Please refer to Class 12 Mathematics Sample Paper with solutions provided below. All sample papers for Mathematics Class 12 have been designed as per the latest paper pattern issued by CBSE for the current academic year. Students should practice these guess papers for Class 12 Mathematics as it will help them to gain more understanding of the type of questions that are expected to be asked in upcoming Class 12 Mathematics exams. Please click on the links below to access free CBSE Sample Papers for Class 12 Mathematics.
Class 12 Mathematics Sample Paper Term 2 With Solutions Set A
Section – A
1. Determine order and degree of given differential equation:
Answer: The given differential equation
The highest order derivative presents in the given differential equation is d2S/dx2 Therefore, its order is two. It is a polynomial equation in d2S/dt2 and dS/dt. The power raised to d2S/dt2 is 1. Hence, its degree is one.
2. Find the coordinates of the foot of the perpendicular from O(0, 0, 0) to the plane 3x – 4y + 5z – 10 = 0.
Answer:Let the coordinates of foot of perpendicular be P(x1, y1, z1) from point O(0, 0, 0). Equation of plane in normal from is given by
3/√50x-4/√50y+5/√50z=√2 [Dividing through by 32+ (-4)2 +52 √50 ]
∴ The direction cosines of the normal drawn from the origin to the given plane are
So coordinate of foot of perpendicular P(3/5,-4/5-,1).
3. Find the angle between unit vectors a∧and b∧ so that √3a − b is also a unit vector.
Answer:a and b are unit vectors and √3a − b is also unit vector To find angle between a and b.
Suppose angle between a and b is θ.
Therefore, the angle between the two unit vectors is π/6·
4. Ten cards numbered 1 to 10 are placed in box, mixed up throughly and then one card is drawn randomly.
If it is known that the number on the drawn card is more than 3, what is the probability that it is an even bumber?
Answer: Let A be the event the number on the card drawn is even and B be the event the number on the card drawn is greater than 3.
Now, the sample space of the experiment is
5. Find:∫sinx- cosx/√1+sin2x dx,0<x<π/2
Find: ∫(log x)2 dx
6. A bag contains 8 black and 5 blue balls three balls are drawn at random without replacements. What is the probability that all drawn balls are black colours?
Answer: 1st black ball drawn at random the probability =8/13
2nd black ball drawn at random the probability = 7/12
3rd black ball drawn at random the probability= 6/11
So, the total probability =28/13x 11=0 .195
Section – B
7. Find the scalar components of a unit vector which is perpendicular to each of the vectors î+ 2ĵ -k̂ and 3î-ĵ -2k̂.
8. Find the shortest distance between the lines l1 and l2 whose vector equations are →r= î= ĵ+λ(2î-ĵ+k̂)
and → r =2î +ĵ -k̂+μ(3î- 5ĵ +2k̂)
Find cartesian equation of a line where the direction ratios of the parallel vector to it are (2, –1, 3) and passes through (5, –2, 4).
Answer: We have given two vector equations
Given, direction ratios of the parallel vector = (2, – 1, 3)
∴ The direction ratios of required line are (2k, – k, 3k)
Also, the given point is (x1, y1, z1) = (5, – 2, 4)
Cartesian equation of line passes through (x1, y1, z1) and with direction ratios a, b, c is given by
9. Find the general solution of the following differential equation :
x dy – (y + 2x2)dx = 0
Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2 y dy = 0, given that
y = π/4 when x = 0.
Section – C
11. Find the coordinates of foot of perpendicular drawn from the point (0, 2, 3) on the line x +3/5=y-1/2=z+4/3.
Also, find the length of perpendicular.
Answer: Given equation of line is
Any random point T on the given line is calculated as follows :
x +3 /5=y-1/2=z+4/3=λ(say)
∴ coordinates of point T are (5λ– 3, 2λ + 1, 3λ – 4). Let point is P (0, 2, 3)
Now, Direction ratios of line PT = (5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3)
= (5λ – 3, 2λ – 1, 3λ – 7)
∴ PT is perpendicular to the given line.
⇒ a1a2 + b1b2 + c1c2 = 0
a1 = 5λ – 3, b1 = 2λ – 1, c1 = 3λ – 7
a2 = 5, b2 = 2, c2 = 3
⇒ 5(5λ – 3) + 2 (2λ – 1) + 3 (3λ – 7) = 0
⇒ 25λ – 15 + 4λ – 2 + 9λ – 21 = 0
⇒ 38λ – 38 = 0
⇒ λ = 1
∴ The foot of perpendicular T = (5λ– 3, 2λ + 1, 3λ – 4)
= (2, 3, –1) (Put l = 1)
Also , length of perpendicular, PT = Distance between point P and T
PT = √(0-2)2 + (2 – 3)2 + (3 + 1)2
= √4 + 1+ 16
= 21 units
12. Find the area under the curve y = 2√ x and between the lines x = 0 and x = 1.
Find the area of parabola y2 = 4ax bounded by its latus rectum.
Answer: Given, y = 2√x
⇒ y2 = 4x
Since, it is a part of a parabola
According to figure, the area of the shaded region
The graph of required parabola and its latus rectum is given alongside. The vertex of parabola y2 = 4ax is at (0, 0). The equation of latus rectum LSL’ is x = a
Also, parabola is symmetric about the X-axis.
∴The required area of the region OLL’O
Answer: Let, I =∫x9/(4x2+1)6dx
14. A doctor is going to visit a patient. From the past experience, it is known that the probabilities that he will come by cab, metro, bike or by other means of transport are respectively 0.3, 0.2, 0.1 and 0.4. The probabilities that he will be late are 0.25, 0.3, 0.35 and 0.1, if he comes by cab, metro, bike and other means of transport respectively.
Based on the above information, answer the following questions :
(i) When the doctor arrives late, what is the probability that he comes by metro?
(ii) When the doctor arrives late, what is the probability that he comes by other means of transport?
Answer: Let E be the event that the doctor visit the patient late and let A1, A2, A3, A4 be the events that the doctor comes by cab, metro, bike and other means of transport respectively.
P(A1) = 0.3, P(A2) = 0.2, P(A3) = 0.1, P(A4) = 0.4
P(E/A1) = Probability that the doctor arriving late when he comes by cab = 0.25
Similarly, P(E/A2) = 0.3, P(E/A3) = 0.35
And P(E/A4) = 0.1
(i) P(A2/E) = Probability that the doctor arriving late when he comes by metro
=P(A2) P (E/ A2)
= (0.2)(0.3)/(0 . 3)( 0. 25) (0 .2 )(0 .3 )+ (0 . 1)(0 . 35)+ (0 .4 )( 0.1)