show 133 is a factor.

Exam Tip: Split powers like \(12^{2k + 3}\) into \(12^2 \cdot 12^{2k + 1}\) and add/subtract coefficients to factor out the divisor 133.

Question 4. (i) If \(P(2n - 1, n) : P(2n + 1, n - 1) = 22 : 7\), find \(n\) 
(ii) If \(P(2n, 3) : P(n, 3) = 11 : 1\), find \(n\)
Answer: (i) We are given:
\(\frac{P(2n - 1, n)}{P(2n + 1, n - 1)} = \frac{22}{7}\)
\(\frac{\frac{(2n - 1)!}{(2n - 1 - n)!}}{\frac{(2n + 1)!}{(2n + 1 - n + 1)!}} = \frac{22}{7}\)
\(\frac{(2n - 1)!}{(n - 1)!} \cdot \frac{(n + 2)!}{(2n + 1)!} = \frac{22}{7}\)
Simplifying factorials:
\(\frac{(n + 2)(n + 1)n}{(2n + 1)(2n)(2n - 1)} = \frac{22}{7}\)
\(\frac{(n + 2)(n + 1)}{2(2n + 1)(2n - 1)} = \frac{22}{7}\)
Solving this quadratic equation gives \(n = 5\).
(ii) Given \(\frac{P(2n, 3)}{P(n, 3)} = \frac{11}{1}\):
\(\frac{\frac{(2n)!}{(2n - 3)!}}{\frac{n!}{(n - 3)!}} = 11\)
\(\frac{2n(2n - 1)(2n - 2)}{n(n - 1)(n - 2)} = 11\)
\(\frac{2n(2n - 1)2(n - 1)}{n(n - 1)(n - 2)} = 11\)
\(\frac{4(2n - 1)}{n - 2} = 11\)
\(8n - 4 = 11n - 22 \implies 3n = 18 \implies n = 6\)
In simple words: Expand the permutations using factorial formulas and solve for n.

Exam Tip: Recall that \(P(n, r) = \frac{n!}{(n-r)!}\) and expand larger factorials until they cancel with smaller ones.

Question 5. How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?
Answer: Numbers between 100 and 1000 are 3-digit numbers, i.e., from 100 to 999.
Let the number be represented by three vacant places: \(\underline{\quad} \; \underline{\quad} \; \underline{\quad}\).
Case 1: 7 is at the hundred's place.
- Hundred's place can be filled in 1 way (only digit 7).
- Ten's place can be filled in 9 ways (0 to 9 except 7).
- Unit's place can be filled in 9 ways (0 to 9 except 7).
Total numbers = \(1 \times 9 \times 9 = 81\).
Case 2: 7 is at the ten's place.
- Hundred's place can be filled in 8 ways (1 to 9 except 7).
- Ten's place can be filled in 1 way (only 7).
- Unit's place can be filled in 9 ways (except 7).
Total numbers = \(8 \times 1 \times 9 = 72\).
Case 3: 7 is at the unit's place.
- Hundred's place can be filled in 8 ways (1 to 9 except 7).
- Ten's place can be filled in 9 ways (except 7).
- Unit's place can be filled in 1 way (only 7).
Total numbers = \(8 \times 9 \times 1 = 72\).
Total numbers with exactly one digit 7 = \(81 + 72 + 72 = 225\).
In simple words: Count the combinations where digit 7 appears in either the hundreds, tens, or units place.

Exam Tip: Be careful with leading zeros; hundred's place cannot start with 0.

Question 6. How many words beginning and ending with a consonant can be formed by using the letters of the word EQUATION?
Answer: The word EQUATION has 8 distinct letters: E, Q, U, A, T, I, O, N.
Vowels = E, U, A, I, O (5 vowels)
Consonants = Q, T, N (3 consonants)
The word must begin and end with a consonant.
- Number of choices for the first letter (consonant) = 3.
- Number of choices for the last letter (consonant) = 2.
- The remaining 6 letters can be arranged in the remaining 6 places in \(6!\) ways.
Total words = \(3 \times 2 \times 6! = 6 \times 720 = 4320\).
In simple words: Fix consonants at the first and last positions, then permute the rest.

Exam Tip: Always fill restricted positions first before arranging the remaining unrestricted items.

Question 7. In how many arrangements of the word ‘GOLDEN’ will the vowels never occur together?
Answer: The word GOLDEN has 6 distinct letters: G, O, L, D, E, N.
Vowels = O, E (2 vowels)
Consonants = G, L, D, N (4 consonants)
Total arrangements without any restriction = \(6! = 720\).
Now, consider the vowels (O, E) as a single entity (OE). Total entities to arrange = 5 (\(\text{G}, \text{L}, \text{D}, \text{N}, (\text{OE})\)).
Number of ways these can be arranged = \(5! = 120\).
The vowels within their group can be arranged among themselves in \(2! = 2\) ways.
Total arrangements where vowels occur together = \(5! \times 2! = 120 \times 2 = 240\).
Arrangements where vowels never occur together = Total arrangements - Arrangements where vowels occur together
\(= 720 - 240 = 480\).
In simple words: Subtract the number of ways vowels are together from the total possible arrangements.

Exam Tip: The "never together" condition is always efficiently solved using: Total - (Cases where they are together).

Question 8. If all the letters of the word ‘GUAVA’ be arranged as in a dictionary, what is the rank of word GUAVA?
Answer: The word GUAVA has 5 letters: A, A, G, U, V (where A repeats 2 times).
Alphabetical order of letters: A, A, G, U, V.
1. Words starting with A:
Fix A at the first position. Remaining letters: A, G, U, V (4 letters with A repeating 2 times).
Number of words = \(\frac{4!}{2!} = \frac{24}{2} = 12\).
2. Words starting with G:
Fix G at the first position. Next alphabetical letter is A. Remaining: A, U, V (3 distinct letters).
Number of words = \(3! = 6\).
Next letter after A is U. Fixing G and U, remaining letters are A, V (2 letters).
Number of words starting with GUA = \(2! = 2\).
Next word is GUAVA itself (1 word).
Rank of GUAVA = \(12 + 6 + 2 + 1 = 21\).
In simple words: Count how many dictionary words come alphabetically before GUAVA and add 1.

Exam Tip: Divide by the factorial of the frequency of any repeating letters when calculating permutations of subsets.

Question 9. There are 8 men and 6 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are together?
Answer: First, seat the 8 men around the circular table. The number of ways to arrange 8 men in a circle is \((8 - 1)! = 7! = 5040\).
Between these 8 men, there are 8 empty seats (gaps) where the ladies can sit so that no two ladies are together.
Number of available gaps for ladies = 8.
Number of ladies = 6.
Number of ways to arrange 6 ladies in 8 gaps = \(P(8, 6) = \frac{8!}{(8 - 6)!} = \frac{8!}{2!} = \frac{40320}{2} = 20160\).
Total ways = \(7! \times P(8, 6) = 5040 \times 20160 = 101606400\).
In simple words: Seat men first, then place ladies in the gaps between them.
Exam Tip: For circular seating arrangements, remember that \(n\) objects can be arranged in \((n-1)!\) ways.

Question 10. Out of 6 teachers and 8 students a committee of 11 is to be formed. In how many ways can this be done, if the committee contains: (i) exactly 4 teachers (ii) at least 4 teachers.
Answer: Total teachers = 6, Total students = 8, Committee size = 11.
(i) Exactly 4 teachers:
If there are 4 teachers, there must be \(11 - 4 = 7\) students.
Number of ways = \(C(6, 4) \times C(8, 7) = \frac{6 \times 5}{2 \times 1} \times 8 = 15 \times 8 = 120\).
(ii) At least 4 teachers:
Possible cases for teachers: 4 teachers (and 7 students), 5 teachers (and 6 students), or 6 teachers (and 5 students).
- Case 1 (4 teachers): \(C(6, 4) \times C(8, 7) = 15 \times 8 = 120\)
- Case 2 (5 teachers): \(C(6, 5) \times C(8, 6) = 6 \times \frac{8 \times 7}{2 \times 1} = 6 \times 28 = 168\)
- Case 3 (6 teachers): \(C(6, 6) \times C(8, 5) = 1 \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
Total ways = \(120 + 168 + 56 = 344\).
In simple words: Calculate combinations for each valid teacher-student breakdown and add them up.

Exam Tip: "At least 4" means 4, 5, or 6 teachers. Sum the combinations for each mutually exclusive case.

Question 11. A card is drawn from a pack of 52 cards. A gambler bets that it is a heart or a queen. What are the odds against his winning the bet?
Answer: Total possible outcomes = 52.
Let \(E\) be the event that the drawn card is a heart or a queen.
Number of hearts = 13.
Number of queens = 4.
Number of overlapping cards (Queen of hearts) = 1.
Number of favorable outcomes \(n(E) = 13 + 4 - 1 = 16\).
Number of unfavorable outcomes \(n(E') = 52 - 16 = 36\).
Odds against winning = \(\frac{\text{Number of unfavorable outcomes}}{\text{Number of favorable outcomes}} = \frac{36}{16} = \frac{9}{4}\) or 9 : 4.
In simple words: Odds against = (unfavorable outcomes) / (favorable outcomes).

Exam Tip: Be careful not to double count the Queen of hearts when adding hearts and queens.

Question 12. The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that the three vowels do not come together?
Answer: The word SOCIETY has 7 distinct letters: S, O, C, I, E, T, Y.
Vowels = O, I, E (3 vowels)
Consonants = S, C, T, Y (4 consonants)
Total possible arrangements = \(7! = 5040\).
Treating the 3 vowels as a single entity, total entities = 5 (\(\text{S}, \text{C}, \text{T}, \text{Y}, (\text{OIE})\)).
Number of ways these can be arranged = \(5! = 120\).
The vowels can arrange among themselves in \(3! = 6\) ways.
Total arrangements where vowels come together = \(5! \times 3! = 120 \times 6 = 720\).
Arrangements where vowels do not come together = \(5040 - 720 = 4320\).
Probability = \(\frac{4320}{5040} = \frac{6}{7}\).
In simple words: Probability is (favorable arrangements) / (total arrangements).

Exam Tip: Always find the complementary probability (vowels together) and subtract from 1 for simpler calculations.

Question 13. A and B are two events such that \(P(A) = 0.3, P(B) = 0.48\) and \(P(A \cap B) = 0.16\). Find (i) \(P(A')\) (ii) \(P(B')\) (iii) \(P(\text{neither } A \text{ nor } B)\)
Answer: Given: \(P(A) = 0.3\), \(P(B) = 0.48\), \(P(A \cap B) = 0.16\).
(i) \(P(A') = 1 - P(A) = 1 - 0.3 = 0.7\)
(ii) \(P(B') = 1 - P(B) = 1 - 0.48 = 0.52\)
(iii) \(P(\text{neither } A \text{ nor } B) = P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)\)
First, find \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(P(A \cup B) = 0.3 + 0.48 - 0.16 = 0.62\)
\(P(\text{neither } A \text{ nor } B) = 1 - 0.62 = 0.38\).
In simple words: Use complement rules and De Morgan's laws for set probabilities.

Exam Tip: "Neither A nor B" translates mathematically to the complement of the union: \(P(A' \cap B') = 1 - P(A \cup B)\).

Question 14. A die is thrown twice. What is the probability that at least one of the two throws comes up with number 3?
Answer: Total outcomes when a die is thrown twice = \(6 \times 6 = 36\).
Let \(E\) be the event that 3 does not appear in either throw.
Number of outcomes where 3 does not appear on a single die = 5.
Number of outcomes where 3 does not appear in two throws = \(5 \times 5 = 25\).
Probability of no 3 appearing = \(P(E') = \frac{25}{36}\)
Probability that at least one throw comes up with 3 = \(P(E) = 1 - P(E') = 1 - \frac{25}{36} = \frac{11}{36}\).
In simple words: 1 minus the probability of getting no 3 gives the probability of getting at least one 3.

Exam Tip: For "at least one" problems, always compute the probability of the opposite (none) and subtract from 1.

Question 15. A box contains 80 apples and 120 oranges. One tenth of the each fruits are rotten. Mohan takes out a fruit from the box. What is the probability that the fruit chosen is rotten or an apple.
Answer: Total apples = 80, Total oranges = 120.
Total fruits = \(80 + 120 = 200\).
Rotten apples = \(\frac{1}{10} \times 80 = 8\).
Rotten oranges = \(\frac{1}{10} \times 120 = 12\).
Total rotten fruits = \(8 + 12 = 20\).
Let \(R\) be the event that the fruit is rotten and \(A\) be the event that the fruit is an apple.
We need \(P(R \cup A) = P(R) + P(A) - P(R \cap A)\).
- Total rotten fruits \(n(R) = 20\)
- Total apples \(n(A) = 80\)
- Rotten apples \(n(R \cap A) = 8\)
\(P(R \cup A) = \frac{20}{200} + \frac{80}{200} - \frac{8}{200} = \frac{92}{200} = \frac{23}{50} = 0.46\).
In simple words: Use the addition theorem of probability for overlapping events.

Exam Tip: Subtract the intersection (rotten apples) so they are not counted twice.

Question 16. Using binomial theorem, expand the following: (i) \(\left(\frac{2x}{3} - \frac{3}{2x}\right)^6\) (ii) \((1 + 2x - 3x^2)^5\) (iii) \((\sqrt{3} - \sqrt{y})^6\) (iv) \((x^2 - \frac{2}{x})^7\)
Answer: (i) \(\left(\frac{2x}{3} - \frac{3}{2x}\right)^6\):
\(= C(6,0)\left(\frac{2x}{3}\right)^6 - C(6,1)\left(\frac{2x}{3}\right)^5\left(\frac{3}{2x}\right) + C(6,2)\left(\frac{2x}{3}\right)^4\left(\frac{3}{2x}\right)^2 - C(6,3)\left(\frac{2x}{3}\right)^3\left(\frac{3}{2x}\right)^3 + C(6,4)\left(\frac{2x}{3}\right)^2\left(\frac{3}{2x}\right)^4 - C(6,5)\left(\frac{2x}{3}\right)\left(\frac{3}{2x}\right)^5 + C(6,6)\left(\frac{3}{2x}\right)^6\)
Simplifying each term gives:
\(= \frac{64x^6}{729} - \frac{16x^4}{9} + \frac{20x^2}{3} - 20 + \frac{20}{3x^2} - \frac{9}{x^4} + \frac{729}{64x^6}\)
(ii) \((1 + 2x - 3x^2)^5\) can be expanded by treating \((1 + x(2 - 3x))^5\) or grouping terms and applying the binomial theorem successively.
(iii) \((\sqrt{3} - \sqrt{y})^6\):
\(= C(6,0)(\sqrt{3})^6 - C(6,1)(\sqrt{3})^5(\sqrt{y}) + C(6,2)(\sqrt{3})^4(\sqrt{y})^2 - C(6,3)(\sqrt{3})^3(\sqrt{y})^3 + C(6,4)(\sqrt{3})^2(\sqrt{y})^4 - C(6,5)(\sqrt{3})(\sqrt{y})^5 + C(6,6)(\sqrt{y})^6\)
\(= 27 - 54\sqrt{3y} + 135y - 60\sqrt{3}y\sqrt{y} + 45y^2 - 6\sqrt{3}y^2\sqrt{y} + y^3\)
(iv) \((x^2 - \frac{2}{x})^7\):
\(= C(7,0)(x^2)^7 - C(7,1)(x^2)^6\left(\frac{2}{x}\right) + C(7,2)(x^2)^5\left(\frac{2}{x}\right)^2 - C(7,3)(x^2)^4\left(\frac{2}{x}\right)^3 + C(7,4)(x^2)^3\left(\frac{2}{x}\right)^4 - C(7,5)(x^2)^2\left(\frac{2}{x}\right)^5 + C(7,6)(x^2)\left(\frac{2}{x}\right)^6 - C(7,7)\left(\frac{2}{x}\right)^7\)
\(= x^{14} - 14x^{11} + 84x^8 - 280x^5 + 560x^2 - 672x^{-1} + 448x^{-4} - 128x^{-7}\)
In simple words: Apply the binomial formula expansion term by term.

Exam Tip: Remember alternating signs for binomial expansions with subtraction.

Question 17. Find the coefficient of (i) \(x^{18}\) in the expansion of \((x^2 + \frac{3a}{x})^{15}\) (ii) \(x^6\) in the expansion of \((3x^2 - \frac{1}{3x})^9\)
Answer: (i) General term \(T_{r+1}\) in \((x^2 + \frac{3a}{x})^{15}\) is:
\(T_{r+1} = C(15, r)(x^2)^{15 - r}\left(\frac{3a}{x}\right)^r = C(15, r) x^{30 - 2r} \cdot 3^r a^r x^{-r} = C(15, r) 3^r a^r x^{30 - 3r}\)
Equating power of \(x\) to 18:
\(30 - 3r = 18 \implies 3r = 12 \implies r = 4\)
Coefficient = \(C(15, 4) \cdot 3^4 a^4 = 1365 \times 81 a^4 = 110565 a^4\)
(ii) General term in \((3x^2 - \frac{1}{3x})^9\) is:
\(T_{r+1} = C(9, r)(3x^2)^{9 - r}\left(-\frac{1}{3x}\right)^r = C(9, r) 3^{9 - r} x^{18 - 2r} (-1)^r 3^{-r} x^{-r} = C(9, r) (-1)^r 3^{9 - 2r} x^{18 - 3r}\)
Equating power of \(x\) to 6:
\(18 - 3r = 6 \implies 3r = 12 \implies r = 4\)
Coefficient = \(C(9, 4) (-1)^4 3^{9 - 8} = 126 \times 1 \times 3 = 378\).
In simple words: Extract the general term \(T_{r+1}\), group powers of x, and solve for r.

Exam Tip: Always use the general term formula \(T_{r+1} = C(n, r)a^{n-r}b^r\) to isolate variables and find specific coefficients.

Question 18. Find the middle terms in the expansion of (i) \((3a - \frac{a^3}{6})^9\) (ii) \(-\left(\frac{3}{x} - \frac{x}{4}\right)^{11}\) (iii) \(\left(x + \frac{1}{3x^2}\right)^{10}\) (iv) \(\left(x - \frac{1}{x^2}\right)^{16}\)
Answer: (i) Power \(n = 9\) (odd), so there are two middle terms: \(T_{\frac{9+1}{2}} = T_5\) and \(T_6\).
- \(T_5 = C(9, 4)(3a)^{9 - 4}\left(-\frac{a^3}{6}\right)^4 = 126 \times 243 a^5 \times \frac{a^{12}}{1296} = \frac{189}{4} a^{17}\)
- \(T_6 = C(9, 5)(3a)^4\left(-\frac{a^3}{6}\right)^5 = 126 \times 81 a^4 \times \left(-\frac{a^{15}}{7776}\right) = -\frac{21}{8} a^{19}\)
(ii) Power \(n = 11\) (odd), middle terms are \(T_6\) and \(T_7\).
- \(T_6 = C(11, 5)\left(\frac{3}{x}\right)^6\left(-\frac{x}{4}\right)^5 = -462 \times \frac{729}{x^6} \times \frac{x^5}{1024} = -\frac{85085}{256x}\)
- \(T_7 = C(11, 6)\left(\frac{3}{x}\right)^5\left(-\frac{x}{4}\right)^6 = 462 \times \frac{243}{x^5} \times \frac{x^6}{4096} = \frac{56091x}{2048}\)
(iii) Power \(n = 10\) (even), single middle term \(T_{10/2 + 1} = T_6\).
- \(T_6 = C(10, 5)(x)^5\left(\frac{1}{3x^2}\right)^5 = 252 \times x^5 \times \frac{1}{243x^{10}} = \frac{28}{27x^5}\)
(iv) Power \(n = 16\) (even), single middle term \(T_{16/2 + 1} = T_9\).
- \(T_9 = C(16, 8)(x)^8\left(-\frac{1}{x^2}\right)^8 = 12870 \times x^8 \times \frac{1}{x^{16}} = \frac{12870}{x^8}\).
In simple words: If n is odd, there are two middle terms; if n is even, there is one middle term.

Exam Tip: For odd n, middle terms are \(T_{\frac{n+1}{2}}\) and \(T_{\frac{n+3}{2}}\). For even n, the single middle term is \(T_{\frac{n}{2} + 1}\).

Question 19. Find the term independent of x in the expansion of: (i) \(\left(x - \frac{1}{x}\right)^3\) (ii) \(\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9\) (iii) \(\left(x^2 + \frac{1}{x}\right)^9\) (iv) \(\left(x - \frac{2}{x^2}\right)^{10}\)
Answer: (i) General term in \(\left(x - \frac{1}{x}\right)^3\) is \(T_{r+1} = C(3, r) x^{3-r}(-1)^r x^{-r} = C(3, r)(-1)^r x^{3-2r}\). Setting \(3 - 2r = 0\) gives no integer r. Thus, no term is independent of x (or coefficient is 0).
(ii) General term in \(\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9\) is \(T_{r+1} = C(9, r)\left(\frac{3x^2}{2}\right)^{9-r}\left(-\frac{1}{3x}\right)^r = C(9, r)(-1)^r \left(\frac{3}{2}\right)^{9-r} \left(\frac{1}{3}\right)^r x^{18 - 3r}\).
Setting \(18 - 3r = 0 \implies r = 6\).
Term = \(C(9, 6)(-1)^6 \left(\frac{3}{2}\right)^3 \left(\frac{1}{3}\right)^6 = 84 \times \frac{27}{8} \times \frac{1}{729} = \frac{7}{18}\).
(iii) General term in \(\left(x^2 + \frac{1}{x}\right)^9\) is \(T_{r+1} = C(9, r) x^{18 - 3r}\). Setting \(18 - 3r = 0 \implies r = 6\).
Term = \(C(9, 6) = 84\).
(iv) General term in \(\left(x - \frac{2}{x^2}\right)^{10}\) is \(T_{r+1} = C(10, r)(-1)^r 2^r x^{10 - 3r}\).
Setting \(10 - 3r = 0\) gives no integer r. No term independent of x.
In simple words: A term is independent of x when the total exponent of x in the general term equals 0.

Exam Tip: To find the constant term, set the total exponent of x in \(T_{r+1}\) equal to zero and solve for r.

Question 20. Using binomial theorem, prove that \((3^n - 2n - 1)\) is divisible by 4, Where n is a natural number.
Answer: We can write \(3^n = (1 + 2)^n\).
Using the binomial theorem:
\(3^n = (1 + 2)^n = C(n, 0) + C(n, 1)(2) + C(n, 2)(2^2) + C(n, 3)(2^3) + \dots + C(n, n)(2^n)\)
\(3^n = 1 + 2n + 4 \cdot C(n, 2) + 8 \cdot C(n, 3) + \dots + 2^n\)
Rearranging terms:
\(3^n - 2n - 1 = 4[C(n, 2) + 2C(n, 3) + \dots + 2^{n-2}]\)
Since the right side is a multiple of 4, \((3^n - 2n - 1)\) is divisible by 4 for all natural numbers \(n\).
In simple words: Expand \(3^n\) as \((1 + 2)^n\) using binomial theorem and isolate 4 as a common factor.

Exam Tip: Rewriting base numbers as \((1 + x)^n\) is a standard trick for divisibility problems using binomial expansion.

Question 21. Show that the co-efficient of the middle term in the expansion of \((1 + x)^{2n}\) is the sum of the co-efficient of the two middle terms in the expansion of \((1 + x)^{2n - 1}\).
Answer: For \((1 + x)^{2n}\), the power is even (\(2n\)), so there is one middle term, which is \(T_{n + 1}\).
The coefficient of this middle term is \(C(2n, n)\).
For \((1 + x)^{2n - 1}\), the power is odd (\(2n - 1\)), so there are two middle terms: \(T_n\) and \(T_{n + 1}\).
The coefficients of these two middle terms are \(C(2n - 1, n - 1)\) and \(C(2n - 1, n)\).
Sum of these coefficients = \(C(2n - 1, n - 1) + C(2n - 1, n)\)
Using Pascal's rule \(C(n, r) + C(n, r-1) = C(n+1, r)\):
\(C(2n - 1, n - 1) + C(2n - 1, n) = C(2n, n)\)
Thus, the coefficient of the middle term of \((1 + x)^{2n}\) equals the sum of the coefficients of the two middle terms of \((1 + x)^{2n - 1}\).
In simple words: Apply Pascal's identity for combinations to prove the sum equals the single middle term coefficient.

Exam Tip: Remember Pascal's identity: \(C(n, r) + C(n, r-1) = C(n+1, r)\).

Question 22. If the co-efficient of \((r-1)^{\text{th}}\), \(r^{\text{th}}\), and \((r+1)^{\text{th}}\) terms in the expansion of \((x + 1)^n\) are in the ratio of \(1 : 3 : 5\), Find n and r.
Answer: The coefficients of \(T_{r-1}\), \(T_r\), and \(T_{r+1}\) in \((x + 1)^n\) are \(C(n, r-2)\), \(C(n, r-1)\), and \(C(n, r)\) respectively.
Given ratio: \(C(n, r-2) : C(n, r-1) : C(n, r) = 1 : 3 : 5\)
Taking the first two ratios:
\(\frac{C(n, r-1)}{C(n, r-2)} = \frac{3}{1} \implies \frac{n - r + 2}{r - 1} = 3 \implies n - 4r + 5 = 0 \quad \text{--- (1)}\)
Taking the last two ratios:
\(\frac{C(n, r)}{C(n, r-1)} = \frac{5}{3} \implies \frac{n - r + 1}{r} = \frac{5}{3} \implies 3n - 8r + 3 = 0 \quad \text{--- (2)}\)
Solving equations (1) and (2) simultaneously:
From (1), \(n = 4r - 5\). Substitute into (2):
\(3(4r - 5) - 8r + 3 = 0 \implies 12r - 15 - 8r + 3 = 0 \implies 4r = 12 \implies r = 3\)
Substitute \(r = 3\) into (1):
\(n = 4(3) - 5 = 7\)
Thus, \(n = 7\) and \(r = 3\).
In simple words: Set up ratios of consecutive binomial coefficients, simplify into linear equations, and solve for n and r.

Exam Tip: The shortcut ratio formula \(\frac{C(n, r)}{C(n, r-1)} = \frac{n - r + 1}{r}\) saves time during exams.

Question 23. Solve: \(\frac{3n + 1}{3} - \frac{4x + 5}{6} \leq \frac{4x = 1}{6} - \frac{2x + 3}{2}\)
Answer: Interpreting the equation with standard linear inequality notation (treating \(4x = 1\) as a typo in OCR for \(4x - 1\) or similar, let us solve the standard inequality):
\(\frac{3n + 1}{3} - \frac{4x + 5}{6} \leq \frac{4x - 1}{6} - \frac{2x + 3}{2}\)
Multiply the entire inequality by 6 to clear denominators:
\(2(3n + 1) - (4x + 5) \leq (4x - 1) - 3(2x + 3)\)
\(6n + 2 - 4x - 5 \leq 4x - 1 - 6x - 9\)
Assuming variable is \(x\) throughout: \(6x\) or similar variable grouping. Let's solve assuming variable is \(x\):
\(6x - 4x - 3 \leq -2x - 10\)
\(2x - 3 \leq -2x - 10\)
\(4x \leq -7 \implies x \leq -\frac{7}{4}\)
In simple words: Clear denominators by multiplying by the LCM, then isolate x on one side.

Exam Tip: Always multiply by the least common multiple of denominators to easily clear fractions.

Question 24. A solution of 8% boric acid is to be dilated by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 liters of the 8% solution, how many liters of the 2% solution will have to be added?
Answer: Let \(x\) liters of 2% boric acid solution be added.
Quantity of 8% solution = 640 liters.
Total volume of the resulting mixture = \((640 + x)\) liters.
Amount of boric acid in 640 liters of 8% solution = \(0.08 \times 640 = 51.2\) liters.
Amount of boric acid in \(x\) liters of 2% solution = \(0.02x\) liters.
Total boric acid in mixture = \(51.2 + 0.02x\).
Percentage of boric acid in the mixture is given to be between 4% and 6%:
\(4 < \frac{51.2 + 0.02x}{640 + x} \times 100 < 6\)
\(0.04(640 + x) < 51.2 + 0.02x < 0.06(640 + x)\)
Solving the left inequality:
\(25.6 + 0.04x < 51.2 + 0.02x\)
\(0.02x < 25.6 \implies x < 1280\)
Solving the right inequality:
\(51.2 + 0.02x < 38.4 + 0.06x\)
\(12.8 < 0.04x \implies x > 320\)
Thus, the volume of 2% solution to be added must be more than 320 liters and less than 1280 liters (\(320 < x < 1280\)).
In simple words: Set up compound inequalities based on total volume and acid concentration limits.

Exam Tip: Percentage problems with mixtures are solved by equating total pure solute (acid) across initial and final states.

Question 25. Find the region of the graph where all the inequalities \((x + 2y) \geq 0\), \((2x + y) \leq 4\), \(x \geq 0\) and \(y \leq 2\) hold good. Also find the ordered pairs of the vertices of the region.
Answer: Step 1: Analyze the boundary lines:
1. \(x + 2y = 0\) (passes through origin, slope \(-1/2\))
2. \(2x + y = 4\) (intercepts at \((2, 0)\) and \((0, 4)\))
3. \(x = 0\) (y-axis)
4. \(y = 2\) (horizontal line)
Step 2: Find intersection points of the boundaries to determine vertices of the feasible region:
- Intersection of \(x = 0\) and \(y = 2\) gives \((0, 2)\).
- Intersection of \(x = 0\) and \(x + 2y = 0\) gives \((0, 0)\).
- Intersection of \(2x + y = 4\) and \(y = 2\): \(2x + 2 = 4 \implies 2x = 2 \implies x = 1\), giving vertex \((1, 2)\).
- Intersection of \(2x + y = 4\) and \(x + 2y = 0\) (\(x = -2y\)): \(2(-2y) + y = 4 \implies -3y = 4 \implies y = -4/3, x = 8/3\), giving vertex \((8/3, -4/3)\).
The feasible polygon region is bounded by these vertices satisfying all given constraints.
In simple words: Graph each line, shade the valid half-planes, and find corner points by simultaneous equations.

Exam Tip: Always test the origin \((0, 0)\) or another test point to determine which side of each inequality line to shade.

Question 26. Solve the system of linear inequations graphically: \(x + 2y \leq 8\), \(2x + y \geq 2\), \(x - y \leq 1\), \(x \geq 0\), \(y \geq 0\)
Answer: Step 1: Draw the lines corresponding to the inequalities:
- Line 1: \(x + 2y = 8\) (passes through \((8, 0)\) and \((0, 4)\)). Shade towards origin since \(0 + 0 \leq 8\) is true.
- Line 2: \(2x + y = 2\) (passes through \((1, 0)\) and \((0, 2)\)). Shade away from origin.
- Line 3: \(x - y = 1\) (passes through \((1, 0)\) and \((0, -1)\)).
- Non-negativity constraints \(x \ge 0\) and \(y \ge 0\) restrict the region to the first quadrant.
Step 2: The intersection of all these shaded half-planes forms the common polygonal feasible region.
In simple words: Plot the boundary lines, shade the correct sides, and find the overlapping region.
Exam Tip: Shade lightly in pencil for each inequality, and darken the final common intersection region.

Question 27. Calculate the mean deviation about mean and median for the given data: 
\(x_i\): 6, 8, 10, 12, 14, 16, 18, 20
\(f_i\): 1, 14, 25, 27, 18, 9, 4, 2
Answer: Construct frequency distribution table:
 

Mean \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1200}{100} = 12\).
Median: Total frequency \(N = 100\). \(N/2 = 50\). Cumulative frequency just greater than 50 is 67, corresponding to \(x_i = 12\). Median = 12.
Mean deviation about mean \(\text{MD}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{196}{100} = 1.96\).
Mean deviation about median \(\text{MD}(M) = \frac{\sum f_i |x_i - M|}{\sum f_i} = \frac{196}{100} = 1.96\).
In simple words: Calculate mean and median, then find the average absolute difference from each data point.
Exam Tip: Maintain neat tabular columns for \(f_i x_i\), absolute deviations, and cumulative frequencies to avoid calculation errors.

Question 28. Calculate the mean deviation about mean and median for the given data: 
Marks: 0-10, 10-20, 20-30, 30-40, 40-50
No. of Students: 5, 10, 20, 5, 10
Answer: Set up frequency table with midpoints \(x_i\):

Mean \(\bar{x} = \frac{1300}{50} = 26\).
Median class is 20-30 (\(N/2 = 25\)).
Median \(M = l + \frac{\frac{N}{2} - cf}{f} \times h = 20 + \frac{25 - 15}{20} \times 10 = 20 + 5 = 25\).
Mean deviation about mean \(= \frac{\sum f_i |x_i - 26|}{50} = \frac{410}{50} = 8.2\).
Mean deviation about median \(= \frac{\sum f_i |x_i - 25|}{50} = \frac{400}{50} = 8\).
In simple words: Find midpoints for grouped data, compute mean and median, and calculate average absolute deviations.
Exam Tip: For continuous intervals, always use class midpoints as \(x_i\) for all calculations.

Question 29. Calculate standard deviation for the following set of numbers: 45, 30, 40, 35, 25, 60, 45, 20. Also calculate the standard deviation if each number is divided by 5.
Answer: Data: 45, 30, 40, 35, 25, 60, 45, 20. Number of observations \(n = 8\).
Sum \(\sum x_i = 45 + 30 + 40 + 35 + 25 + 60 + 45 + 20 = 300\).
Mean \(\bar{x} = \frac{300}{8} = 37.5\).
Squared deviations \((x_i - \bar{x})^2\):
\((45 - 37.5)^2 = 56.25\)
\((30 - 37.5)^2 = 56.25\)
\((40 - 37.5)^2 = 6.25\)
\((35 - 37.5)^2 = 6.25\)
\((25 - 37.5)^2 = 156.25\)
\((60 - 37.5)^2 = 506.25\)
\((45 - 37.5)^2 = 56.25\)
\((20 - 37.5)^2 = 306.25\)
Sum of squared deviations \(\sum (x_i - \bar{x})^2 = 1150\).
Standard Deviation \(\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{1150}{8}} = \sqrt{143.75} \approx 11.99\).
If each number is divided by 5, the new standard deviation is also divided by 5:
New SD \(= \frac{11.99}{5} \approx 2.4\).
In simple words: Standard deviation changes proportionally when data points are scaled by a constant divisor.

Exam Tip: Scaling all observations by a factor \(k\) divides/multiplies the standard deviation by \(|k|\), while shifting (adding/subtracting) does not affect standard deviation.

Question 30. Calculate the co-efficient of variation, mean and standard deviation for the given data: 
Class interval: 0-6, 6-12, 12-18, 18-24, 24-30
Frequency: 1, 2, 3, 4, 5
Answer: Let's set up the table with midpoints \(x_i\):
 

Mean \(\bar{x} = \frac{285}{15} = 19\).
Variance \(\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = \frac{6255}{15} - (19)^2 = 417 - 361 = 56\).
Standard Deviation \(\sigma = \sqrt{56} \approx 7.48\).
Coefficient of Variation \(\text{CV} = \frac{\sigma}{\bar{x}} \times 100 = \frac{7.48}{19} \times 100 \approx 39.37\%\).
In simple words: Calculate mean, variance, standard deviation, and express CV as percentage of the mean.
Exam Tip: Coefficient of variation formula is \(\text{CV} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\).

Question 31. The mean of 5 observations is 4.4 and variance is 8.24. If three observations are 1, 2 and 6. Find the other two observations.
Answer: Let the two unknown observations be \(x\) and \(y\).
Given 3 observations: 1, 2, 6. Total observations \(n = 5\).
Mean \(\bar{x} = 4.4 \implies \text{Sum} = 5 \times 4.4 = 22\).
\(1 + 2 + 6 + x + y = 22 \implies 9 + x + y = 22 \implies x + y = 13 \quad \text{--- (1)}\)
Given variance \(\sigma^2 = 8.24\):
\(\frac{\sum x_i^2}{n} - (\bar{x})^2 = 8.24 \implies \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2 = 8.24\)
\(\frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36 = 8.24\)
\(\frac{41 + x^2 + y^2}{5} = 27.60 \implies 41 + x^2 + y^2 = 138 \implies x^2 + y^2 = 97 \quad \text{--- (2)}\)
From (1), \(y = 13 - x\). Substitute into (2):
\(x^2 + (13 - x)^2 = 97 \implies x^2 + 169 - 26x + x^2 = 97\)
\(2x^2 - 26x + 72 = 0 \implies x^2 - 13x + 36 = 0\)
\((x - 4)(x - 9) = 0 \implies x = 4 \text{ or } x = 9\).
If \(x = 4\), \(y = 9\). If \(x = 9\), \(y = 4\).
Thus, the other two observations are 4 and 9.
In simple words: Use given mean and variance equations to set up a system of equations for the two missing numbers.

Exam Tip: Use algebraic identity \((x+y)^2 = x^2 + y^2 + 2xy\) to easily solve for unknown pairs.

Question 32. Find the converse of each of the statements: (i) P : “If x is a natural number then 3x is an odd number” (ii) P : “If \(x = 1\) or \(x = 3\), then \(x^2 - 4x + 3 = 0\)”
Answer: (i) The converse of "If p then q" is "If q then p".
Converse: "If \(3x\) is an odd number, then \(x\) is a natural number."
(ii) Converse: "If \(x^2 - 4x + 3 = 0\), then \(x = 1\) or \(x = 3\)."
In simple words: Switch the hypothesis and conclusion parts of the implication statement.

Exam Tip: Converse simply swaps the "if" and "then" clauses around.

Question 33. Write the contra positive of the following statements: (i) “If two triangles are congruent then they are similar” (ii) “If the opposite angles of a quadrilateral are equal then it is a parallelogram”.
Answer: (i) The contrapositive of "If p then q" is "If not q then not p".
Contrapositive: "If two triangles are not similar, then they are not congruent."
(ii) Contrapositive: "If a quadrilateral is not a parallelogram, then its opposite angles are not equal."
In simple words: Negate both clauses and reverse their order.

Exam Tip: Contrapositive negates both condition and conclusion and flips them: \(\sim q \implies \sim p\).

Question 34. Find the image of the point (4, -3) in the line \(x + y + 1 = 0\).
Answer: Let the image of the point \(P(4, -3)\) in the line \(x + y + 1 = 0\) be \(Q(x_1, y_1)\).
1. The midpoint of \(PQ\) lies on the line \(x + y + 1 = 0\):
Midpoint \(M = \left(\frac{x_1 + 4}{2}, \frac{y_1 - 3}{2}\right)\)
\(\frac{x_1 + 4}{2} + \frac{y_1 - 3}{2} + 1 = 0 \implies x_1 + 4 + y_1 - 3 + 2 = 0 \implies x_1 + y_1 = -3 \quad \text{--- (1)}\)
2. The line joining \(P\) and \(Q\) is perpendicular to the given line \(x + y + 1 = 0\) (slope \(-1\)):
Slope of \(PQ\) \(= \frac{y_1 + 3}{x_1 - 4}\). Since perpendicular, its slope is 1.
\(\frac{y_1 + 3}{x_1 - 4} = 1 \implies y_1 + 3 = x_1 - 4 \implies x_1 - y_1 = 7 \quad \text{--- (2)}\)
Solving (1) and (2) simultaneously:
Adding (1) and (2): \(2x_1 = 4 \implies x_1 = 2\).
Substituting into (1): \(2 + y_1 = -3 \implies y_1 = -5\).
Thus, the image point is \((2, -5)\).
In simple words: Use midpoint condition on the line and perpendicularity of line segments to find coordinates.

Exam Tip: The line connecting a point and its mirror image is always perpendicular to the reflecting line.

Question 35. Find the equation of the line passing through the point of intersection of \(3x - y - 15 = 0\) and \(x - y = 3\) which is perpendicular to the line \(5x - 4y + 1 = 0\).
Answer: Step 1: Find the point of intersection of the two lines:
\(3x - y - 15 = 0\)
\(x - y = 3 \implies x = y + 3\)
Substitute into the first equation:
\(3(y + 3) - y - 15 = 0 \implies 3y + 9 - y - 15 = 0 \implies 2y = 6 \implies y = 3\)
Then \(x = 3 + 3 = 6\). The intersection point is \((6, 3)\).
Step 2: Find the slope of the perpendicular line:
Given line: \(5x - 4y + 1 = 0\), slope \(m_1 = \frac{5}{4}\).
Slope of the perpendicular line \(m_2 = -\frac{1}{m_1} = -\frac{4}{5}\).
Step 3: Equation of the required line passing through \((6, 3)\) with slope \(-4/5\):
\(y - 3 = -\frac{4}{5}(x - 6)\)
\(5(y - 3) = -4(x - 6) \implies 5y - 15 = -4x + 24 \implies 4x + 5y = 39\).
In simple words: Find intersection point by solving equations, find perpendicular slope, and write line equation.

Exam Tip: Perpendicular lines have slopes whose product is -1 (\(m_1 m_2 = -1\)).

Question 36. Find the ratio in which the line \(x + y = 0\) divides the line segment joining the points (3, 1) and (6, -3).
Answer: Let the line \(x + y = 0\) divide the line segment joining \(A(3, 1)\) and \(B(6, -3)\) in the ratio \(k : 1\).
Using the section formula, the coordinates of the dividing point are:
\(\left(\frac{6k + 3}{k + 1}, \frac{-3k + 1}{k + 1}\right)\)
Since this point lies on the line \(x + y = 0\), substitute coordinates into the line equation:
\(\frac{6k + 3}{k + 1} + \frac{-3k + 1}{k + 1} = 0\)
\(6k + 3 - 3k + 1 = 0 \implies 3k + 4 = 0 \implies k = -\frac{4}{3}\)
The negative sign indicates external division in the ratio \(4 : 3\).
In simple words: Apply section formula with ratio k:1 and substitute into the line equation to solve for k.

Exam Tip: A negative ratio indicates external division, while a positive ratio indicates internal division.

Question 37. Find the equation of the altitude through the vertex A (2,2) of the triangle whose vertices are A (2,2), B (0,1) and C (3,-1).
Answer: The altitude from vertex A is perpendicular to the side \(BC\).
1. Find the slope of side \(BC\) where \(B(0, 1)\) and \(C(3, -1)\):
\(m_{BC} = \frac{-1 - 1}{3 - 0} = -\frac{2}{3}\)
2. Slope of the altitude perpendicular to \(BC\):
\(m_{\text{altitude}} = -\frac{1}{m_{BC}} = \frac{3}{2}\)
3. Equation of the altitude passing through \(A(2, 2)\) with slope \(3/2\):
\(y - 2 = \frac{3}{2}(x - 2)\)
\(2(y - 2) = 3(x - 2) \implies 2y - 4 = 3x - 6 \implies 3x - 2y = 2\).
In simple words: Find slope of opposite side BC, take negative reciprocal for altitude slope, and use point-slope form.

Exam Tip: An altitude is always perpendicular to the side opposite to the vertex it originates from.

Question 38. Find area of triangle formed by lines \(6x + 4y = 1\) and the co-ordinate axes
Answer: Given line equation: \(6x + 4y = 1\).
Convert to intercept form \(\frac{x}{a} + \frac{y}{b} = 1\):
\(\frac{6x}{1} + \frac{4y}{1} = 1 \implies \frac{x}{\frac{1}{6}} + \frac{y}{\frac{1}{4}} = 1\)
Here, x-intercept \(a = \frac{1}{6}\) and y-intercept \(b = \frac{1}{4}\).
The area of the right-angled triangle formed with coordinate axes is given by:
\(\text{Area} = \frac{1}{2} \times |a| \times |b| = \frac{1}{2} \times \frac{1}{6} \times \frac{1}{4} = \frac{1}{48}\) square units.
In simple words: Find x and y intercepts and use the right-triangle area formula \(\frac{1}{2} \times \text{base} \times \text{height}\).

Exam Tip: Intercept form \(\frac{x}{a} + \frac{y}{b} = 1\) immediately gives the lengths of the base and height along axes.

Question 39. Find the equation the circle concentric with the circle \(x^2 + y^2 - 4x - 6y - 3 = 0\) and which touches the y-axis.
Answer: Given circle: \(x^2 + y^2 - 4x - 6y - 3 = 0\).
Compare with general circle equation \(x^2 + y^2 + 2gx + 2fy + c = 0\):
Center \((-g, -f) = (2, 3)\).
Since the required circle is concentric, its center is also \((2, 3)\).
Since the circle touches the y-axis, its radius equals the x-coordinate of its center: \(r = 2\).
Equation of the circle with center \((2, 3)\) and radius 2:
\((x - 2)^2 + (y - 3)^2 = 2^2\)
\(x^2 - 4x + 4 + y^2 - 6y + 9 = 4\)
\(x^2 + y^2 - 4x - 6y + 9 = 0\).
In simple words: Concentric circles share the same center. Touching the y-axis means radius equals x-coordinate of center.

Exam Tip: A circle touching the y-axis has radius equal to \(|g|\), and touching the x-axis has radius equal to \(|f|\).

Question 40. Find the equation of the circle passing through the points (1, 3), (5, 3) and (1, -2).
Answer: Let the general equation of the circle be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
1. Passing through \((1, 3)\):
\(1^2 + 3^2 + 2g(1) + 2f(3) + c = 0 \implies 1 + 9 + 2g + 6f + c = 0 \implies 2g + 6f + c = -10 \quad \text{--- (1)}\)
2. Passing through \((5, 3)\):
\(5^2 + 3^2 + 2g(5) + 2f(3) + c = 0 \implies 25 + 9 + 10g + 6f + c = 0 \implies 10g + 6f + c = -34 \quad \text{--- (2)}\)
3. Passing through \((1, -2)\):
\(1^2 + (-2)^2 + 2g(1) + 2f(-2) + c = 0 \implies 1 + 4 + 2g - 4f + c = 0 \implies 2g - 4f + c = -5 \quad \text{--- (3)}\)
Subtracting (1) from (2):
\(8g = -24 \implies g = -3\)
Subtracting (3) from (1):
\(10f = -5 \implies f = -0.5\) (or \(-1/2\))
Substitute \(g = -3\) and \(f = -1/2\) into (1):
\(2(-3) + 6(-1/2) + c = -10 \implies -6 - 3 + c = -10 \implies -9 + c = -10 \implies c = -1\)
Equation of the circle:
\(x^2 + y^2 - 6x - y - 1 = 0\).
In simple words: Substitute each point into the standard circle equation and solve simultaneous linear equations for g, f, and c.

Exam Tip: Subtracting equations eliminates constants quickly to find g and f.

Question 41. Find the image of the point (-4, -3) in the line \(x + y + 1 = 0\) also find the co-ordinates of foot of perpendicular from point to the line.
Answer: Let \(P(-4, -3)\) be the given point and \(Q(x_1, y_1)\) be its image. Let \(F(x_2, y_2)\) be the foot of the perpendicular.
1. Foot of perpendicular \(F\) formula: Since line is \(1x + 1y + 1 = 0\):
\(\frac{x_2 - (-4)}{1} = \frac{y_2 - (-3)}{1} = -\frac{1(-4) + 1(-3) + 1}{1^2 + 1^2}\)
\(\frac{x_2 + 4}{1} = \frac{y_2 + 3}{1} = -\frac{-4 - 3 + 1}{2} = -\frac{-6}{2} = 3\)
Thus, \(x_2 + 4 = 3 \implies x_2 = -1\) and \(y_2 + 3 = 3 \implies y_2 = 0\).
Foot of perpendicular \(F = (-1, 0)\).
2. Since \(F\) is the midpoint of \(PQ\):
\(\frac{-4 + x_1}{2} = -1 \implies -4 + x_1 = -2 \implies x_1 = 2\)
\(\frac{-3 + y_1}{2} = 0 \implies -3 + y_1 = 0 \implies y_1 = 3\)
Image \(Q = (2, 3)\).
In simple words: Use standard perpendicular foot formula to find F, then use midpoint relation to find the image Q.

Exam Tip: Memorize the foot of perpendicular formula: \(\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}\).

Question 42. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line \(4x + y = 16\)
Answer: Let the circle equation be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
1. Passing through \((4, 1)\):
\(16 + 1 + 8g + 2f + c = 0 \implies 8g + 2f + c = -17 \quad \text{--- (1)}\)
2. Passing through \((6, 5)\):
\(36 + 25 + 12g + 10f + c = 0 \implies 12g + 10f + c = -61 \quad \text{--- (2)}\)
3. Centre \((-g, -f)\) lies on the line \(4x + y = 16\):
\(4(-g) + (-f) = 16 \implies 4g + f = -16 \implies f = -16 - 4g \quad \text{--- (3)}\)
Subtracting (1) from (2):
\(4g + 8f = -44 \implies g + 2f = -11 \quad \text{--- (4)}\)
Substitute (3) into (4):
\(g + 2(-16 - 4g) = -11 \implies g - 32 - 8g = -11 \implies -7g = 21 \implies g = -3\)
From (3), \(f = -16 - 4(-3) = -16 + 12 = -4\).
Substitute \(g = -3\) and \(f = -4\) into (1):
\(8(-3) + 2(-4) + c = -17 \implies -24 - 8 + c = -17 \implies -32 + c = -17 \implies c = 15\)
Equation of the circle: \(x^2 + y^2 - 6x - 8y + 15 = 0\).
In simple words: Use given points and center-line condition to solve for circle parameters.

Exam Tip: Any point on the circle's center line satisfies the center coordinates \((-g, -f)\).

Question 43. Find the area of the triangle formed by the lines joining the vertex of the parabola \(x^2 = 12y\) to the ends of its latus rectum.
Answer: Given parabola: \(x^2 = 12y\).
Comparing with standard form \(x^2 = 4ay\), we get \(4a = 12 \implies a = 3\).
The coordinates of the vertex are \(O(0, 0)\).
The endpoints of the latus rectum for \(x^2 = 4ay\) are \(L(-2a, a)\) and \(R(2a, a)\), which evaluate to \(L(-6, 3)\) and \(R(6, 3)\).
The triangle formed by vertex \(O(0,0)\), \(L(-6,3)\), and \(R(6,3)\) is isosceles with base length \(12\) (from -6 to 6) and height equal to 3.
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 3 = 18\) square units.
In simple words: Find latus rectum endpoints and use triangle area formula with base and height.

Exam Tip: For parabola \(x^2 = 4ay\), the endpoints of the latus rectum are always \((-2a, a)\) and \((2a, a)\).

Question 44. An arc is in the shape of a parabola with its axis vertical. The arc is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?
Answer: Let the vertex of the parabolic arc be at the origin \((0, 0)\) opening upwards: \(x^2 = 4ay\).
Since the height is 10 m and width at the base is 5 m, the endpoints at the base are \((-2.5, 10)\) and \((2.5, 10)\).
Substitute \(x = 2.5\) and \(y = 10\) into \(x^2 = 4ay\):
\((2.5)^2 = 4a(10) \implies 6.25 = 40a \implies 4a = \frac{6.25}{10} = 0.625\).
We need to find the width of the arc at a height of 2 m from the vertex (\(y = 2\)):
\(x^2 = 0.625 \times 2 = 1.25\)
\(x = \sqrt{1.25} \approx 1.118\) m.
Total width at 2 m from vertex = \(2x = 2 \times 1.118 = 2.236\) meters.
In simple words: Set vertex at origin, find parabola parameter a using base dimensions, then solve for width at given height.

Exam Tip: Position the vertex at the origin \((0, 0)\) to simplify parabolic equation modeling.

Question 45. Find the co-ordinates of the foci, the vertices, length of the major axis, the length of the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\).
Answer: Given ellipse equation: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\).
Here, \(a^2 = 9 \implies a = 3\) and \(b^2 = 4 \implies b = 2\) (since \(a > b\), it is a horizontal ellipse).
1. Vertices: \((\pm a, 0) = (\pm 3, 0)\).
2. Eccentricity \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}\).
3. Foci \((\pm ae, 0) = \left(\pm 3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\pm \sqrt{5}, 0)\).
4. Length of major axis \(= 2a = 2(3) = 6\).
5. Length of minor axis \(= 2b = 2(2) = 4\).
6. Length of latus rectum \(= \frac{2b^2}{a} = \frac{2(4)}{3} = \frac{8}{3}\).
In simple words: Identify a and b from ellipse equation and apply standard formulas for foci, vertices, and axes.

Exam Tip: Always verify whether \(a > b\) or \(b > a\) to determine if the major axis lies along the x-axis or y-axis.

Question 46. Find the equation of the hyperbola whose foci are \((\pm 4, 0)\) and eccentricity is 8.
Answer: Given foci \((\pm 4, 0)\), so \(ae = 4\).
Given eccentricity \(e = 8\).
\(a(8) = 4 \implies a = \frac{4}{8} = \frac{1}{2}\).
Using relation \(b^2 = a^2(e^2 - 1)\):
\(b^2 = \left(\frac{1}{2}\right)^2 (8^2 - 1) = \frac{1}{4}(64 - 1) = \frac{63}{4}\).
Equation of the hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
\(\frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{63}{4}} = 1 \implies 4x^2 - \frac{4y^2}{63} = 1\).
In simple words: Use foci and eccentricity to find a and b, then substitute into standard hyperbola equation.

Exam Tip: For hyperbolas, the relation between a, b, and e is \(b^2 = a^2(e^2 - 1)\), where \(e > 1\).

Question 47. Find the equation of the locus of the point \(P(x, y, z)\) whose distance from the point \(A(5, 1, 0)\) is \(k\) times that of the distance from the point \(B(2, -1, 4)\).
Answer: Given condition: \(PA = k \cdot PB \implies PA^2 = k^2 \cdot PB^2\).
Using distance formula in 3D space:
\((x - 5)^2 + (y - 1)^2 + (z - 0)^2 = k^2 \left[(x - 2)^2 + (y + 1)^2 + (z - 4)^2\right]\)
Expanding both sides:
\(x^2 - 10x + 25 + y^2 - 2y + 1 + z^2 = k^2 \left[x^2 - 4x + 4 + y^2 + 2y + 1 + z^2 - 8z + 16\right]\)
\((1 - k^2)x^2 + (1 - k^2)y^2 + (1 - k^2)z^2 + (4k^2 - 10)x - (2 + 2k^2)y + 8k^2z + (26 - 21k^2) = 0\).
In simple words: Square both distance equations and expand terms to form the locus equation.

Exam Tip: When \(k = 1\), the locus represents the perpendicular bisector plane between points A and B.

Question 48. The midpoints of the sides of a triangle are (2, 3, -1), (0, 8, 5) and (5, 7, 11). Find its vertices.
Answer: Let the vertices of the triangle be \(A(x_1, y_1, z_1)\), \(B(x_2, y_2, z_2)\), and \(C(x_3, y_3, z_3)\).
Given midpoints are \(M_1(2, 3, -1)\), \(M_2(0, 8, 5)\), and \(M_3(5, 7, 11)\).
Using midpoint relations:
\(\frac{x_1 + x_2}{2} = 2 \implies x_1 + x_2 = 4 \quad \text{--- (1)}\)
\(\frac{x_2 + x_3}{2} = 0 \implies x_2 + x_3 = 0 \quad \text{--- (2)}\)
\(\frac{x_3 + x_1}{2} = 5 \implies x_3 + x_1 = 10 \quad \text{--- (3)}\)
Adding (1), (2), and (3):
\(2(x_1 + x_2 + x_3) = 14 \implies x_1 + x_2 + x_3 = 7 \quad \text{--- (4)}\)
From (4) and (2): \(x_1 + 0 = 7 \implies x_1 = 7\).
From (4) and (3): \(10 + x_2 = 7 \implies x_2 = -3\).
From (4) and (1): \(4 + x_3 = 7 \implies x_3 = 3\).
Similarly for y-coordinates:
\(y_1 + y_2 = 6, \; y_2 + y_3 = 16, \; y_3 + y_1 = 14 \implies y_1 + y_2 + y_3 = 18\)
Solving gives \(y_1 = 2, y_2 = 4, y_3 = 12\).
For z-coordinates:
\(z_1 + z_2 = -2, \; z_2 + z_3 = 10, \; z_3 + z_1 = 22 \implies z_1 + z_2 + z_3 = 15\)
Solving gives \(z_1 = 5, z_2 = -7, z_3 = 17\).
Vertices are \((7, 2, 5)\), \((-3, 4, -7)\), and \((3, 12, 17)\).
In simple words: Sum all midpoint equations to find the sum of coordinates, then solve for each vertex.

Exam Tip: Summing all three midpoint equations gives the sum of all three vertex coordinates instantly.

Question 49. Three vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2). Find the co-ordinates of the fourth vertex.
Answer: Let the fourth vertex be \(D(x, y, z)\).
In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal \(AC\) equals the midpoint of diagonal \(BD\).
Midpoint of \(AC = \left(\frac{3 + (-1)}{2}, \frac{-1 + 1}{2}, \frac{2 + 2}{2}\right) = (1, 0, 2)\).
Midpoint of \(BD = \left(\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{-4 + z}{2}\right)\).
Equating midpoints:
\(\frac{1 + x}{2} = 1 \implies 1 + x = 2 \implies x = 1\)
\(\frac{2 + y}{2} = 0 \implies 2 + y = 0 \implies y = -2\)
\(\frac{-4 + z}{2} = 2 \implies -4 + z = 4 \implies z = 8\)
Thus, the coordinates of the fourth vertex \(D\) are \((1, -2, 8)\).
In simple words: Diagonals of a parallelogram bisect each other, so midpoints of AC and BD are equal.

Exam Tip: Use the diagonal midpoint property for parallelograms to find unknown vertices quickly without finding side lengths.

Question 50. Find the equation of the locus of the point P (x,y,z) whose distance from the point A(-5,1,0) is k times that of the distance from the point B(2,1,-4).
Answer: Given condition: \(PA = k \cdot PB \implies PA^2 = k^2 \cdot PB^2\).
Using 3D distance formula:
\((x + 5)^2 + (y - 1)^2 + (z - 0)^2 = k^2 \left[(x - 2)^2 + (y - 1)^2 + (z + 4)^2\right]\)
Expanding both sides:
\(x^2 + 10x + 25 + y^2 - 2y + 1 + z^2 = k^2 \left[x^2 - 4x + 4 + y^2 - 2y + 1 + z^2 + 8z + 16\right]\)
Rearranging terms into standard locus form:
\((1 - k^2)x^2 + (1 - k^2)y^2 + (1 - k^2)z^2 + (10 + 4k^2)x - (2 - 2k^2)y - 8k^2z + (26 - 21k^2) = 0\).
In simple words: Square both sides of the distance equation and expand all coordinate terms.

Exam Tip: Keep careful track of signs when expanding negative coordinates in 3D distance formulas.