Class 12 Mathematics Sample Paper Term 2 With Solutions Set B

Please refer to Class 12 Mathematics Sample Paper Term 2 With Solutions Set B 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 Term 2 CBSE Sample Papers for Mathematics in Standard 12.

Sample Paper Term 2 Class 12 Mathematics With Solutions Set B

SECTION – A

1. Find ∫𝑙𝑜𝑔𝑥 /(1+𝑙𝑜𝑔𝑥)²𝑑𝑥
Answer : 

OR

Find  ∫𝑠𝑖𝑛2𝑥 / √9−𝑐𝑜𝑠⁴𝑥𝑑𝑥
Answer : 

2. Write the sum of the order and the degree of the following differential equation: d/dx(dy/dx) = 5
Answer : Solution: Order = 2
Degree = 1
Sum = 3

3. If 𝑎̂ and 𝑏̂ are unit vectors, then prove that
|𝑎̂+𝑏̂|=2𝑐𝑜𝑠 θ/2 where 𝜃 is the angle between them.
Answer : (𝑎̂+𝑏̂).(𝑎̂+𝑏̂)=|𝑎̂|²+|𝑏̂|²+2(𝑎̂.𝑏̂)
|𝑎̂+𝑏̂|²=1+1+2𝑐𝑜𝑠𝜃
=2(1+𝑐𝑜𝑠𝜃)=4𝑐𝑜𝑠²𝜃/2
∴|𝑎̂+𝑏̂|=2𝑐𝑜𝑠𝜃/2

4. Find the direction cosines of the following line:
3–x/–1 = 2y–1/2 = z/4
Answer : The given line is 𝑥−3/1=𝑦−1/2/1=𝑧/4
Its direction ratios are <1, 1, 4>
Its direction cosines are 〈1/3√2, 1/3√2, 4/3√2〉

5. A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-by-one without replacement.
Answer : Let X be the random variable defined as the number of red balls.
Then X = 0, 1
P(X=0) = 3/4 × 2/3 = 6/12 = 1/2
P(X=1) =1/4 × 3/3 + 3/4 × 1/3 = 6/12 = 1/2
Probability Distribution Table:

6. Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack?
Answer : The required probability = P((The first is a red jack card and The second is a jack card) or (The first is a red non-jack card and The second is a jack card))
= 2/52 × 3/51 + 24/52 × 4/51 = 1/26

SECTION – B

7. Find: 

Answer : 

8. Find the general solution of the following differential equation: 
𝑥𝑑𝑦/𝑑𝑥=𝑦−𝑥𝑠𝑖𝑛(𝑦/𝑥)
Answer : We have the differential equation: 𝑑𝑦/𝑑𝑥=𝑦𝑥−𝑠𝑖𝑛(𝑦/𝑥)
The equation is a homogeneous differential equation.
Putting 𝑦 = 𝑣𝑥 ⇒ 𝑑𝑦/𝑑𝑥 = 𝑣 + 𝑥𝑑𝑣/𝑑𝑥
The differential equation becomes 𝑣+𝑥𝑑𝑣/𝑑𝑥=𝑣−𝑠𝑖𝑛𝑣
⇒ 𝑑𝑣/𝑠𝑖𝑛𝑣 = −𝑑𝑥/𝑥 ⇒ 𝑐𝑜𝑠𝑒𝑐𝑣𝑑𝑣 =−𝑑𝑥/𝑥
Integrating both sides, we get
𝑙𝑜𝑔|𝑐𝑜𝑠𝑒𝑐𝑣−𝑐𝑜𝑡𝑣|=−𝑙𝑜𝑔|𝑥|+𝑙𝑜𝑔𝐾,𝐾>0 (Here, 𝑙𝑜𝑔𝐾 is an arbitrary constant.)
⇒𝑙𝑜𝑔|(𝑐𝑜𝑠𝑒𝑐𝑣−𝑐𝑜𝑡𝑣)𝑥|=𝑙𝑜𝑔𝐾
⇒|(𝑐𝑜𝑠𝑒𝑐𝑣−𝑐𝑜𝑡𝑣)𝑥|=𝐾
⇒(𝑐𝑜𝑠𝑒𝑐𝑣−𝑐𝑜𝑡𝑣)𝑥=±𝐾
⇒(𝑐𝑜𝑠𝑒𝑐𝑦/𝑥−𝑐𝑜𝑡𝑦/𝑥)𝑥=𝐶, which is the required general solution.

OR 

Find the particular solution of the following differential equation, given that y = 0 when x = 𝜋/4:
dy/dx + ycot x = 2/1+sin x
Answer : The differential equation is a linear differential equation
I F = 𝑒^{∫𝑐𝑜𝑡𝑥𝑑𝑥} = 𝑒^{𝑙𝑜𝑔𝑠𝑖𝑛𝑥} = 𝑠𝑖𝑛𝑥
The general solution is given by 

9. If 𝑎⃗≠0,⃗⃗⃗ 𝑎⃗.𝑏⃗⃗=𝑎⃗.𝑐⃗,𝑎⃗×𝑏⃗⃗=𝑎⃗×𝑐⃗, then show that 𝑏⃗⃗=𝑐⃗.
Answer : We have a̅.(b̅−c̅)=0
⇒(b̅−c̅)=0⃗⃗ or a̅⊥(b̅−c̅)
⇒b̅=c̅ or a̅⊥(b̅−c̅)
Also, a̅×(b̅−c̅)=0⃗⃗
⇒(b̅−c̅)=0⃗⃗ or a̅∥(b̅−c̅)
⇒b̅=c̅ or a̅∥(b̅−c̅)
a̅ 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑏𝑒 𝑏𝑜𝑡ℎ 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 (b̅−c̅) 𝑎𝑛𝑑 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 (b̅−c̅)
Hence, b̅=c̅.

10. Find the shortest distance between the following lines:
𝑟⃗=(𝑖̂+𝑗̂−𝑘̂)+𝑠(2𝑖̂+𝑗̂+𝑘̂)
𝑟⃗=(𝑖̂+𝑗̂+2𝑘̂)+𝑡(4𝑖̂+2𝑗̂+2𝑘̂)
Answer :   

OR

Find the vector and the cartesian equations of the plane containing the point 𝑖̂+2𝑗̂−𝑘̂ and parallel to the lines
𝑟⃗=(𝑖̂+2𝑗̂+2𝑘̂)+𝑠(2𝑖̂−3𝑗̂+2𝑘̂)=0 and 𝑟⃗=(3𝑖̂+𝑗̂−2𝑘̂)+𝑡(𝑖̂−3𝑗̂+𝑘̂)=0
Answer : Since, the plane is parallel to the given lines, the cross product of the vectors 2𝑖̂−3𝑗̂+2𝑘̂ and 𝑖̂−3𝑗̂+𝑘̂ will be a normal to the plane  

The vector equation of the plane is 𝑟⃗.(3𝑖̂−3𝑘̂)=(𝑖̂+2𝑗̂−𝑘̂).(3𝑖̂−3𝑘)̂
or, 𝑟⃗.(𝑖̂−𝑘̂)=2
and the cartesian equation of the plane is x – z – 2 = 0

SECTION – C

11. Evaluate: ∫|𝑥³−3𝑥²+2𝑥|𝑑𝑥.
Answer : The given definite integral = ∫₋₁²|𝑥(𝑥−1)(𝑥−2)|𝑑𝑥 

12. Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y 2 = x and the x-axis.
Answer :  

OR

Using integration, find the area of the region {(𝑥,𝑦):0≤𝑦≤√3𝑥,𝑥²+𝑦²≤4}
Answer : Solving 𝑦=√3𝑥 𝑎𝑛𝑑 𝑥²+𝑦²=4 , we get the points of intersection as (1, √3) and (-1, −√3)  

13. Find the foot of the perpendicular from the point (1, 2, 0) upon the plane
x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane.
Answer : The equation of the line perpendicular to the plane and passing through the point (1, 2, 0) is
𝑥−1/1=𝑦−2/−3=𝑧/2
The coordinates of the foot of the perpendicular are (𝜇+1,−3𝜇+2,2𝜇) for some 𝜇
These coordinates will satisfy the equation of the plane. Hence, we have 𝜇+1−3(−3𝜇+2)+2(2𝜇)=9 ⇒𝜇=1
The foot of the perpendicular is (2, -1, 2).
Hence, the required distance = √(1−2)²+(2+1)²+(0−2)²=√14 𝑢𝑛𝑖𝑡𝑠

14. CASE-BASED/DATA-BASED

Fig 3 An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at sometime within a fixed one-year period with probability 0.6, whereas this probability is 0.2 for a person who is not accident prone. The company knows that 20 percent of the population is accident prone.

Based on the given information, answer the following questions.

(i) what is the probability that a new policyholder will have an accident within a year of purchasing a policy?
Answer : Let E₁ = The policy holder is accident prone.
E₂ = The policy holder is not accident prone.
E = The new policy holder has an accident within a year of purchasing a policy.
(i) P(E)= P(E₁)× P(E∕E₁) + P(E₂)× P(E∕E₂)
= 20/100 × 6/10 + 80/100 × 2/10 = 7/25

(ii) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone?
Answer :