ISC Class 12 Mathematics Sample Paper 2022 with Solutions

SECTION A - 32 MARKS

 

Question 1
Choose the correct option for the following questions.

 

(i) If \(\int \frac{(\log x)^{2}}{x} dx = \frac{(\log x)^{k}}{k} + c\), then the value of \(k\) is: [1]
(A) 3
(B) 2
(C) 1
(D) None of the above options

Answer: (A) 3

Let \(\log x = t\), then \(\frac{1}{x} dx = dt\). The integral becomes \(\int t^{2} dt = \frac{t^{3}}{3} + c = \frac{(\log x)^{3}}{3} + c\), giving \(k = 3\).

Teacher's Note:
a) Use substitution method by setting the inner logarithmic function as \(t\).
b) Do not confuse the power of the function with the denominator constant during integration.

 

(ii) \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(k - x) dx\), then the value of \(k\) is: [1]
(A) \(a\)
(B) \(2a\)
(C) Independent of \(a\)
(D) None of the above options

Answer: (B) 2a

Using standard definite integral properties, \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(2a - x) dx\), hence \(k = 2a\).

Teacher's Note:
a) Recall the property \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(2a - x) dx\).
b) Students often mistakenly write \(a\) instead of \(2a\) for the upper limit replacement.

 

(iii) The degree of the differential equation \(\frac{d^{2}y}{dx^{2}} + 3\left(\frac{dy}{dx}\right)^{2} = x^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{2}\) is: [1]
(A) 1
(B) 2
(C) 3
(D) 4

Answer: (B) 2

The highest order derivative is \(\frac{d^{2}y}{dx^{2}}\) (order 2), and its highest power in the given equation is 2, so the degree is 2.

Teacher's Note:
a) Degree is the highest power of the highest order derivative when the equation is polynomial in derivatives.
b) Do not sum the powers of different derivative terms; look strictly at the highest order term.

 

(iv) Given \(\int e^{x}\left(\frac{x - 1}{x^{2}}\right) dx = e^{x}f(x) + c\). Then \(f(x)\) satisfying the equation is: [1]
(A) \(x\)
(B) \(x^{2}\)
(C) \(\frac{1}{x}\)
(D) None of the above options

Answer: (C) \(\frac{1}{x}\)

Split the integral into \(\int e^{x}\left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx\). Since \(\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^{2}}\), the result is \(e^{x}\left(\frac{1}{x}\right) + c\).

Teacher's Note:
a) Apply the standard form \(\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + c\).
b) Identify \(f(x) = \frac{1}{x}\) and verify its derivative correctly.

 

(v) Two cards are drawn out randomly from a pack of 52 cards one after the other, without replacement. The probability of first card being a king and second card not being a king is: [1]
(A) \(\frac{48}{663}\)
(B) \(\frac{24}{663}\)
(C) \(\frac{12}{663}\)
(D) \(\frac{4}{663}\)

Answer: (A) \(\frac{48}{663}\)

Probability = \(\frac{4}{52} \times \frac{48}{51} = \frac{192}{2652} = \frac{48}{663}\).

Teacher's Note:
a) Account for dependent events due to drawing without replacement by reducing total and specific card counts.
b) Simplify fractions carefully to match the given options.

 

(vi) If two balls are drawn from a bag containing 3 white, 4 black and 5 red balls. Then, the probability that the drawn balls are of different colours is: [1]
(A) \(\frac{1}{66}\)
(B) \(\frac{3}{66}\)
(C) \(\frac{19}{66}\)
(D) \(\frac{47}{66}\)

Answer: (D) \(\frac{47}{66}\)

Total ways = \({}^{12}C_{2} = 66\). Different color ways = \({}^{3}C_{1}{}^{4}C_{1} + {}^{4}C_{1}{}^{5}C_{1} + {}^{5}C_{1}{}^{3}C_{1} = 12 + 20 + 15 = 47\). Probability = \(\frac{47}{66}\).

Teacher's Note:
a) Consider all pairs of different colors: white-black, black-red, and red-white.
b) Ensure all combinations are summed correctly before dividing by the total sample space.

 

Question 2

(a) Evaluate : \(\int \frac{x^{3} - x^{2} + x - 1}{x - 1} dx\) [2]

Answer:
\(\int \frac{x^{2}(x - 1) + 1(x - 1)}{x - 1} dx = \int (x^{2} + 1) dx = \frac{x^{3}}{3} + x + c\).

Teacher's Note:
a) Factorize the numerator by grouping terms to cancel out the denominator factor \((x - 1)\).
b) Remember to include the constant of integration \(c\) at the end.

OR

(b) Evaluate : \(\int \log_{10} x dx\) [2]

Answer:
\(\int \log_{10} x dx = \frac{1}{\log_{e} 10} \int \log_{e} x dx = \frac{1}{\log_{e} 10} [x(\log_{e} x - 1)] + c = x(\log_{e} x - 1)\log_{10} e + c\).

Teacher's Note:
a) Change the base of the logarithm to natural log before integrating by parts.
b) Treat \(\log_{e} x\) as the first function and 1 as the second function during integration by parts.

 

Question 3

(a) Solve the differential equation : \(\csc^{3} x dy - \csc y dx = 0\) [2]

Answer:
Separating variables: \(\frac{dy}{\csc y} = \frac{dx}{\csc^{3} x} \implies \sin y dy = \sin^{3} x dx\). Integrating both sides gives \(-\cos y = -\cos x + \frac{\cos^{3} x}{3} + c\), or \(\cos y = \cos x - \frac{\cos^{3} x}{3} + c\).

Teacher's Note:
a) Convert cosecant terms into sine functions using reciprocal identities.
b) Use substitution \(t = \cos x\) to solve \(\int (1 - t^{2}) \sin x dx\).

OR

(b) Solve the differential equation : \(\frac{dy}{dx} = 2^{-y}\) [2]

Answer:
Separating variables: \(2^{y} dy = dx\). Integrating both sides gives \(\frac{2^{y}}{\log 2} = x + c_{1}\), which simplifies to \(2^{y} = x \log 2 + c\).

Teacher's Note:
a) Separate variables so that all terms involving \(y\) are on one side and \(x\) on the other.
b) Recall that \(\int a^{y} dy = \frac{a^{y}}{\log a}\).

 

Question 4

Evaluate : \(\int_{2}^{8} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} dx\) [2]

Answer:
Let \(I = \int_{2}^{8} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} dx \ldots (1)\). Using property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\), we get \(I = \int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10 - x} + \sqrt{x}} dx \ldots (2)\). Adding (1) and (2) gives \(2I = \int_{2}^{8} 1 dx = [x]_{2}^{8} = 6\), so \(I = 3\).

Teacher's Note:
a) Use the definite integral property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx\).
b) Adding both equations eliminates the square root terms in the numerator.

 

Question 5

(a) A bag contains 6 red and 5 blue balls and another bag contains 5 red and 8 blue balls. A ball is drawn from the first bag and without noticing its colour is placed in the second bag. If a ball is drawn from the second bag, then find the probability that the drawn ball is red in colour. [4]

Answer:
Let \(E_{1}\) be drawing a red ball from the first bag, \(E_{2}\) be drawing a blue ball from the first bag, and \(A\) be drawing a red ball from the second bag.
\(P(E_{1}) = \frac{6}{11}\), \(P(A|E_{1}) = \frac{6}{14}\)
\(P(E_{2}) = \frac{5}{11}\), \(P(A|E_{2}) = \frac{5}{14}\)
\(P(A) = P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2}) = \left(\frac{6}{11} \times \frac{6}{14}\right) + \left(\frac{5}{11} \times \frac{5}{14}\right) = \frac{36}{154} + \frac{25}{154} = \frac{61}{154}\).

Teacher's Note:
a) Apply the Law of Total Probability considering the two mutually exclusive cases of ball transfer.
b) Recalculate the composition of the second bag correctly after the ball is transferred.

OR

(b) A bag contains 3 red and 4 white balls and another bag contains 2 red and 3 white balls. If one ball is drawn from the first bag and 2 balls are drawn from the second bag, then find the probability that all three balls are of the same colour. [4]

Answer:
Probability = P(1 red from bag I and 2 red from bag II) + P(1 white from bag I and 2 white from bag II)
\(= \frac{{}^{3}C_{1}}{{}^{7}C_{1}} \times \frac{{}^{2}C_{2}}{{}^{5}C_{2}} + \frac{{}^{4}C_{1}}{{}^{7}C_{1}} \times \frac{{}^{3}C_{2}}{{}^{5}C_{2}} = \frac{3}{7} \times \frac{1}{10} + \frac{4}{7} \times \frac{3}{10} = \frac{3}{70} + \frac{12}{70} = \frac{15}{70} = \frac{3}{14}\).

Teacher's Note:
a) Break down the requirement into mutually exclusive cases for all red or all white selections.
b) Use combinations accurately for drawing multiple balls from the second bag.

 

Question 6

Evaluate : \(\int \frac{1 + \sin x}{1 - \sin x} dx\) [4]

Answer:
Rationalize the denominator: \(\int \frac{(1 + \sin x)^{2}}{1 - \sin^{2} x} dx = \int \frac{1 + \sin^{2} x + 2\sin x}{\cos^{2} x} dx = \int (\sec^{2} x + \tan^{2} x + 2\sec x \tan x) dx\)
\(= \int (\sec^{2} x + \sec^{2} x - 1 + 2\sec x \tan x) dx = \int (2\sec^{2} x - 1 + 2\sec x \tan x) dx = 2\tan x - x + 2\sec x + c\).

Teacher's Note:
a) Multiply numerator and denominator by \((1 + \sin x)\) to convert the denominator into \(\cos^{2} x\).
b) Split the resulting expression into standard integrable trigonometric integrals.

 

Question 7

In a bolt factory, machines X, Y and Z manufacture 20%, 35% and 45% respectively of the total output. Of their output 8%, 6% and 5% respectively are defective bolts. One bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured in the machine Y? [6]

Answer:
Let \(E_{1}, E_{2}, E_{3}\) be events of bolts produced by machines X, Y, Z, and \(D\) be the event of drawing a defective bolt.
\(P(E_{1}) = 0.20\), \(P(E_{2}) = 0.35\), \(P(E_{3}) = 0.45\)
\(P(D|E_{1}) = 0.08\), \(P(D|E_{2}) = 0.06\), \(P(D|E_{3}) = 0.05\)
By Bayes' Theorem, \(P(E_{2}|D) = \frac{P(E_{2})P(D|E_{2})}{P(E_{1})P(D|E_{1}) + P(E_{2})P(D|E_{2}) + P(E_{3})P(D|E_{3})}\)
\(= \frac{0.35 \times 0.06}{(0.20 \times 0.08) + (0.35 \times 0.06) + (0.45 \times 0.05)} = \frac{0.0210}{0.0160 + 0.0210 + 0.0225} = \frac{0.0210}{0.0595} = \frac{210}{595} = \frac{6}{17}\).

Teacher's Note:
a) Define respective machine probabilities and conditional defective probabilities clearly.
b) Simplify the final fraction accurately by dividing numerator and denominator by their greatest common divisor.

 

Question 8

(a) Evaluate : \(\int \frac{dx}{\sin x + \sin 2x}\) [6]

Answer:
\(\int \frac{dx}{\sin x(1 + 2\cos x)} = \int \frac{\sin x dx}{\sin^{2} x(1 + 2\cos x)} = \int \frac{\sin x dx}{(1 - \cos^{2} x)(1 + 2\cos x)}\).
Let \(\cos x = t \implies -\sin x dx = dt\). Integral becomes \(-\int \frac{dt}{(1 - t)(1 + t)(1 + 2t)}\).
Using partial fractions: \(\frac{-1}{(1-t)(1+t)(1+2t)} = \frac{-1/6}{1-t} + \frac{1/2}{1+t} + \frac{-4/3}{1+2t}\).
Integrating gives \(-\frac{1}{6}\log(1 - \cos x) + \frac{1}{2}\log(1 + \cos x) - \frac{2}{3}\log(1 + 2\cos x) + c\).

Teacher's Note:
a) Multiply numerator and denominator by \(\sin x\) to set up substitution using \(\cos x\).
b) Apply partial fraction decomposition carefully to resolve the rational expression.

OR

(b) Evaluate : \(\int_{0}^{\pi/4} \log(1 + \tan x) dx\) [6]

Answer:
Let \(I = \int_{0}^{\pi/4} \log(1 + \tan x) dx \ldots (1)\).
Using property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\),
\(I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right) dx = \int_{0}^{\pi/4} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) dx = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan x}\right) dx\)
\(I = \int_{0}^{\pi/4} \log 2 dx - \int_{0}^{\pi/4} \log(1 + \tan x) dx = \int_{0}^{\pi/4} \log 2 dx - I\)
\(2I = [\log 2 \cdot x]_{0}^{\pi/4} = \frac{\pi}{4}\log 2 \implies I = \frac{\pi}{8}\log 2\).

Teacher's Note:
a) This is a standard definite integral property problem solved by adding the original and transformed integrals.
b) Remember to split the logarithmic quotient into difference of integrals before combining with \(2I\).

 

SECTION B - 8 MARKS

 

Question 9
Choose the correct option for the following questions.

 

(i) The equation of the plane which is parallel to \(2x - 3y + z = 0\) and which passes through \((1, -1, 2)\) is: [1]
(A) \(2x - 3y + z - 7 = 0$
(B) \(2x - 3y + z + 7 = 0$
(C) \(2x - 3y + z - 8 = 0$
(D) \(2x - 3y + z + 6 = 0\)

Answer: (A) \(2x - 3y + z - 7 = 0\)

Parallel plane equation is \(2x - 3y + z = d\). Substituting \((1, -1, 2)\) gives \(2(1) - 3(-1) + 2 = d \implies d = 7\), so \(2x - 3y + z - 7 = 0\).

Teacher's Note:
a) Parallel planes share identical coefficients of \(x\), \(y\), and \(z\).
b) Substitute the given point coordinates to determine the constant \(d\).

 

(ii) The intercepts made on the coordinate axes by the plane \(2x + y - 2z = 3\) are: [1]
(A) \(-\frac{3}{2}, -3, -\frac{3}{2}\)
(B) \(\frac{3}{2}, 3, -\frac{3}{2}\)
(C) \(\frac{3}{2}, -3, -\frac{3}{2}\)
(D) \(\frac{3}{2}, 3, \frac{3}{2}\)

Answer: (B) \(\frac{3}{2}, 3, -\frac{3}{2}\)

Divide the equation by 3: \(\frac{2x}{3} + \frac{y}{3} - \frac{2z}{3} = 1 \implies \frac{x}{3/2} + \frac{y}{3} + \frac{z}{-3/2} = 1\). Intercepts are \(\frac{3}{2}, 3, -\frac{3}{2}\).

Teacher's Note:
a) Reduce the plane equation to intercept form \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
b) Pay close attention to the negative sign associated with the \(z\)-intercept denominator.

 

Question 10
Find the equation of the plane passing through the point \((1, 1, 1)\) and is perpendicular to the line \(\frac{x - 1}{3} = \frac{y - 2}{0} = \frac{z - 3}{4}\). Also, find the distance of this plane from the origin. [2]

Answer:
Direction ratios of the line are \((3, 0, 4)\), which serve as normal vector components for the plane. Equation: \(3(x - 1) + 0(y - 1) + 4(z - 1) = 0 \implies 3x + 4z - 7 = 0\).
Perpendicular distance from origin \((0,0,0)\) = \(\frac{|3(0) + 4(0) - 7|}{\sqrt{3^{2} + 0^{2} + 4^{2}}} = \frac{7}{5}\) units.

Teacher's Note:
a) Direction ratios of a perpendicular line become the normal vector coefficients of the plane.
b) Use the standard point-normal form for plane equations and the point-to-plane distance formula.

 

Question 11
Using integration, find the area of the region bounded between the line \(x = 4\) and the parabola \(y^{2} = 16x\). [4]

Answer:

[Figure: Symmetric parabolic curve opening rightwards about the X-axis, intersecting the vertical line \(x = 4\) at points \((4, 8)\) and \((4, -8)\), with the shaded region enclosed between the curve and the line.]

Area \(= 2 \int_{0}^{4} y dx = 2 \int_{0}^{4} \sqrt{16x} dx = 8 \int_{0}^{4} \sqrt{x} dx = 8 \left[\frac{x^{3/2}}{3/2}\right]_{0}^{4} = \frac{16}{3}(4)^{3/2} = \frac{128}{3}\) sq. units.

Teacher's Note:
a) Exploit the symmetry about the X-axis to integrate from 0 to 4 and multiply by 2.
b) Substitute limits correctly to evaluate fractional powers of 4.

 

SECTION C - 8 MARKS

 

Question 12
Choose the correct option for the following questions.

 

(i) If the regression line of \(x\) on \(y\) is, \(9x + 3y - 46 = 0\) and \(y\) on \(x\) is, \(3x + 12y - 7 = 0\), then the correlation coefficient 'r' is equal to: [1]
(A) \(-\frac{1}{12}\)
(B) \(\frac{1}{12}\)
(C) \(-\frac{1}{2\sqrt{3}}\)
(D) \(\frac{1}{2\sqrt{3}}\)

Answer: (C) \(-\frac{1}{2\sqrt{3}}\)

From \(9x + 3y - 46 = 0 \implies x = -\frac{1}{3}y + \frac{46}{9}\), so \(b_{xy} = -\frac{1}{3}\). From \(3x + 12y - 7 = 0 \implies y = -\frac{1}{4}x + \frac{7}{12}\), so \(b_{yx} = -\frac{1}{4}\). Thus \(r = -\sqrt{b_{xy}b_{yx}} = -\sqrt{\left(-\frac{1}{3}\right)\left(-\frac{1}{4}\right)} = -\frac{1}{2\sqrt{3}}\).

Teacher's Note:
a) Extract regression coefficients \(b_{xy}\) and \(b_{yx}\) from the respective slope-intercept forms.
b) Assign the negative sign to \(r\) since both regression coefficients are negative.

 

(ii) If \(\overline{X} = 40, \overline{Y} = 6, \sigma_{x} = 10, \sigma_{y} = 1.5\) and \(r = 0.9\) for the two sets of data X and Y, then the regression line of X on Y will be : [1]
(A) \(x - 6y - 4 = 0$
(B) \(x + 6y - 4 = 0$
(C) \(x - 6y + 4 = 0$
(D) \(x + 6y + 4 = 0\)

Answer: (A) \(x - 6y - 4 = 0\)

Regression line of X on Y: \(x - \overline{X} = r\frac{\sigma_{x}}{\sigma_{y}}(y - \overline{Y}) \implies x - 40 = 0.9\left(\frac{10}{1.5}\right)(y - 6) \implies x - 40 = 6(y - 6) \implies x - 6y - 4 = 0\).

Teacher's Note:
a) Use the standard formula for the regression line of \(X\) on \(Y\).
b) Simplify carefully to match the linear equation format given in the options.

 

Question 13
For 5 observations of pairs \((x, y)\) of variables X and Y, the following results are obtained: \(\sum x = 15, \sum y = 25, \sum x^{2} = 55, \sum y^{2} = 135, \sum xy = 83\). Calculate the value of \(b_{xy}\) and \(b_{yx}\). [2]

Answer:
\(b_{xy} = \frac{n\sum xy - (\sum x)(\sum y)}{n\sum y^{2} - (\sum y)^{2}} = \frac{5(83) - (15)(25)}{5(135) - (25)^{2}} = \frac{415 - 375}{675 - 625} = \frac{40}{50} = 0.8\).
\(b_{yx} = \frac{n\sum xy - (\sum x)(\sum y)}{n\sum x^{2} - (\sum x)^{2}} = \frac{5(83) - (15)(25)}{5(55) - (15)^{2}} = \frac{415 - 375}{275 - 225} = \frac{40}{50} = 0.8\).

Teacher's Note:
a) Apply the standard formula for regression coefficients using summation values.
b) Double-check the denominator calculations to avoid simple arithmetic mistakes.

 

Question 14
A manufacturer wishes to produce two commodities A and B. The number of units of material, labour and equipment needed to produce one unit of each commodity is shown in the table given below. Also shown is the available number of units of each item, material, labour, and equipment. Find the maximum profit if each unit of commodity A earns a profit of Rs. 2 and each unit of B earns a profit of Rs. 3. [4]

Items	Commodity A	Commodity B	Available no. of Units
Material	1	2	8
Labour	3	2	12
Equipment	1	1	10

Answer:
Let \(x\) and \(y\) be units of commodity A and B respectively. Constraints:
\(x + 2y \le 8\)
\(3x + 2y \le 12\)
\(x + y \le 10\)
\(x \ge 0, y \ge 0\)
Objective function: Maximize \(Z = 2x + 3y\).
Corner points of feasible region: \(O(0, 0)\), \(A(4, 0)\), \(B(2, 3)\), \(C(0, 4)\).
Evaluating \(Z\) at corner points:
At \(O(0,0)\): \(Z = 0\)
At \(A(4,0)\): \(Z = 2(4) + 3(0) = 8\)
At \(B(2,3)\): \(Z = 2(2) + 3(3) = 13\) (Maximum)
At \(C(0,4)\): \(Z = 2(0) + 3(4) = 12\)
Maximum profit is Rs. 13.

Teacher's Note:
a) Formulate the linear programming problem correctly by identifying constraints and the objective function.
b) Test all boundary intersection vertices to determine the absolute maximum profit point.