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# CBSE Class 12 Maths Sample Papers Term 2 with Solutions

## Class 12 Maths Sample Papers Term 2

Class 12 Maths Sample Paper Term 2: The Central Board of Secondary Education is going to conduct the CBSE Term 2 Exam from 26th April 2022 onwards. Central Board of Secondary Education has released CBSE Term 2 Class 12 Sample paper on its official website to give students an idea about the CBSE Term 2 exam. To check their preparedness, students must solve the CBSE Term 2 Sample Paper of Mathematics provided on this page. The students appearing in Class 12 and 10 exams conducted by the Central Board of Secondary Education must read the whole article and bookmark this page to get all the latest updates from the Central Board of Secondary Education.

Latest News: CBSE Launch Additional Question papers for Class 12 Physics

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## Maths Class 12 Sample Paper Term 2: Exam Pattern

Students taking the CBSE Term 2 exam must be familiar with the CBSE Term 2 exam format. Unlike the CBSE Term 1 exam, the CBSE Term 2 exam is subjective, therefore students must prepare differently for the CBSE Term 2 exam. Students will have two hours to complete the CBSE Term 2 exam paper. Check out the exam pattern of Class 12 Term 2 mathematics given below:

Section A: This section has very short type questions. Each question holds 2 marks.

Section B: This section has short-type questions. Each question holds 3 marks.

Section C: This section has long-type questions. Each question holds 4 marks.

## Class 12 Maths Sample Papers Term 2 with Solutions

Q. Find ∫ 𝑙𝑜𝑔𝑥/(1+𝑙𝑜𝑔𝑥) 2 𝑑x

Solution:

∫ 𝑙𝑜𝑔𝑥/(1+𝑙𝑜𝑔𝑥)2 𝑑𝑥

= ∫ 𝑙𝑜𝑔𝑥+1−1/(1+𝑙𝑜𝑔𝑥) 2 𝑑𝑥

= ∫ 1/1+𝑙𝑜𝑔𝑥 𝑑𝑥 − ∫ 1/(1+𝑙𝑜𝑔𝑥) 2 𝑑𝑥

= 1/1+𝑙𝑜𝑔𝑥 × 𝑥 − ∫ −1/(1+𝑙𝑜𝑔𝑥)2 × 1 𝑥 × 𝑥𝑑𝑥 − ∫ 1 (1+𝑙𝑜𝑔𝑥)2 𝑑𝑥

= 𝑥 /1+𝑙𝑜𝑔𝑥 + c

Q. Find ∫ 𝑠𝑖𝑛2𝑥/√9−𝑐𝑜𝑠4𝑥 𝑑x

Solution: Put 𝑐𝑜𝑠2𝑥 = 𝑡

⇒ −2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥𝑑𝑥 = 𝑑𝑡

⇒ 𝑠𝑖𝑛2𝑥𝑑𝑥 = −𝑑𝑡 The given integral

= − ∫ 𝑑𝑡 /√3 2−𝑡 2 = −sin−1 𝑡 /3 + 𝑐

= −sin−1 𝑐𝑜𝑠2𝑥/3 + 𝑐

Q. Write the sum of the order and the degree of the following differential equation: 𝑑/𝑑𝑥 ( 𝑑𝑦/𝑑𝑥) = 5

Solution: Order = 2

Degree = 1

Sum = 3

Q. If 𝑎̂ and 𝑏̂ are unit vectors, then prove that |𝑎̂ + 𝑏̂| = 2𝑐𝑜𝑠 𝜃 2 , where 𝜃 is the angle between them.

Solution: (𝑎̂ + 𝑏̂). (𝑎̂ + 𝑏̂) = |𝑎̂| 2 + |𝑏̂| 2 + 2(𝑎̂. 𝑏̂)

|𝑎̂ + 𝑏̂| 2 = 1 + 1 + 2𝑐𝑜𝑠𝜃 = 2(1 + 𝑐𝑜𝑠𝜃)

= 4𝑐𝑜𝑠2 𝜃 2

∴ |𝑎̂ + 𝑏̂| = 2𝑐𝑜𝑠 𝜃/2

Q. Find the direction cosines of the following line: 3 − 𝑥 /−1 = 2𝑦 − 1/ 2 = 𝑧/ 4

Solution: 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〉

Q. 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-byone without replacement.

Solution: 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:

 X 0 1 P(X) 1/2 1/2

Q. 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?

Solution:

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

Q. ∫ 𝑥+1/(𝑥 2+1)𝑥 𝑑x

Solution: Let 𝑥+1 (𝑥 2+1)𝑥 = 𝐴𝑥+𝐵 𝑥 2+1 + 𝐶 𝑥

= (𝐴𝑥+𝐵)𝑥+𝐶(𝑥 2+1) (𝑥 2+1)𝑥

⇒ 𝑥 + 1 = (𝐴𝑥 + 𝐵)𝑥 + 𝐶(𝑥 2 + 1) (An identity) Equating the coefficients, we get B = 1, C = 1, A + C = 0

Hence, A = -1, B = 1, C = 1

The given integral

= ∫ −𝑥+1 𝑥 2+1 𝑑𝑥 + ∫ 1 𝑥 𝑑𝑥 = −1 2 ∫ 2𝑥 − 2 𝑥 2 + 1 𝑑𝑥 + ∫ 1 𝑥 𝑑𝑥

= −1 2 ∫ 2𝑥 𝑥 2 + 1 𝑑𝑥 + ∫ 1 𝑥 2 + 1 𝑑𝑥 + ∫ 1 𝑥 𝑑𝑥

= −1 2 log(𝑥 2 + 1) + tan−1 𝑥 + 𝑙𝑜𝑔|𝑥| + c

Q. Find the general solution of the following differential equation: 𝑥/𝑑𝑦/𝑑𝑥 = 𝑦 − 𝑥𝑠𝑖𝑛( 𝑦/𝑥 )

Solution: We have the differential equation: 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥 − 𝑠𝑖𝑛( 𝑦/𝑥 )

The equation is a homogeneous differential equation. Putting 𝑦 = 𝑣𝑥

⇒ 𝑑𝑦/𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑣/𝑑𝑥

The differential equation becomes

𝑣 + 𝑥 𝑑𝑣/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

⇒ (𝑐𝑜𝑠𝑒𝑐 𝑦 𝑥 − 𝑐𝑜𝑡 𝑦 𝑥 ) 𝑥 = 𝐶, which is the required general solution.

Q. Find the particular solution of the following differential equation, given that y = 0 when  x = 𝜋/4 : 𝑑𝑦/𝑑𝑥 + 𝑦𝑐𝑜𝑡𝑥 = 2/1 + 𝑠𝑖𝑛𝑥

Solution:

The differential equation is a linear differential equation

If = 𝑒 ∫ 𝑐𝑜𝑡𝑥𝑑𝑥 = 𝑒 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 = 𝑠𝑖𝑛𝑥

The general solution is given by

𝑦𝑠𝑖𝑛𝑥 = ∫ 2 𝑠𝑖𝑛𝑥 1 + 𝑠𝑖𝑛𝑥 𝑑𝑥

⇒ 𝑦𝑠𝑖𝑛𝑥 = 2/∫ 𝑠𝑖𝑛𝑥 + 1 − 1/1 + 𝑠𝑖𝑛𝑥 𝑑𝑥 = 2/∫[1 − 1/1 + 𝑠𝑖𝑛𝑥 ]𝑑𝑥

⇒ 𝑦𝑠𝑖𝑛𝑥 = 2/∫[1 − 1/1 + cos ( 𝜋 2 − 𝑥) ]𝑑𝑥

⇒ 𝑦𝑠𝑖𝑛𝑥 = 2 ∫[1 − 1 2𝑐𝑜𝑠2 ( 𝜋 4 − 𝑥 2 ) ]𝑑𝑥

⇒ 𝑦𝑠𝑖𝑛𝑥 = 2 ∫[1 − 1 2 𝑠𝑒𝑐2 ( 𝜋 4 − 𝑥 2 )]𝑑𝑥

⇒ 𝑦𝑠𝑖𝑛𝑥 = 2[𝑥 + tan ( 𝜋 4 − 𝑥 2 )] + 𝑐

Given that y = 0, when x = 𝜋/4

Hence, 0 = 2[ 𝜋 4 + 𝑡𝑎𝑛 𝜋 8 ] + 𝑐

⇒ 𝑐 = − 𝜋 2 − 2𝑡𝑎𝑛 𝜋/8

Hence, the particular solution is 𝑦 = 𝑐𝑜𝑠𝑒𝑐𝑥[2 {𝑥 + tan ( 𝜋/4 − 𝑥/2 )} − ( 𝜋/2 + 2𝑡𝑎𝑛 𝜋/8 )]

Q. If 𝑎⃗ ≠ 0, ⃗⃗⃗ 𝑎⃗. 𝑏⃗⃗ = 𝑎⃗. 𝑐⃗, 𝑎⃗ × 𝑏⃗⃗ = 𝑎⃗ × 𝑐⃗, then show that 𝑏⃗⃗ = 𝑐⃗.

Solution: We have 𝑎⃗. (𝑏⃗⃗ − 𝑐⃗) = 0

⇒ (𝑏⃗⃗ − 𝑐⃗) = ⃗0⃗ or 𝑎⃗ ⊥ (𝑏⃗⃗ − 𝑐⃗)

⇒ 𝑏⃗⃗ = 𝑐⃗ or 𝑎⃗ ⊥ (𝑏⃗⃗ − 𝑐⃗)

Also, 𝑎⃗ × (𝑏⃗⃗ − 𝑐⃗) = ⃗0⃗

⇒ (𝑏⃗⃗ − 𝑐⃗) = ⃗0⃗ or 𝑎⃗ ∥ (𝑏⃗⃗ − 𝑐⃗)

⇒ 𝑏⃗⃗ = 𝑐⃗ or 𝑎⃗ ∥ (𝑏⃗⃗ − 𝑐⃗) 𝑎⃗ 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑏𝑒 𝑏𝑜𝑡ℎ 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 (𝑏⃗⃗ − 𝑐⃗) 𝑎𝑛𝑑 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 (𝑏⃗⃗ − 𝑐⃗)

Hence, 𝑏⃗⃗ = 𝑐⃗.

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

Solution:

Here, the lines are parallel. The shortest distance

(3𝑘̂) × (2𝑖̂+ 𝑗̂+ 𝑘̂) = | 𝑖̂ 𝑗̂ 𝑘̂ 0 0 3 2 1 1 | = −3𝑖̂+ 6𝑗̂

Hence, the required shortest distance = 3√5/√6 units

Q. 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) and 𝑟⃗ = (3𝑖̂+ 𝑗̂− 2𝑘̂) + 𝑡(𝑖̂− 3𝑗̂+ 𝑘̂)

Solution: 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

(2𝑖̂− 3𝑗̂+ 2𝑘̂) × (𝑖̂− 3𝑗̂+ 𝑘̂) = | 𝑖̂ 𝑗̂ 𝑘̂ 2 −3 2 1 −3 1 | = 3𝑖̂− 3𝑘̂

The vector equation of the plane is 𝑟⃗. (3𝑖̂− 3𝑘̂) =(𝑖̂+ 2𝑗̂− 𝑘̂). (3𝑖̂− 3𝑘̂) or, 𝑟⃗. (𝑖̂− 𝑘̂) = 2

the cartesian equation of the plane is x – z – 2 = 0

Q. Evaluate: ∫ |𝑥 3 − 3𝑥 2 + 2𝑥|𝑑𝑥 2 −1

Solution: The given definite integral = ∫ |𝑥(𝑥 − 1)(𝑥 − 2)|𝑑𝑥 2 −1

= ∫ |𝑥(𝑥 − 1)(𝑥 − 2)|𝑑𝑥 0 −1 + ∫ |𝑥(𝑥 − 1)(𝑥 − 2)|𝑑𝑥 + ∫ |𝑥(𝑥 − 1)(𝑥 − 2)|𝑑𝑥 2 1/1 0

= − ∫ (𝑥 3 − 3𝑥 2 + 2𝑥)𝑑𝑥 0 −1 + ∫ (𝑥 3 − 3𝑥 2 + 2𝑥)𝑑𝑥 1 0 − ∫ (𝑥 3 − 3𝑥 2 + 2𝑥)𝑑𝑥 2 1

= −[ 𝑥 4 4 − 𝑥 3 + 𝑥 2 ]−1 0 + [ 𝑥 4 4 − 𝑥 3 + 𝑥 2 ]0 1 − [ 𝑥 4 4 − 𝑥 3 + 𝑥 2 ]1 2

= 9 4 + 1 4 + 1 4 = 11

Q. 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.

Solution: Solving x + y = 2 and y 2 = x simultaneously, we get the points of intersection as (1, 1) and (4, -2).

The required area = the shaded area = ∫ √𝑥 1 0 𝑑𝑥 + ∫ (2 − 𝑥)𝑑𝑥 2 1

= 2 3 [𝑥 3 2]0 1 + [2𝑥 − 𝑥 2 2 ]1 2

= 2 3 + 1 2 = 7 6 square units

Q. Using integration, find the area of the region: {(𝑥, 𝑦): 0 ≤ 𝑦 ≤ √3𝑥, 𝑥 2 + 𝑦 2 ≤ 4}

Solution: Solving 𝑦 = √3𝑥 𝑎𝑛𝑑 𝑥 2 + 𝑦 2 = 4 , we get the points of intersection as (1, √3) and (-1, −√3)

The required area = the shaded area = ∫ √3𝑥 1 0 𝑑𝑥 + ∫ √4 − 𝑥 2𝑑𝑥 2 1

= √3 2 [𝑥 2 ]0 1 + 1 2 [𝑥√4 − 𝑥 2 + 4 sin−1 𝑥 2 ] 1 2

= √3 2 + 1 2 [2𝜋 − √3 − 2 𝜋 3 ]

= 2𝜋/3 square units

Q. 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.

Solution: 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

Q. 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?

(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?

Solution: 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) By Bayes’ Theorem, 𝑃(𝐸1⁄𝐸) = 𝑃(𝐸1 )×𝑃(𝐸/𝐸1 )/𝑃(𝐸)

= 20/100 × 6/10/280/1000 = 3/7

## Class 12 Maths Term 2 Sample Paper Video Solution

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Related Post:

## Class 12 Maths Term 2 Sample Paper Solutions: FAQs

Q. Where can I get CBSE Class 12th Mathematics Sample Paper 2022?

On this page, you will get CBSE Class 12th Mathematics Sample Paper 2022.

Q. What is the duration to solve the CBSE Term 2 Mathematics paper?

The students will get 2 hrs to solve the CBSE Term 2 Mathematics paper

Q. When will CBSE conduct CBSE Term 2 Examination?

The Central Board of Secondary Education will conduct the Term 2 exam from 26th April 2022.

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