Class 12 Mathematics Sample Paper With Solutions Set H

Please refer to Class 12 Mathematics Sample Paper With Solutions Set H below. These Class 12 Mathematics Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Mathematics Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all CBSE Sample Papers for Mathematics in Standard 12.

Sample Paper Class 12 Mathematics With Solutions Set H

SECTION – A

1. If A and B are square matrices of the same order 3, such that |A| = 2 and AB = 2I , write the value of B .
Sol. As AB = 2I ⇒ |AB| = |2I| ⇒ |A||B| = 2³|I| ⇒ 2|B| = 8×1 ∴ B = 4 .
Note that the order of A and B is 3 and AB = 2I so, I is also of order 3.

2. If f(x) = x+1, find d/dx (fof )(x).
Sol. We have (fof )(x) = f (f(x)) = f (x+1) = x+1+1= x+2 ∴ d/dx (fof )(x) = d/dx(x+2)=1

3. Find the order and degree of the differential equation

Class 12 Mathematics Sample Paper With Solutions Set H

Sol. Order is 2 and degree is 1.

4. If a line makes angles 90°, 135°, 45° with the x, y and z axes respectively, find its direction cosines.
Sol. Here α = 90° , β =135° , γ = 45° .
Then direction cosines of the line are cos90° , cos135° , cos 45° i.e., 0,−1/√2 , 1/√2

Find the vector equation of the line which passes through the point (3, 4, 5) and is parallel to the vector 2î + 2ĵ−3k̂ .
Sol. The vector equation of the line is r̅ = 3î + 4ĵ+ 5k̂ + λ(2î + 2ĵ−3k̂ ) .
Note we have used r̅ = a̅ + λb̅

SECTION – B

5. Examine whether the operation * defined on R by a*b = ab +1 is
(i) a binary or not?
(ii) if a binary operation, is it associative or not?
Sol. Since ab +1 ∈ R ∀ a, b ∈ R . Therefore a *b ∈ R ∀ a, b ∈ R .
(i) Hence * is binary operation.
(ii) Let a, b,c ∈ R .
Now a *(b*c) = a *(bc +1) = abc + a +1 and (a *b)*c = (ab +1)*c = abc + c +1.
As a *(b*c) ≠ (a *b)*c , so * isn’t associative.

6. Find a matrix A such that 2A − 3B + 5C = O, where

Sol.

7. Find :

Sol.

8. Find :

Sol.

Find : ∫ sin⁻¹(2x)dx .
Sol.

9. Form the differential equation representing the family of curves y = e^2x (a + bx) , where ‘a’ and ‘b’ are arbitrary constants.
Sol. We have y = e^2x (a + bx) ⇒ e^−2xy = a + bx
⇒ e^−2xy’+ ye^−2x (−2) = 0 + b×1 ⇒ e^−2x{y’ − 2y} = b
⇒ e^−2x{y’− 2y’}+ e^−2x (−2){y’− 2y} = 0 ∴ y’ − 4y’ + 4y = 0 .

10. If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is √3.
Sol.

If a̅ = 2î + 3ĵ+ k̂, b̅ = î − 2ĵ+ k̂ and c̅ = −3î + ĵ+ 2k̂ , find [a b c].
Sol.

11. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event “number is even” and B be the event “number is marked with red”.
Find whether the events A and B are independent or not.
Sol. Here A = {2, 4, 6} and B = {1, 2, 3} and A∩B ={2}
Now P(A) = 3/6 = 1/2 , P(B) = 3/6 = 1/2 and P(A∩B) = 1/6 ≠ P(A)P(B)
Hence A and B are not independent.

12. A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of (i) 5 successes? (ii) at most 5 successes?
Sol. Here n = 6, p : probability of getting an odd number on the die. So, p = 3/6 =1/2, q =1/2 .
As P(X=r) = ⁶C_r (1/2)^r (1/2)^6−r = ⁶C_r × 1/2⁶
(i) P(X = 5) = ⁶C₅ × 1/2⁶= 6/64
(ii) P(X ≤ 5) = 1−P(X>5) = 1−P(X=6) = 1−⁶C₆ × 1/2⁶ = 1−1/64 = 63/64

The random variable X has a probability distribution P(X) of the following form, where ‘k’ is some number.

Determine the value of k.
Sol. Since ∑P(X) =1 ∴ P(X = 0) + P(X =1) + P(X = 2) + P(X = 3) +… =1
⇒ k + 2k + 3k + 0 +… =1 ∴ k =1/6 .

SECTION – C

13. Show that the relation R on R defined as R = {(a, b) : a ≤ b}, is reflexive, and transitive but not symmetric.
Sol. Here (a, a) ∈ R ∀ a ∈ R as a ≤ a is true . So, R is reflexive.
Let a, b,c ∈ R. Let (a, b) ∈ R and (b,c) ∈ R .
That is, a ≤ b and b ≤ c , which clearly, implies a ≤ c . Hence, (a,c) ∈ R . So, R is transitive.
Now let a =1,b = 2 .
We can notice that (1,2) ∈ R as 1≤ 2 is true but, (2,1) ∉ R as 2 ≤1 is false .
So, R isn’t symmetric, as (a, b) ∈ R does not imply (b,a) ∈ R .

Prove that the function f : N → N, defined by f (x) = x2 + x +1 is one-one but not onto.
Find inverse of f : N →S, where S is range of f.
Sol. Here f : N → N, defined by f (x) = x² + x +1.
Let x₁ , x₂ ∈ N and f (x₁) = f (x₂) . That is, x²₁+ x₁ +1 = x²₂ + x₂ +1 ⇒ x²₁− x²₂ = x₂ − x₁
⇒ (x₁ − x₂ )(x₁ + x₂ ) + (x₁ − x₂ ) = 0 ⇒ (x₁ − x₂ )(x₁ + x₂ +1) = 0
⇒ (x₁ − x₂) = 0 ∴ x₁ = x₂ [∵ x₁ + x₂ +1 ≠ 0 ∀ x₁, x₂ ∈ N
So, f is one-one.
Let y = f (x) and y ∈ N.

14. Solve : tan⁻¹ 4x tan⁻¹6x = π/4.
Sol.

As x = −1/2 doesn’t satisfy the equation. Hence, x =1/12 is the only solution.

15. Using properties of determinants, prove that

Sol.

16. If log(x² + y² ) = 2 tan⁻¹ (y/x) , show that dy/dx = x+y/x−y
Sol.

If x^y − y^x = a^b , find dy/dx.
Sol.

17.

Sol.

18. Find the equation of tangent to the curve y = √3x − 2 which is parallel to the line 4x − 2y + 5 = 0 . Also, write the equation of normal to the curve at the point of contact.
Sol.

19. Find :

Sol.

20. Prove that

Sol.

21. Solve the differential equation :

given that y = 0 when x =1.
Sol.

Solve the differential equation :

subject to the initial condition y(0) = 0.
Sol.

22. If î + ĵ+ k̂, 2î + 5ĵ, 3î + 2ĵ− 3k̂ and î − 6ĵ− k̂ respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether A̅B̅ and C̅D̅ are collinear or not.
Sol. Given O̅A̅ = î + ĵ+ k̂, O̅B̅ = 2î + 5ĵ, O̅C̅ = 3î + 2ĵ−3k̂ and O̅D̅ = î − 6ĵ− k̂.
The vectors parallel to the lines AB and CD are respectively, given by A̅B̅ and C̅D̅.
Also the angle between lines AB and CD will be the same as the angle between A̅B̅ and C̅D̅.
Now A̅B̅ = î + 4ĵ− k̂, C̅D̅= −2î −8ĵ+ 2k̂.

So, the required angle between lines AB and CD is π i.e., 180° .
Hence, the vectors A̅B̅ and C̅D̅ are collinear.

23. Find the value of λ , so that the lines

are at right angle. Also, find whether the lines are intersecting or not.
Sol.

SECTION – D

24.

find A^–1. Hence, solve the following system of equations :
x + y + z = 6, x + 2z = 7, 3x + y + z =12 .
Sol.

Find the inverse of the following matrix using elementary operations,

Sol.

25. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per square metre for the base and Rs 45 per square metre for the sides, what is the cost of least expensive tank?
Sol. Given volume of the box of height (depth) 2 m is, Lbh = 8 ⇒ Lb × 2 = 8 ⇒ b = 4/L
The cost of tank, C = 70(L×b) + 45(2×L×h) + 45(2×b×h) = 280 +180(L + b)

26. Using integration, find the area of the triangle ABC, whose vertices are A(2, 5), B(4, 7) and C(6, 2).
Sol.

Find the area of the region lying about x-axis and included between the circle x² + y² = 8x and inside of the parabola y² = 4x .
Sol.

27. Find the vector and Cartesian equations of the plane passing through the points (2, 2, –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.
Sol. Eq. of plane passing through the points (2, 2, –1), (3, 4, 2) and (7, 0, 6) is :

⇒ 20(x − 2) +8(y − 2) −12(z +1) = 0 ⇒ 5x + 2y −3z =17…(i)
And, vector equation is r̅.(5î + 2ĵ−3k̂) =17 .
The d.r.’s of the normal to the plane (i) are 5, 2, –3.
As the d.r.’s of the plane through (4, 3, 1) and parallel to (i) will be proportional to the d.r.’s of the plane (i).
So, the eq. of this plane is 5(x − 4) + 2(y −3) −3(z −1) = 0 ⇒5x + 2y − 3z = 23
Therefore, the required vector eq. is r̅.(5î + 2ĵ− 3k̂) = 23 .

Find the vector equation of the plane that contains the lines r̅ = (î + ĵ) + λ(î + 2ĵ− k̂) and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained.
Sol. As the required plane π (say) contains the line r̅ = (î + ĵ) + λ(î + 2ĵ− k̂) so, the point (1, 1, 0) on the line will also lie on this plane. Also the plane π contains (–1, 3, –4).
Let A(1, 1, 0) and B(–1, 3, –4). So, A̅B̅ = −2î + 2ĵ− 4k̂.
Normal vector to the plane π can be obtained by (î + 2ĵ− k̂ )×A̅B̅

28. A manufacturer has three machines operators A, B and C. The first operator A produces 1% of defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job 30% of the time and C on the job for 20% of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by A?
Sol. Let E : the chosen item is defective.
Also let E₁, E₂, E₃ be the events that the item was produced by operators A, B, C respectively.
We have P(E₁) = 50%, P(E₂) = 30%, P(E₃ ) = 20%,
1 2 3 P(E | E₁ ) =1%, P(E | E₂) = 5%, P(E | E₃) = 7%.

29. A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer’s profit on an item if model A is Rs 15 and on an item of model B is Rs 10. Howmany of items of each model should be made per day in order to maximize daily profit?
Formulate the above LPP and solve it graphically and find the maximum profit.
Sol. Let number of items of model A and B made per day be x and y, respectively.
It’s given that no man is expected to work more than 8 hours per day. Also 5 skilled men and 10 semi-skilled men are available so, at most 40 hours and 80 hours respectively time per day is available.