CBSE Class 12 Applied Mathematics Question Paper 2026 Solved Code 465

SECTION A

1. The real x for which \( 2(2x + 3) - 10 \lt 6(x - 2) \) is : [1 Mark]
(A) \( x \gt 2 \)
(B) \( x \gt 3 \)
(C) \( x \gt 4 \)
(D) \( x \gt -4 \)

Answer: (C) \( x \gt 4 \)

Teacher's Note:
a) Expand both sides: \( 4x + 6 - 10 \lt 6x - 12 \Rightarrow 4x - 4 \lt 6x - 12 \).
b) Rearrange: \( 8 \lt 2x \Rightarrow x \gt 4 \).
c) The sign of the inequality stays the same because we only divide by a positive number (2).

2. The number of all possible matrices of order \( 3 \times 2 \) with each entry 1 or 2 is : [1 Mark]
(A) 6
(B) 16
(C) 24
(D) 64

Answer: (D) 64

Teacher's Note:
a) A \( 3 \times 2 \) matrix has \( 3 \times 2 = 6 \) entries.
b) Each entry has 2 choices, so the number of matrices is \( 2^{6} = 64 \).

3. If AB = A and BA = B, then \( (B^{2} + B) \) is equal to : [1 Mark]
(A) 2A
(B) O
(C) 2I
(D) 2B

Answer: (D) 2B

Teacher's Note:
a) \( B^{2} = B \cdot B = (BA)B = B(AB) = BA = B \).
b) So \( B^{2} + B = B + B = 2B \).
c) Use the given relations step by step; matrix multiplication is associative but not commutative.

4. The value of \( \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} \) is : [1 Mark]
(A) 5
(B) \( -7 \)
(C) 9
(D) 0

Answer: (D) 0

Teacher's Note:
a) Expand along \( R_1 \): \( 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0 \).
b) Quick check: \( R_3 - R_2 = R_2 - R_1 \), so the rows are dependent and the determinant is 0.

5. General solution of differential equation \( y \log y \, dx - x \, dy = 0 \) is : [1 Mark]
(A) \( y = \log |cx| \)
(B) \( y = e^{|cx|} \)
(C) \( y = e^{c + x} \)
(D) \( \log y = |c + x| \)

Answer: (B) \( y = e^{|cx|} \)

Teacher's Note:
a) Separate the variables: \( \frac{dy}{y \log y} = \frac{dx}{x} \).
b) Integrate: \( \log |\log y| = \log |x| + \log |c| \Rightarrow \log y = cx \Rightarrow y = e^{cx} \).
c) For \( \int \frac{dy}{y \log y} \), put \( \log y = t \), so \( \frac{dy}{y} = dt \).

6. If \( \int_{0}^{40} \frac{dx}{2x + 1} = \log k \), then the value of k is : [1 Mark]
(A) 3
(B) 9
(C) \( \frac{9}{2} \)
(D) \( \frac{3}{2} \)

Answer: (B) 9

Teacher's Note:
a) \( \int_{0}^{40} \frac{dx}{2x + 1} = \frac{1}{2} \left[ \log |2x + 1| \right]_{0}^{40} = \frac{1}{2} (\log 81 - \log 1) \).
b) \( \frac{1}{2} \log 81 = \log 81^{\frac{1}{2}} = \log 9 \), so \( k = 9 \).
c) Do not forget the factor \( \frac{1}{2} \) that comes from the coefficient of x.

7. If \( x = t^{2} \) and \( y = t^{3} \), then \( \frac{d^{2}y}{dx^{2}} \) is equal to : [1 Mark]
(A) \( \frac{3}{2} \)
(B) \( \frac{3}{4t} \)
(C) \( \frac{1}{2t^{2}} \)
(D) \( \frac{3}{2t} \)

Answer: (B) \( \frac{3}{4t} \)

Teacher's Note:
a) \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^{2}}{2t} = \frac{3t}{2} \).
b) \( \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\left( \frac{3t}{2} \right) \times \frac{dt}{dx} = \frac{3}{2} \times \frac{1}{2t} = \frac{3}{4t} \).
c) Common mistake: forgetting to multiply by \( \frac{dt}{dx} \) in the second derivative.

8. The rate of change of the area of a circle with respect to its radius r (in cm²/s), when r = 6 cm, is : [1 Mark]
(A) \( 10 \pi \)
(B) \( 12 \pi \)
(C) \( 8 \pi \)
(D) \( 11 \pi \)

Answer: (B) \( 12 \pi \)

Teacher's Note:
a) \( A = \pi r^{2} \Rightarrow \frac{dA}{dr} = 2 \pi r \).
b) At \( r = 6 \), \( \frac{dA}{dr} = 2 \pi \times 6 = 12 \pi \).

9. Let X be a discrete random variable, then the variance of X is : : [1 Mark]
(A) \( E(X^{2}) \)
(B) \( E(X^{2}) - [E(X)]^{2} \)
(C) \( E(X^{2}) + [E(X)]^{2} \)
(D) \( \sqrt{E(X^{2}) - [E(X)]^{2}} \)

Answer: (B) \( E(X^{2}) - [E(X)]^{2} \)

Teacher's Note:
a) Variance is the mean of squares minus the square of the mean.
b) Option (D) is the standard deviation, not the variance - do not mix them up.

10. If 'm' is the mean of a Poisson distribution, then its variance is given by : [1 Mark]
(A) \( m^{2} \)
(B) \( \sqrt{m} \)
(C) m
(D) \( \frac{m}{2} \)

Answer: (C) m

Teacher's Note:
a) In a Poisson distribution, mean = variance = m.
b) So the standard deviation is \( \sqrt{m} \), which is option (B) - a common trap.

11. The total area under a standard normal curve is : [1 Mark]
(A) 1
(B) \( \sqrt{2} \)
(C) 2
(D) \( \frac{1}{2} \)

Answer: (A) 1

Teacher's Note:
a) The area under any probability density curve equals the total probability, which is 1.
b) The area on each side of the mean \( z = 0 \) is \( \frac{1}{2} \).

12. For the purpose of t-test of significance, if a random sample of size 34 is drawn from a normal population, then the degree of freedom \( (\nu) \) is : [1 Mark]
(A) 32
(B) 33
(C) 34
(D) 35

Answer: (B) 33

Teacher's Note:
a) For a single sample t-test, degrees of freedom \( \nu = n - 1 \).
b) Here \( \nu = 34 - 1 = 33 \).

13. The range of variable t of the t-distribution is : [1 Mark]
(A) \( (0, 1) \)
(B) \( (1, 2) \)
(C) \( (-1, 1) \)
(D) \( (-\infty, \infty) \)

Answer: (D) \( (-\infty, \infty) \)

Teacher's Note:
a) The t-curve is symmetric about \( t = 0 \) and extends on both sides without end.
b) Like the normal curve, it never touches the horizontal axis.

14. The present value of a perpetuity of Rs. R payable at the end of each period, when the money is worth \( i \) per period is : [1 Mark]
(A) \( Ri \)
(B) \( R + \frac{R}{i} \)
(C) \( \frac{R}{i} \)
(D) \( R - Ri \)

Answer: (C) \( \frac{R}{i} \)

Teacher's Note:
a) For an ordinary perpetuity (payment at the end of each period), \( PV = \frac{R}{i} \).
b) For a perpetuity due (payment at the beginning), \( PV = R + \frac{R}{i} \), which is option (B).

15. Using flat rate method, the EMI to repay a loan of Rs. 20,000 in \( 2\frac{1}{2} \) years at an interest rate of 8% per annum is : [1 Mark]
(A) Rs. 700
(B) Rs. 800
(C) Rs. 900
(D) Rs. 100

Answer: (B) Rs. 800

Teacher's Note:
a) Interest \( = \frac{20000 \times 8 \times 2.5}{100} = \) Rs. 4,000, so total amount \( = 20000 + 4000 = \) Rs. 24,000.
b) Number of months \( = 2.5 \times 12 = 30 \), so EMI \( = \frac{24000}{30} = \) Rs. 800.

16. If an investment of Rs. 10,000 becomes Rs. 60,000 in 4 years, then the Compound Annual Growth Rate (CAGR) is : [1 Mark]
(A) \( \frac{\sqrt[4]{6} - 1}{100} \)
(B) \( \frac{\sqrt[4]{6} + 1}{100} \)
(C) \( (\sqrt[4]{6} - 1) \times 100 \)
(D) \( (\sqrt[4]{6} + 1) \times 100 \)

Answer: (C) \( (\sqrt[4]{6} - 1) \times 100 \)

Teacher's Note:
a) CAGR \( = \left[ \left( \frac{\text{Final value}}{\text{Initial value}} \right)^{\frac{1}{n}} - 1 \right] \times 100 \).
b) \( \left( \frac{60000}{10000} \right)^{\frac{1}{4}} - 1 = \sqrt[4]{6} - 1 \); multiply by 100 to get a percentage.

17. A machine costs Rs. 45,000 with an estimated useful life of 5 years and a scrap value of Rs. 10,000. The annual depreciation of the machine is : [1 Mark]
(A) Rs. 8,000
(B) Rs. 7,000
(C) Rs. 6,000
(D) Rs. 5,000

Answer: (B) Rs. 7,000

Teacher's Note:
a) Straight line method: annual depreciation \( = \frac{\text{Cost} - \text{Scrap value}}{\text{Useful life}} \).
b) \( \frac{45000 - 10000}{5} = \frac{35000}{5} = \) Rs. 7,000.

18. The maximum value of the function \( z = 7x + 5y \), subject to the constraints \( x \le 3, y \le 2, x \ge 0, y \ge 0 \) is : [1 Mark]
(A) 10
(B) 21
(C) 31
(D) 29

Answer: (C) 31

Teacher's Note:
a) The feasible region is a rectangle with corner points \( (0, 0), (3, 0), (3, 2), (0, 2) \).
b) Values of z are 0, 21, 31 and 10; the maximum is \( z = 7(3) + 5(2) = 31 \) at \( (3, 2) \).

19. Assertion (A) : \( \int \frac{1}{\sqrt{9 - x^{2}}} \, dx = \sin^{-1} \frac{x}{3} + C \)
Reason (R) : \( \int \frac{1}{\sqrt{a^{2} - x^{2}}} \, dx = \sin^{-1} \frac{x}{a} + C \) [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) The Reason is the standard formula \( \int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1} \frac{x}{a} + C \).
b) Putting \( a = 3 \) (since \( 9 = 3^{2} \)) gives the Assertion directly, so R explains A.

20. Assertion (A) : In sinking fund, a fixed amount at regular intervals is deposited.
Reason (R) : In Savings Bank Account, any amount, any time can be deposited. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) A sinking fund is built by depositing a fixed sum at regular intervals to meet a future liability, so A is true.
b) A savings bank account allows deposits of any amount at any time, so R is true.
c) R talks about a different account, so it does not explain A.

SECTION B

21. A man in a boat goes 12 km downstream and comes back to the starting point by rowing non-stop in a total time of 3 hours. If the speed of the stream is 3 km/h, find the speed with which the man can row the boat in still water. [2 Marks]

Answer:
1. Let the speed of the boat in still water be \( x \) km/h.
2. Speed downstream \( = (x + 3) \) km/h and speed upstream \( = (x - 3) \) km/h.
3. Total time: \( \frac{12}{x + 3} + \frac{12}{x - 3} = 3 \).
4. \( 12(x - 3) + 12(x + 3) = 3(x^{2} - 9) \Rightarrow 24x = 3x^{2} - 27 \Rightarrow 3x^{2} - 24x - 27 = 0 \).
5. Dividing by 3: \( x^{2} - 8x - 9 = 0 \Rightarrow (x - 9)(x + 1) = 0 \Rightarrow x = 9 \) or \( x = -1 \).
6. Rejecting the negative value, \( x = 9 \).
Hence, the speed of the boat in still water is 9 km/h.

Teacher's Note:
a) Downstream speed = boat speed + stream speed; upstream speed = boat speed - stream speed.
b) Use time = distance \( \div \) speed for each leg and add them.
c) Always reject the negative root, as speed cannot be negative.

22. (a) Solve the following differential equation :
\( \frac{dy}{dx} = \frac{2 - y}{x + 1} \) [2 Marks]

Answer:
1. Separating the variables: \( \frac{dy}{2 - y} = \frac{dx}{x + 1} \).
2. Integrating both sides: \( \int \frac{dy}{2 - y} = \int \frac{dx}{x + 1} \).
3. \( -\log |2 - y| = \log |x + 1| - \log c \).
4. \( \log |2 - y| + \log |x + 1| = \log c \Rightarrow \log |(2 - y)(x + 1)| = \log c \).
5. Hence the general solution is \( |(2 - y)(x + 1)| = c \), i.e. \( (2 - y)(x + 1) = C \).

Teacher's Note:
a) \( \int \frac{dy}{2 - y} = -\log |2 - y| \); the minus sign comes from the coefficient of y.
b) Writing the constant as \( \log c \) makes it easy to combine the logarithms.

OR

(b) If \( \int_{a}^{b} x^{3} \, dx = 0 \) and \( \int_{a}^{b} x^{2} \, dx = \frac{2}{3} \), then find the values of 'a' and 'b'. [2 Marks]

Answer:
1. \( \left[ \frac{x^{4}}{4} \right]_{a}^{b} = 0 \Rightarrow \frac{1}{4}(b^{4} - a^{4}) = 0 \Rightarrow b^{4} = a^{4} \) ... (i)
2. \( \left[ \frac{x^{3}}{3} \right]_{a}^{b} = \frac{2}{3} \Rightarrow \frac{1}{3}(b^{3} - a^{3}) = \frac{2}{3} \Rightarrow b^{3} - a^{3} = 2 \) ... (ii)
3. From (i), \( b = a \) or \( b = -a \). If \( b = a \), then (ii) gives \( 0 = 2 \), which is impossible. So \( b = -a \).
4. Putting \( b = -a \) in (ii): \( -a^{3} - a^{3} = 2 \Rightarrow a^{3} = -1 \Rightarrow a = -1 \).
5. Hence \( a = -1 \) and \( b = 1 \).

Teacher's Note:
a) Evaluate each definite integral first to get two equations in a and b.
b) Check the answer: \( \int_{-1}^{1} x^{3} dx = 0 \) (odd function) and \( \int_{-1}^{1} x^{2} dx = \frac{2}{3} \).

23. (a) For a Poisson distribution, if mean (m) = 1, then find P(r = 1). [2 Marks]

Answer:
1. For a Poisson distribution, \( P(X = r) = \frac{e^{-m} m^{r}}{r!} \).
2. With \( m = 1 \) and \( r = 1 \): \( P(r = 1) = \frac{e^{-1} (1)^{1}}{1!} \).
3. \( P(r = 1) = e^{-1} = \frac{1}{e} \).

Teacher's Note:
a) Write the Poisson formula first; it carries a mark.
b) Leave the answer as \( e^{-1} \) or \( \frac{1}{e} \) since calculators are not allowed.

OR

(b) Find the mean and standard deviation of the Binomial distribution \( B\left( 4, \frac{1}{3} \right) \). [2 Marks]

Answer:
1. Here \( n = 4 \), \( p = \frac{1}{3} \), so \( q = 1 - \frac{1}{3} = \frac{2}{3} \).
2. Mean \( = np = 4 \times \frac{1}{3} = \frac{4}{3} \).
3. Variance \( = npq = 4 \times \frac{1}{3} \times \frac{2}{3} = \frac{8}{9} \).
4. Standard deviation \( = \sqrt{npq} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3} \).

Teacher's Note:
a) In \( B(n, p) \), the first number is n (number of trials) and the second is p.
b) Standard deviation is the square root of the variance \( npq \); do not stop at variance.

24. Calculate 3-yearly moving averages for the following data :
Years (t) | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 | 2020
Variables (x) | 3 | 5 | 7 | 10 | 12 | 14 | 15 | 16 [2 Marks]

Answer:
Year (t) | Variable (x) | 3-yearly moving total | 3-yearly moving average
2013 | 3 | - | -
2014 | 5 | \( 3 + 5 + 7 = 15 \) | \( \frac{15}{3} = 5 \)
2015 | 7 | \( 5 + 7 + 10 = 22 \) | \( \frac{22}{3} = 7.33 \)
2016 | 10 | \( 7 + 10 + 12 = 29 \) | \( \frac{29}{3} = 9.67 \)
2017 | 12 | \( 10 + 12 + 14 = 36 \) | \( \frac{36}{3} = 12 \)
2018 | 14 | \( 12 + 14 + 15 = 41 \) | \( \frac{41}{3} = 13.67 \)
2019 | 15 | \( 14 + 15 + 16 = 45 \) | \( \frac{45}{3} = 15 \)
2020 | 16 | - | -

Teacher's Note:
a) Each moving total is placed against the middle year of its group of three.
b) The first and last years have no 3-yearly average; show them with a dash.
c) Quick method: next total = previous total - first value dropped + new value added.

25. A man takes a personal loan of Rs. 2,00,000 at an interest rate of 15% p.a. compounded monthly, to be repaid by equal monthly instalments in 4 years. Calculate the EMI, using reducing balance method.
[Given : \( (1.0125)^{-48} = 0.55 \)] [2 Marks]

Answer:
1. Formula: \( EMI = \frac{P i}{1 - (1 + i)^{-n}} \).
2. Here \( P = \) Rs. 2,00,000, \( i = \frac{15}{1200} = 0.0125 \), \( n = 4 \times 12 = 48 \).
3. \( EMI = \frac{200000 \times 0.0125}{1 - (1.0125)^{-48}} = \frac{2500}{1 - 0.55} \).
4. \( EMI = \frac{2500}{0.45} = \) Rs. 5,555.56 (approx.)

Teacher's Note:
a) For monthly compounding, divide the annual rate by 12 and multiply the years by 12.
b) Writing the formula and substituting correctly carries most of the marks.

SECTION C

26. (a) (i) Apply addition modulo to positive integers 17 and 13 for modulo 30.
(ii) Find subtraction modulo 8 for numbers 11 and 3. [3 Marks]

Answer:
(i) \( (17 + 13) \bmod 30 = 30 \bmod 30 \).
Since 30 leaves remainder 0 when divided by 30, \( 17 +_{30} 13 = 0 \).
(ii) \( (11 - 3) \bmod 8 = 8 \bmod 8 \).
Since 8 leaves remainder 0 when divided by 8, \( 11 -_{8} 3 = 0 \).

Teacher's Note:
a) \( a \bmod n \) is the remainder when a is divided by n.
b) First add or subtract the numbers, then take the remainder with the given modulus.

OR

(b) Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank. [3 Marks]

Answer:
1. Part of the tank filled by A, B and C in 1 hour \( = \frac{1}{8} \).
2. Part filled by A, B and C in 2 hours \( = \frac{2}{8} = \frac{1}{4} \).
3. Remaining part \( = 1 - \frac{1}{4} = \frac{3}{4} \).
4. A and C fill \( \frac{3}{4} \) in 9 hours, so in 1 hour they fill \( \frac{3}{4} \times \frac{1}{9} = \frac{1}{12} \).
5. Part filled by B alone in 1 hour \( = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24} \).
6. Hence pipe B alone can fill the tank in 24 hours.

Teacher's Note:
a) Always work with the part of the tank filled in 1 hour.
b) B's 1-hour work = (A + B + C)'s 1-hour work - (A + C)'s 1-hour work.

27. Using Cramer's rule, show that the following system of linear equations is consistent and hence solve it :
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \) [3 Marks]

Answer:
1. \( D = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix} = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 \).
2. Since \( D \neq 0 \), the system has a unique solution, so it is consistent.
3. \( D_1 = \begin{vmatrix} 11 & -3 & 5 \\ -5 & 2 & -4 \\ -3 & 1 & -2 \end{vmatrix} = 11(-4 + 4) + 3(10 - 12) + 5(-5 + 6) = 0 - 6 + 5 = -1 \).
4. \( D_2 = \begin{vmatrix} 2 & 11 & 5 \\ 3 & -5 & -4 \\ 1 & -3 & -2 \end{vmatrix} = 2(10 - 12) - 11(-6 + 4) + 5(-9 + 5) = -4 + 22 - 20 = -2 \).
5. \( D_3 = \begin{vmatrix} 2 & -3 & 11 \\ 3 & 2 & -5 \\ 1 & 1 & -3 \end{vmatrix} = 2(-6 + 5) + 3(-9 + 5) + 11(3 - 2) = -2 - 12 + 11 = -3 \).
6. \( x = \frac{D_1}{D} = \frac{-1}{-1} = 1 \), \( y = \frac{D_2}{D} = \frac{-2}{-1} = 2 \), \( z = \frac{D_3}{D} = \frac{-3}{-1} = 3 \).
Hence \( x = 1, y = 2, z = 3 \).

Teacher's Note:
a) \( D_1, D_2, D_3 \) are formed by replacing the x, y, z column of D by the constants column.
b) Verify in the first equation: \( 2(1) - 3(2) + 5(3) = 11 \).

28. (a) Find the intervals in \( \mathbb{R} \) for which the function \( f(x) = x^{4} - 2x^{2} \) is increasing or decreasing. [3 Marks]

Answer:
1. \( f'(x) = 4x^{3} - 4x = 4x(x^{2} - 1) = 4x(x - 1)(x + 1) \).
2. \( f'(x) = 0 \Rightarrow x = -1, 0, 1 \). These points divide \( \mathbb{R} \) into four intervals.
3. Sign of \( f'(x) \): in \( (-\infty, -1) \) it is negative; in \( (-1, 0) \) it is positive; in \( (0, 1) \) it is negative; in \( (1, \infty) \) it is positive.
4. Hence \( f(x) \) is decreasing in \( (-\infty, -1) \cup (0, 1) \) and increasing in \( (-1, 0) \cup (1, \infty) \).

Teacher's Note:
a) Factorise \( f'(x) \) fully and mark the critical points on a number line.
b) Test one value in each interval, e.g. \( f'(2) = 24 \gt 0 \), to fix the signs.
c) Closed intervals such as \( [-1, 0] \cup [1, \infty) \) are also accepted.

OR

(b) Find : \( \int \frac{2x + 1}{18 - 4x - x^{2}} \, dx \) [3 Marks]

Answer:
1. \( I = -\int \frac{2x + 1}{x^{2} + 4x - 18} \, dx \). Write \( 2x + 1 = (2x + 4) - 3 \), where \( 2x + 4 = \frac{d}{dx}(x^{2} + 4x - 18) \).
2. \( I = -\int \frac{2x + 4}{x^{2} + 4x - 18} \, dx + \int \frac{3}{x^{2} + 4x - 18} \, dx \).
3. First integral: \( -\log |x^{2} + 4x - 18| \).
4. For the second, \( x^{2} + 4x - 18 = (x + 2)^{2} - 22 = (x + 2)^{2} - (\sqrt{22})^{2} \).
5. Using \( \int \frac{dx}{X^{2} - a^{2}} = \frac{1}{2a} \log \left| \frac{X - a}{X + a} \right| \): \( \int \frac{3 \, dx}{(x + 2)^{2} - (\sqrt{22})^{2}} = \frac{3}{2\sqrt{22}} \log \left| \frac{x + 2 - \sqrt{22}}{x + 2 + \sqrt{22}} \right| \).
6. Hence \( I = -\log |x^{2} + 4x - 18| + \frac{3}{2\sqrt{22}} \log \left| \frac{x + 2 - \sqrt{22}}{x + 2 + \sqrt{22}} \right| + C \).

Teacher's Note:
a) Split the numerator as A(derivative of denominator) + B, then integrate each part.
b) Complete the square carefully: \( x^{2} + 4x - 18 = (x + 2)^{2} - 22 \).
c) Do not forget the constant of integration C.

29. A soap manufacturing company was distributing a particular brand of a soap through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozen. After the campaign, a sample of 26 shops was taken and mean sales was found to be 147 dozen with standard deviation 16. Can you consider the advertisement campaign effective ? [Given \( t_{25} (0.05) = 2.06 \)] [3 Marks]

Answer:
1. Given: \( \bar{x} = 147 \), \( \mu = 140 \), \( n = 26 \), \( s = 16 \).
2. Null hypothesis \( H_0 \): there is no significant difference between \( \bar{x} \) and \( \mu \) (the campaign is not effective).
Alternate hypothesis \( H_1 \): there is a significant difference between \( \bar{x} \) and \( \mu \).
3. Test statistic: \( t = \frac{\bar{x} - \mu}{s / \sqrt{n - 1}} = \frac{147 - 140}{16 / \sqrt{25}} = \frac{7}{16/5} = \frac{35}{16} = 2.187 \).
4. Degrees of freedom \( = n - 1 = 25 \), and \( t_{25}(0.05) = 2.06 \).
5. Since \( |t| = 2.187 \gt 2.06 \), the null hypothesis is rejected.
Hence the advertisement campaign is effective.

Teacher's Note:
a) State both hypotheses clearly before calculating; they carry marks.
b) If \( t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = 2.23 \) is used, it is also accepted and gives the same conclusion.
c) Reject \( H_0 \) when the calculated \( |t| \) is greater than the table value.

30. A company XYZ Ltd. has issued a bond having a face value of Rs. 10,000 paying annual dividend at \( 8.5\% \) p.a. The bond will be redeemed at par at the end of 10 years. Find the purchase value of this bond, if the investor wishes a yield rate of 8%. [Given : \( (1.08)^{-10} = 0.46319349 \)] [3 Marks]

Answer:
1. Here \( F = \) Rs. 10,000, \( n = 10 \), \( i = 0.08 \).
2. Annual dividend \( R = 8.5\% \) of 10000 \( = \) Rs. 850; redemption value \( C = F = \) Rs. 10,000.
3. Purchase price \( V = R \left[ \frac{1 - (1 + i)^{-n}}{i} \right] + C(1 + i)^{-n} \).
4. \( V = 850 \left[ \frac{1 - (1.08)^{-10}}{0.08} \right] + 10000 (1.08)^{-10} = 850 \times \frac{1 - 0.46319349}{0.08} + 10000 \times 0.46319349 \).
5. \( V = 850 \times \frac{0.53680651}{0.08} + 4631.93 = 850 \times 6.71008 + 4631.93 = 5703.57 + 4631.93 \).
6. \( V = \) Rs. 10,335.50 (approx.)

Teacher's Note:
a) Purchase price = present value of all dividends + present value of the redemption amount.
b) Dividend is calculated on the face value, but discounting uses the yield rate.
c) Since the dividend rate (8.5%) is more than the yield rate (8%), the price is above par - a quick check.

31. Solve the following Linnear Programming Problem (LPP) graphically :
Maximize \( z = 3x + 5y \)
subject to the constraints :
\( x + 2y \le 2000 \)
\( x + y \le 1500 \)
\( y \le 600 \)
\( x \ge 0, y \ge 0 \) [3 Marks]

Answer:
1. Draw the lines \( x + 2y = 2000 \) (through \( (2000, 0) \) and \( (0, 1000) \)), \( x + y = 1500 \) (through \( (1500, 0) \) and \( (0, 1500) \)) and \( y = 600 \). The feasible region lies in the first quadrant, below all three lines, and is the bounded polygon OABCD.
2. Corner points: O(0, 0); A(0, 600); B(800, 600) from \( y = 600 \) and \( x + 2y = 2000 \); C(1000, 500) from \( x + 2y = 2000 \) and \( x + y = 1500 \); D(1500, 0).
3. Value of \( z = 3x + 5y \) at the corner points:
O(0, 0) : 0
A(0, 600) : 3000
B(800, 600) : \( 2400 + 3000 = 5400 \)
C(1000, 500) : \( 3000 + 2500 = 5500 \)
D(1500, 0) : 4500
4. Hence the maximum value of z is 5500 at \( x = 1000, y = 500 \).

Teacher's Note:
a) Solve pairs of boundary lines to get the exact corner points, e.g. subtract \( x + y = 1500 \) from \( x + 2y = 2000 \) to get \( y = 500 \).
b) A neat, correctly shaded graph carries half the marks.
c) Evaluate z at every corner point in a table before choosing the maximum.

SECTION D

32. Solve the following inequation :
\( \frac{2x - 1}{12} - \frac{x - 11}{3} \lt \frac{3x + 1}{4}, x \in R \) [5 Marks]

Answer:
1. The LCM of 12, 3 and 4 is 12. Multiplying both sides by 12 (a positive number, so the sign does not change):
\( (2x - 1) - 4(x - 11) \lt 3(3x + 1) \).
2. Expanding: \( 2x - 1 - 4x + 44 \lt 9x + 3 \Rightarrow -2x + 43 \lt 9x + 3 \).
3. Transposing: \( 43 - 3 \lt 9x + 2x \Rightarrow 40 \lt 11x \Rightarrow 11x \gt 40 \).
4. Dividing by 11: \( x \gt \frac{40}{11} \).
5. Hence the solution set is \( \left( \frac{40}{11}, \infty \right) \), i.e. all real x greater than \( \frac{40}{11} \).

Teacher's Note:
a) Clear the fractions first by multiplying with the LCM; this avoids calculation errors.
b) Take care with the sign: \( -4(x - 11) = -4x + 44 \).
c) Write the final answer in interval form; the endpoint \( \frac{40}{11} \) is not included.

33. (a) Find the consumer's surplus for the demand function \( p = 25 - x - x^{2} \), where the prevailing market price \( p_0 = 19 \). [5 Marks]

Answer:
1. At equilibrium, \( p = p_0 = 19 \) and \( x = x_0 \): \( 25 - x_0 - x_0^{2} = 19 \Rightarrow x_0^{2} + x_0 - 6 = 0 \).
2. \( (x_0 + 3)(x_0 - 2) = 0 \Rightarrow x_0 = 2 \) (rejecting the negative value \( x_0 = -3 \)).
3. \( p_0 x_0 = 19 \times 2 = 38 \).
4. Consumer's surplus \( CS = \int_{0}^{x_0} p \, dx - p_0 x_0 = \int_{0}^{2} (25 - x - x^{2}) \, dx - 38 \).
5. \( = \left[ 25x - \frac{x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{2} - 38 = \left( 50 - 2 - \frac{8}{3} \right) - 38 \).
6. \( = 10 - \frac{8}{3} = \frac{22}{3} \).
Hence the consumer's surplus is \( \frac{22}{3} \) (about 7.33) units.

Teacher's Note:
a) First find the equilibrium quantity \( x_0 \) by putting \( p = p_0 \) in the demand function.
b) CS = area under the demand curve from 0 to \( x_0 \) minus \( p_0 x_0 \).
c) Reject the negative root since quantity cannot be negative.

OR

(b) Solve the following initial value differential equation :
\( (x - 1) \frac{dy}{dx} = 2xy \), when \( y(2) = 1 \). [5 Marks]

Answer:
1. Separating the variables: \( \frac{dy}{y} = \frac{2x}{x - 1} \, dx \).
2. Write \( \frac{2x}{x - 1} = \frac{2(x - 1) + 2}{x - 1} = 2\left( 1 + \frac{1}{x - 1} \right) \).
3. Integrating: \( \int \frac{dy}{y} = 2 \int \left( 1 + \frac{1}{x - 1} \right) dx \Rightarrow \log |y| = 2(x + \log |x - 1|) + c \).
4. So \( \log |y| - \log (x - 1)^{2} = 2x + c \).
5. Using \( y(2) = 1 \): \( \log 1 - \log 1 = 4 + c \Rightarrow 0 = 4 + c \Rightarrow c = -4 \).
6. Hence the particular solution is \( \log |y| - \log (x - 1)^{2} = 2x - 4 \), i.e. \( y = (x - 1)^{2} e^{2x - 4} \).

Teacher's Note:
a) Divide the numerator by the denominator before integrating \( \frac{2x}{x - 1} \).
b) Substitute the initial condition only after integrating, to find c.
c) Check: at \( x = 2 \), \( y = (1)^{2} e^{0} = 1 \).

34. (a) Two cards are drawn at random and one by one with replacement from a well-shuffled pack of 52 playing cards. Find the probability distribution of the number of aces. Also, find its mean and variance. [5 Marks]

Answer:
1. Let X denote the number of aces. X can take the values 0, 1 and 2.
2. Probability of an ace in one draw \( p = \frac{4}{52} = \frac{1}{13} \), so \( q = 1 - \frac{1}{13} = \frac{12}{13} \). Draws are with replacement, so they are independent.
3. \( P(X = 0) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169} \).
\( P(X = 1) = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169} \).
\( P(X = 2) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169} \).
4. Probability distribution:
X | 0 | 1 | 2
P(X) | \( \frac{144}{169} \) | \( \frac{24}{169} \) | \( \frac{1}{169} \)
5. Mean \( E(X) = np = 2 \times \frac{1}{13} = \frac{2}{13} \).
6. Variance \( = npq = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169} \).

Teacher's Note:
a) Since the draws are with replacement, X follows a binomial distribution with \( n = 2 \), \( p = \frac{1}{13} \).
b) Check that the probabilities add to 1: \( \frac{144 + 24 + 1}{169} = 1 \).
c) Mean can also be found as \( \sum xP(x) = 0 + \frac{24}{169} + \frac{2}{169} = \frac{26}{169} = \frac{2}{13} \).

OR

(b) It is given that 2% of the screws manufactured by a company are defective. Use Poisson distribution to find the probability that a packet of 100 screws contains (i) no defective screw, (ii) one defective screw. [5 Marks]

Answer:
1. Let X be the number of defective screws in a packet. Probability of a defective screw \( p = \frac{2}{100} = 0.02 \), and \( n = 100 \).
2. Mean \( m = np = 100 \times 0.02 = 2 \).
3. Poisson formula: \( P(X = r) = \frac{e^{-m} m^{r}}{r!} \).
4. (i) No defective screw: \( P(X = 0) = \frac{e^{-2} \, 2^{0}}{0!} = e^{-2} \).
5. (ii) One defective screw: \( P(X = 1) = \frac{e^{-2} \, 2^{1}}{1!} = 2e^{-2} \).

Teacher's Note:
a) Poisson is used when n is large and p is small; find \( m = np \) first.
b) Remember \( 0! = 1 \) and \( m^{0} = 1 \).
c) Answers may be left in terms of e since calculators are not allowed.

35. Mr. Arya wants to know the amount he should pay for a gold mine expected to yield an annual return of Rs. 4 lakh for the next 10 years, after which it will be worthless. Find the amount he should pay for the mine, if he wants to yield 18% annual return on his investment and also set up a sinking fund to replace the purchase price. Assume that the sinking fund earns 10% annually. [Use \( (1.1)^{10} = 2.5937 \)] [5 Marks]

Answer:
1. Let the purchase price of the gold mine be Rs. \( x \).
2. Return on investment \( = 18\% \) of \( x = 0.18x \).
3. The rest of the annual return goes to the sinking fund: annual deposit \( = \) Rs. \( (400000 - 0.18x) \).
4. The sinking fund must grow to \( x \) in 10 years at 10%. Annual deposit \( A = \frac{S r}{(1 + r)^{n} - 1} \), so \( 400000 - 0.18x = \frac{0.1x}{(1.1)^{10} - 1} = \frac{0.1x}{2.5937 - 1} = \frac{0.1x}{1.5937} \).
5. \( 400000 = x \left( \frac{0.1}{1.5937} + 0.18 \right) = x(0.0627 + 0.18) = 0.2427x \).
6. \( x = \frac{400000}{0.2427} = 1648125.26 \) (approx.)
Hence Mr. Arya should pay about Rs. 16,48,125 for the gold mine.

Teacher's Note:
a) Annual income = return on investment + sinking fund deposit; set up this equation first.
b) The sinking fund deposit formula is \( A = \frac{S r}{(1 + r)^{n} - 1} \), where S is the purchase price to be replaced.
c) Small differences due to rounding (for example Rs. 16,47,806 without rounding 0.0627) are acceptable.

SECTION E

Case Study - 1

36. Three schools A, B and C organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand-made fans, mats and plates from recycled material at a cost of Rs. 25, Rs. 50 and Rs. 10 each respectively. The number of articles sold are given below :
Articles / School | A | B | C
Hand-made fans | 50 | 30 | 35
Mats | 60 | 35 | 40
Plates | 40 | 50 | 25
Based on the above information, answer the following questions :

(i) What is the price matrix ? [1 Mark]

Answer: The prices of a fan, a mat and a plate are Rs. 25, Rs. 50 and Rs. 10, so the price matrix is \( P = \begin{bmatrix} 25 & 50 & 10 \end{bmatrix} \).

Teacher's Note:
a) Write the prices as a row matrix in the same order as the rows of the sales table (fans, mats, plates).
b) A \( 1 \times 3 \) row matrix can be multiplied by the \( 3 \times 3 \) sales matrix.

(ii) What is the sales matrix ? [1 Mark]

Answer: Taking rows as fans, mats, plates and columns as schools A, B, C:
\( S = \begin{bmatrix} 50 & 30 & 35 \\ 60 & 35 & 40 \\ 40 & 50 & 25 \end{bmatrix} \)

Teacher's Note:
a) Copy the table exactly into a \( 3 \times 3 \) matrix, keeping rows and columns in order.
b) Each column shows the sales of one school.

(iii) (a) What is the matrix of funds collected by School B ? [2 Marks]

Answer:
1. Funds collected by School B \( = \begin{bmatrix} 25 & 50 & 10 \end{bmatrix} \begin{bmatrix} 30 \\ 35 \\ 50 \end{bmatrix} \).
2. \( = \begin{bmatrix} 25 \times 30 + 50 \times 35 + 10 \times 50 \end{bmatrix} = \begin{bmatrix} 750 + 1750 + 500 \end{bmatrix} \).
3. \( = \begin{bmatrix} 3000 \end{bmatrix} \), i.e. School B collected Rs. 3,000.

Teacher's Note:
a) Multiply the price row matrix by the column of School B from the sales matrix.
b) A \( 1 \times 3 \) matrix times a \( 3 \times 1 \) matrix gives a \( 1 \times 1 \) matrix.

OR

(iii) (b) What is the total amount of funds collected by Schools A and C ? [2 Marks]

Answer:
1. Funds collected by School A \( = \begin{bmatrix} 25 & 50 & 10 \end{bmatrix} \begin{bmatrix} 50 \\ 60 \\ 40 \end{bmatrix} = \begin{bmatrix} 1250 + 3000 + 400 \end{bmatrix} = \) Rs. 4,650.
2. Funds collected by School C \( = \begin{bmatrix} 25 & 50 & 10 \end{bmatrix} \begin{bmatrix} 35 \\ 40 \\ 25 \end{bmatrix} = \begin{bmatrix} 875 + 2000 + 250 \end{bmatrix} = \) Rs. 3,125.
3. Total funds collected by Schools A and C \( = 4650 + 3125 = \) Rs. 7,775.

Teacher's Note:
a) Find each school's funds separately using matrix multiplication, then add.
b) Check each product term carefully, e.g. \( 50 \times 60 = 3000 \) for mats in School A.

Case Study - 2

37. When observed over a long period of time, a time series data can predict trends that can forecast increase or decrease or stagnation of a variable under consideration. Such analytical studies can benefit a business for forecasting or prediction of future estimated sales or production. Mathematically, for finding a line of best-fit to represent a trend, many methods are available. Methods like moving averages and least squares are some of the techniques to predict such trends.
Mr. Nitin runs a soap-making factory and the record of his sales of soaps for the period 2018 - 2024 is as follows :
Year | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 | 2024
Sales (in Rs. thousands) | 80 | 90 | 92 | 83 | 94 | 99 | 92
Based on the above information, answer the following questions :

(i) Obtain the trend line to the given data. [1 Mark]

Answer:
1. Take origin at the middle year 2021, so \( X = x_i - 2021 \); X takes values \( -3, -2, -1, 0, 1, 2, 3 \).
2. Working table:
Year | Y | X | \( X^{2} \) | XY
2018 | 80 | \( -3 \) | 9 | \( -240 \)
2019 | 90 | \( -2 \) | 4 | \( -180 \)
2020 | 92 | \( -1 \) | 1 | \( -92 \)
2021 | 83 | 0 | 0 | 0
2022 | 94 | 1 | 1 | 94
2023 | 99 | 2 | 4 | 198
2024 | 92 | 3 | 9 | 276
Total: \( \sum Y = 630 \), \( \sum X = 0 \), \( \sum X^{2} = 28 \), \( \sum XY = 56 \)
3. \( a = \frac{\sum Y}{n} = \frac{630}{7} = 90 \) and \( b = \frac{\sum XY}{\sum X^{2}} = \frac{56}{28} = 2 \).
4. Trend line: \( Y_t = a + bX = 90 + 2X \), where \( X = \) year \( - 2021 \).

Teacher's Note:
a) With an odd number of years, take the middle year as origin so that \( \sum X = 0 \).
b) Then the normal equations reduce to \( a = \frac{\sum Y}{n} \) and \( b = \frac{\sum XY}{\sum X^{2}} \).

(ii) Find the average change in the sales. [1 Mark]

Answer: The average (annual) change in sales is the slope \( b = 2 \) thousand, i.e. \( 2 \times 1000 = \) Rs. 2,000 per year.

Teacher's Note:
a) The slope b of the trend line gives the average change per unit of time.
b) Remember to convert to rupees, since sales are given in thousands.

(iii) (a) Find the sum of the differences between the actual sales and the trend values (for 2018 - 2024). [2 Marks]

Answer:
1. Trend values \( Y_t = 90 + 2X \): 2018: 84, 2019: 86, 2020: 88, 2021: 90, 2022: 92, 2023: 94, 2024: 96.
2. Differences \( Y - Y_t \): \( 80 - 84 = -4 \), \( 90 - 86 = 4 \), \( 92 - 88 = 4 \), \( 83 - 90 = -7 \), \( 94 - 92 = 2 \), \( 99 - 94 = 5 \), \( 92 - 96 = -4 \).
3. \( \sum (Y - Y_t) = -4 + 4 + 4 - 7 + 2 + 5 - 4 = 0 \).

Teacher's Note:
a) For a least squares line, the sum of deviations of actual values from trend values is always zero.
b) This is a quick check that the trend line has been calculated correctly.

OR

(iii) (b) What are the expected sales for the year 2025 ? [2 Marks]

Answer:
1. For 2025, \( X = 2025 - 2021 = 4 \).
2. Trend value \( Y_t = 90 + 2 \times 4 = 98 \) (in Rs. thousands).
3. Hence the expected sales for 2025 are Rs. 98,000.

Teacher's Note:
a) Convert the year to the coded value X using the same origin (2021).
b) State the final answer in rupees, not just as 98.

Case Study - 3

38. A man has Rs. 15,000 for purchasing rice and wheat. A bag of rice and a bag of wheat cost Rs. 1,800 and Rs. 1,200 respectively. He has a storage capacity of 10 bags. He earns a profit of Rs. 100 and Rs. 90 per bag of rice and wheat respectively. Assuming that he can sell all the items that he can buy, he purchases x bags of rice and y bags of wheat.
Based on the above information and by formulation of Linear Programming Problem (LPP), answer the following questions :

(i) Write the objective function which represents the total profit from the sale of total bags of both types. [1 Mark]

Answer: Profit on x bags of rice is Rs. 100x and on y bags of wheat is Rs. 90y, so the objective function is: Maximise \( Z = 100x + 90y \).

Teacher's Note:
a) The objective function uses profit per bag, not cost per bag.
b) Mention that Z is to be maximised.

(ii) Write the constraints that relate the total cost of both types of bags. [1 Mark]

Answer:
1. Cost constraint: \( 1800x + 1200y \le 15000 \), i.e. \( 3x + 2y \le 25 \).
2. Storage constraint: \( x + y \le 10 \).
3. Non-negativity: \( x \ge 0, y \ge 0 \).

Teacher's Note:
a) Divide \( 1800x + 1200y \le 15000 \) by 600 to simplify it to \( 3x + 2y \le 25 \).
b) Do not forget the non-negativity constraints.

(iii) (a) How many bags of each type should the man buy to get maximum profit ? [2 Marks]

Answer:
1. Draw \( x + y = 10 \) (through \( (10, 0) \) and \( (0, 10) \)) and \( 3x + 2y = 25 \) (through \( \left( \frac{25}{3}, 0 \right) \) and \( (0, 12.5) \)). The feasible region is in the first quadrant below both lines.
2. Solving \( x + y = 10 \) and \( 3x + 2y = 25 \): \( 3x + 2(10 - x) = 25 \Rightarrow x = 5, y = 5 \).
3. Corner points and values of \( Z = 100x + 90y \):
O(0, 0) : 0
A(0, 10) : 900
B(5, 5) : \( 500 + 450 = 950 \)
C\( \left( \frac{25}{3}, 0 \right) \) : \( \frac{2500}{3} \approx 833.33 \)
4. Z is maximum (950) at B(5, 5). So the man should buy 5 bags of rice and 5 bags of wheat.

Teacher's Note:
a) Find the intersection of the two boundary lines to get the key corner point.
b) The point \( (10, 0) \) is not feasible since \( 3(10) = 30 \gt 25 \).

OR

(iii) (b) Find the profit that the man can earn by selling all the bags. [2 Marks]

Answer:
1. Feasible region: \( x + y \le 10 \), \( 3x + 2y \le 25 \), \( x \ge 0, y \ge 0 \), with corner points \( (0, 0), (0, 10), (5, 5), \left( \frac{25}{3}, 0 \right) \).
2. Values of \( Z = 100x + 90y \): \( (0, 0) : 0 \); \( (0, 10) : 900 \); \( (5, 5) : 950 \); \( \left( \frac{25}{3}, 0 \right) : \frac{2500}{3} \).
3. The maximum value of Z is 950 at \( (5, 5) \).
Hence the maximum profit the man can earn is Rs. 950.

Teacher's Note:
a) Evaluate Z at every corner point of the feasible region in a table.
b) The profit asked is the maximum value of the objective function.

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