# CBSE Class 12 Mathematics Sample Paper Set I

See below CBSE Class 12 Mathematics Sample Paper Set I with solutions. We have provided CBSE Sample Papers for Class 12 Mathematics as per the latest paper pattern issued by CBSE for the current academic year. All sample papers provided by our Class 12 Mathematics teachers are with answers. You can see the sample paper given below and use them for more practice for Class 12 Mathematics examination.

SECTION A

1. Let A = {1, 2, 3, 4}. Let R be the equivalence relation on A×A defined by (a, b) R (c, d) iff a + d = b + c . Find the equivalence class [(1, 3)].
Sol. Let (1, 3) R (x, y) for all (x, y) ∈ A × A .
That implies, 1+ y = 3+ x i.e., y − x = 2 .
So (x, y) may be (1, 3), (2, 4).
Hence [(1,3)] = {(1,3), (2, 4)}.

2. If A = [aij] is a matrix of order 2× 2 , such that |A| = –15 and Cij represents the cofactor of aij, then find a21 C21 + a22 C22.
Sol.

3. Give an example of vectors a̅ and b̅ such that |a̅| = |b̅| but a̅ ≠ b̅
Sol. Let a̅ = î and b̅ = ĵ . Note that | a̅ | =1 = | b̅ | but a̅ ≠ b.
Other correct examples should be given full marks.

4. Determine whether the binary operation * on the set N of natural numbers defined by a*b = 2ab is associative or not.
Sol. 1*(2*3) =1*26 =1*64 = 264 , (1*2)*3 = 22 *3 = 4*3 = 212 .
Clearly 1*(2*3) ≠ (1*2)*3. Hence * is not associative.

SECTION B

5. If 4sin−1 x + cos−1 x = π , then find the value of x.
Sol. 4sin−1 x + cos−1 x = π     ⇒  3sin−1 x + sin−1 x + cos−1 x = π     ⇒ 3sin−1 x = π − π/2
⇒ sin−1 x = π/6      ⇒  x = sin π/6      ∴ = x = 1/2.

6.

Sol.

7. Prove that

Sol.

8. Find the approximate change in the value of 1/x, when x changes from x = 2 to x = 2.002.
Sol.

Hence y is decreased by 0.0005.
Note that x = 2 to x = 2.002 so, Δx ≈ dx = 2.002 − 2 = 0.002 .

9. Find

Sol.

10. Verify that ax2 + by2 =1 is a solution of the differential equation x(yy2+ y12 ) = yy1.
Sol.

11. Find the projection (vector) of 2î − ĵ,+ k̂ on î − 2ĵ,+ k̂ .
Sol.

12. If A and B are two events such that P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, then find P(A|B).
Sol. As P(B|A) = P(A∩B)/P(A)  ⇒ P(A∩B) = P(B|A)P(A) = 0.6×0.4 = 0.24
∴ P(A|B) = P(A∩B)/P(B) = 0.24/0.8 = 0.3 or 3/10.

SECTION C

13.

Sol.

14. Find ‘a’ and ‘b’, if the function given by

Sol. As f is differentiable at x =1 so, it is continuous at x =1 as well.

OR

Determine the values of ‘a’ and ‘b’ s. t. the following function is continuous at x = 0 :

Sol. It is given that f (x) is continuous at x = 0 .

Hence a = 0 and b may be any real number except 0 i.e., b ∈ R −{0}.

15.

Sol.

16. Find the equation (s) of the tangent (s) to the curve y = (x3 −1)(x − 2) at the points where the curve intersects the x-axis.
Sol. When the curve cuts x-axis, y = 0 i.e., (x3 −1)(x − 2) = 0
⇒ (x −1)(x2 + x +1)(x − 2) = 0 ∴ x =1, 2 (as x2 + x +1 ≠ 0 ∀ x ∈ R .
Hence points of contact are A(1, 0) and B(2, 0).
Now dy/dx = (x3−1)+3(x−2)x2 = 4x3 − 6x2 −1              ∴ (dy/dx)at A = −3, (dy/dx) at B = 7
Now equation of tangent at A : y − 0 = −3(x −1) i.e., 3x + y = 3.
And equation of tangent at B : y − 0 = 7(x − 2) i.e., 7x − y =14.

OR

Find the intervals in which the function f (x) = −3log(1 + x) + 4log(2 + x) − 4/2+x is  strictly  increasing or strictly decreasing.
Sol.

17. A person wants to plant some trees in his community park. The local nursery has to perform this task. It charges the cost of planting trees by the following formula :
C(x) = x3 − 45x2 + 600x , where x is the number of trees and C(x) is the cost of planting x trees in rupees. The local authority has imposed a restriction that it can plant 10 to 20 trees in one community park for a fair distribution. For how many trees should the person place the order so that he has to spend the least amount? How much is the least amount? Use calculus to answer these questions. Which value is being exhibited by the person?
Sol. We’ve C(x) = x3 − 45x2 + 600x , 10 ≤ x ≤ 20 .
For the time being we may assume that the function C(x) is continuous at all the points in the interval [10, 20].
Now C'(x) = 3x2 −90x + 600 = 3(x −10)(x − 20) , C'(x) = 6x − 90
For C'(x) = 3(x −10)(x − 20) = 0         ⇒ x =10, 20 .
Note that C'(10) = −30 < 0 and C'(20) = 30 > 0 .
So C(x) is minimum at x = 20 and maximum at x = 10.
Hence C(10) = 2500, C(20) = 2000 .
Therefore the person must place the order for 20 trees in order to spend the least amount which
is Rs 2000.
Value Exhibited: Concern of the person for a healthy environment despite having economic constraints.

18. Find

Sol.

19. Find the particular solution of the differential equation : yeydx = (y3 + 2xey )dy, y(0) =1.
Sol.

OR

Show that (x − y)dy = (x + 2y)dx is a homogeneous differential equation. Also find the general solution of the given differential equation.
Sol.

20. If a̅, b̅, c̅ are three vectors such that a̅ + b̅ + c̅ = 0 , then prove that a̅× b̅ = b̅×c̅ = c̅×a̅,, and hence show that [a̅ b̅ c̅] = 0.
Sol. Given a̅ + b̅ + c̅ = 0        ⇒ a̅ = −b̅ − c̅        ⇒ a̅ × b̅ = (−b̅ − c̅)× b̅ = −b̅×b̅ − c̅×b̅
⇒ a̅ ×b̅ = −0 + b̅× c̅              ⇒ a̅ ×b̅ = b̅×c̅…(i)
That is a̅ × b̅ = (−a̅ − c̅)× c̅                     [ ∵ a̅ + b̅ + c̅ = 0        ⇒ b̅ = −a̅ − c̅
⇒ a̅ × b̅ = −a̅ ×c̅ − c̅ × c̅        ⇒ a̅ × b̅ = c̅ × a̅ − 0‾         ⇒ a̅ × b̅ = c̅ × a̅ …(ii)
by (i) and (ii), a̅× b̅ = b̅×c̅ = c̅×a̅
Now [a̅ b̅ c̅] = (a̅× b̅).c̅ = (b̅×c̅).c̅ = [b̅ c̅ c̅] = 0       [by (i)
As the scalar triple product of three vectors is 0, if any two of them are equal vectors.

21.

Sol.

22. Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is the probability that they came from Bag III?
Sol. Let E1, E2 and E3 be the events that the bag I, bag II and bag III is chosen respectively.
Let E : the two balls drawn from the chosen bag are white and red.

23. Four bad oranges are accidently mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot. Find the mean and variance of the distribution.
Sol. Let X denotes the random variable. Then X can take values 0, 1, 2.
The probability distribution is as follow :

SECTION D

24. If the function f :R→R be defined by f (x) = 2x − 3 and g :R→R by g(x) = x3 + 5, then find fog and show that fog is invertible. Also find (fog)–1, hence find (fog)–1 (9).
Sol. fog :R→R defined by fog(x) = f (g(x)) = f (x3 + 5) = 2×3 + 7 = h(x) say .
Now h(x) = 2x3 + 7 .
Let x1, x2 , ∈ R (Domain of h(x)) such that  h(x1) = h(x2)
i.e., 2x31 + 7 = 2x32 + 7  ⇒ x1= x2. Hence h (x) is one-one.
Let y = h(x) such that y ∈ R (Codomain of h(x)).

OR

A binary operation * is defined on the set R of real numbers by

If at least one of a and b is 0, then prove that a * b = b * a. Check whether * is commutative. Find the identity element for *, if it exists.
Sol. et a, b ∈ R such that a = 0, b ≠ 0 .
Then a *b = 0*b = |0| +b = 0 + b = b, b*a = b*0 = b        ∴  a *b = b*a .
Again let a, b ∈ R such that a ≠ 0, b = 0 .
Then a *b = a *0 = a, b*a = 0*a = |0| +a = 0 + a = a       ∴  a *b = b*a .
For commutativity, let’s check when a ≠ 0, b ≠ 0 .
Then a *b = |a|+ b, b*a = |b|+ a . Clearly a *b may not be equal to b*a .
For example, (−1)*3 = | −1| +3 = 4, 3*(−1) = | 3| +(−1) = 2 , i.e., (−1)*3 ≠ 3*(−1) ,
∴ a *b ≠ b*a ∀ a, b ∈ R . Hence * is not commutative.
Let e be the identity element for *.
Then a * e = e*a = a for all a ∈ R .
Now a * e = a if e = 0, and e*a = a if e = 0 .
(As 0*0 = 0 and 0*a = | 0 | +a = a for a ≠ 0)
Hence e = 0 is the identity element for *.

25.

then find A–1 and hence solve the following system of equations :
3x + 4y + 7z =14, 2x − y + 3z = 4, x + 2y −3z = 0 .
Sol.

OR

find the inverse of A using elementary row transformations and hence solve the matrix equation : XA = [1 0 1] .
Sol.

26. Using integration, find the area in the first quadrant bounded by the curve y = x | x |, the circle x2 + y2 = 2 and the y-axis.
Sol.

27. Evaluate

Sol.

OR

Evaluate

Sol.

28. Find the distance of point −2î + 3ĵ,− 4k̂ from the line r̅ = î + 2ĵ,− k̂ + λ(î + 3ĵ,−9k̂) measured parallel to the plane x − y + 2z − 3 = 0 .
Sol.

29. A company produces two different products. One of them needs 1/4 of an hour of assembly work per unit, 1/8 of an hour in quality control work and Rs 1.2 in raw materials. The other product requires 1/3 of an hour of assembly work per unit, 1/3 of an hour in quality control work and Rs 0.9 in raw materials. Given the current availability of staff in the company, each day there is at most a total of 90 hours available for assembly and 80 hours for quality control. The first product described has a market value (sale price) of Rs 9 per unit and the second product described has a market value (sale price) of Rs 8 per unit. In addition, the maximum amount of daily sales for the first product is  estimated to be 200 units, without there being a maximum limit of daily sales for the second product. Formulate and solve graphically the LPP and find the maximum profit.
Sol. Let x and y be the no. of units of Product I and II to be produced daily, respectively.
To maximize : Z = Rs{(9 −1.2)x + (8− 0.9)y} = Rs( 7.8x + 7.1y ).