See below CBSE Class 12 Mathematics Sample Paper Set F 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. Find the area of a parallelogram whose adjacent sides are represented by the vectors 2î – 3k̂and 4ĵ + 2k̂ .
Solution:
2. Find the sum of the intercepts cut off by the plane 2x + y – z = 5 on the coordinate axes.
Solution: Given plane 2x + y – z = 5 i.e., x/5/2 + y/5 + z/-5 = 1
On comparing to x/a + y/b + z/c = 1 , we have x, y and z-intercepts as 5/2, 5, –5 respectively.
∴ sum of the intercepts = 5/2 + 5 + (-5) = 5/2 .
3. Find the unit vector in the direction of the sum of the vectors 2î + 3ĵ -k̂and 4î – 3ĵ + 2k̂ .
Solution:
4. Write the sum of the order and degree of the differential equation (d2y/dx2)2 + (dy/dx)3 + x4 = 0 .
Solution: Order and degree of the differential equation (d2y/dx2)2 + (dy/dx)3 + x4 = 0 is 2 and 2 respectively.
∴ required sum of order and degree = 2 + 2 = 4 .
5. Write the solution of the differential equation dy/dx = 2-y .
Solution:
6. If
then write the cofactor of the element a21 of its 2nd row.
Solution: Cofactor of element a21 , C21 = – [6 x 3 -(-7)(-3)] = 3 .
SECTION – B
7. Find the point on the curve 9y2 = x3 , where the normal to the curve makes equal intercepts on the axes.
Solution: We have 9y2 = x3 . Let the required point be P (α,β) . So, 9β2 = α3 ….(i)
On differentiating the eq. of curve w.r.t. x, 18y dy/dx = 3x2 ⇒ dy/dx = x2/6y
8. If y = [x+√1+x2]n , then show that (1+x2) d2y/dx2 + x dy/dx = n2y .
Solution:
9. Find whether the following function is differentiable at x =1 and x = 2 or not :
Solution:
Hence f (x) is differentiable at x = 2.
10. In a parliament election, a political party hired a public relation firm to promote its candidates in 3 ways – telephone, house calls and letters. The cost per contact (in paise) is given in matrix A as
The number of contacts of each type made in two cities X and Y is given in the matrix B as
Find the total amount spent by the party in the two cities.
What should one consider before casting his/her vote – party’s promotional activity or their social activities?
Solution: The cost per contact (in paise) is given in matrix A as
The number of contacts of each type made in two cities X and Y is given in the matrix B as
The total amount spent by the party in the two cities = BA
Hence, the party spent 990000 paise (or, ₹ 9900) in the City X and, 2120000 paise (or, ₹ 21200) in the City Y.
One should consider party’s social activities instead of promotional activities of the party before casting his/her vote.
11. Evaluate : ∫e2x . sin(3x+1)dx.
Solution:
12. Evaluate :
Solution:
13. Three machines E1, E2 and E3 in a certain factory producing electric bulbs, produces 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by each of machines E1 and E2 are defective and that 5% of those produced by machine E3 are defective. If one bulb is picked up at random from a day’s production, calculate the probability that it is defective.
Solution: Let E1, E2 and E3 denote the events that bolts produced by machines E1, E2 and E3 respectively.
Let A be the event that the selected bulb is defective.
OR
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6, and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.
Solution: Since X denotes the larger of the two numbers obtained from 2, 3, 4, 5, 6 and 7.
So values of X : 3, 4, 5, 6, 7.
14. The two vectors ĵ + k̂and 3î – ĵ + 4k̂ represent the two sides vectors Α̅Β̅ and Α̅C̅ respectively of triangle ABC. Find the length of the median through A.
Solution:
15. Find the equation of a plane which passes through the point (3, 2, 0) and contains the line x-3/1 = y-6/5 = z-4/4 .
Solution: The equation of plane through (3, 2, 0) is A(x -3) + B(y – 2) + C(z – 0) = 0…(i) where A, B, C are the d.r.’s of the normal to the required plane.
As plane (i) contains the line x-3/1 = y-6/5 = z-4/4 with d.r.’s as 1, 5, 4 so,
A + 5B + 4C = 0…(ii) .
Also (3, 6, 4) lies on the given line and plane (i) as well so, A(3-3) + B(6 – 2) + C(4 – 0) = 0
i.e., 0.A+ 4B+ 4C = 0 ⇒ 0A+ B+ C = 0…(iii)
Solving (ii) and (iii), we get A/1 = B/-1 = C/1 i.e., d.r.’s of the normal to plane (i) are 1, –1, 1.
Hence equation of plane is : 1.(x – 3) -1.(y – 2) +1.z = 0 i.e., x – y + z =1.
16. If 2 tan-1(cos θ) = tan-1(2 cosec θ) , (θ ≠ 0) , then find the value of θ .
Solution:
OR
If tan-1(1/1+2.2) + tan-1 (1/1+2.3) +….+ tan-1(1/1+n.(n+1) = tan-1 θ , then find the value of θ .
Solution:
17. If
and I is the identity matrix of order 2, then show that A2 = 4A- 3I . Hence find A-1 .
Solution:
OR
If
and (A+B)2 = A2 + B2 , then find the values of a and b.
Solution: We’ve (A+B)2 = A2 + B2 ⇒ (A+B)(A+B) = A2+B2
⇒ A.A + A.B + B.A + B.B = A.A + B.B ⇒ A.B = -B.A
By equality of matrices, we get :
a – b = -a – 2, 2 = a + 1, 2a – b = 2 – b,3 = b -1
On solving these equations, we get : a =1, b = 4 .
18. Using properties of determinants, prove the following :
Solution:
19. Evaluate :
Solution:
OR
Evaluate :
Solution:
SECTION – C
20. Solve the given differential equation : [y – x cos(y/x)] dy + [y cos(y/x) -2x sin (y/x)] dx = 0 .
Solution:
OR
Solve the given differential equation : √1+x2 + y2 + x2 y2 dx + xydy = 0 .
Solution:
21. Find the probability distribution of the number of doublets in four throws of a pair of dice. Also find the mean and variance of this distribution.
Solution: Let E : getting a doublet on the pair of dice ∴ P(E) = 1/6 , P(E̅) = 5/6 .
Let X : Number of doublets in four throws of a pair of dice. So values of X are 0, 1, 2, 3, 4.
22. Let f : N → R be a function defined as f (x) = 4x2 + 12x + 15 . Show that f : N → S, where S is the range of f, is invertible. Also find the inverse of f.
Solution: Here f : N → R, f (x) = 4x2 + 12x +15
Let y be an arbitrary element of range S of function f. Then y = 4x2 + 12x + 15, for some x in N, which implies that y = (2x + 3)2 + 6.
23. Using integration, find the area of the region bounded by the line x – y + 2 = 0 , the curve x =√y and y-axis.
Solution: We have x – y +2 = 0…(i) and x =√y ….(ii)
24. Find the distance of the point P(1, –2, 3) from the plane x – y + z = 5 measured parallel to the line whose direction cosines are proportional to 2, 3, –6.
Solution: Equation of a line through P(1,–2,3) & parallel to a line whose d.r.’s are proportional to 2, 3, –6 is : x-1/2 = y+2/3 = z-3/-6 = λ ……(i) .
Any random point on line (i) is Q(2λ +1,3λ – 2,3 – 6λ) .
If Q lies on the given equation of plane x – y + z = 5 then,
25. Maximise z = 8x + 9y , subject to the constraints given below :
2x + 3y ≤ 6 ,
3x – 2y ≤ 6 ,
y ≤ 1,
x, y ≥ 0 .
Solution: To Maximise z = 8x + 9y ,
Subject to the constraints given below :
2x + 3y ≤ 6 ,
3x – 2y ≤ 6 ,
y ≤ 1; x, y ≥ 0 .
So maximum value of z is attained at (30/13 , 6/13) and maximum value is 22,8/13 .
26. Find the minimum value of (ax + by), where xy = c2 .
Solution: Given xy + c2…(i)
OR
Find the coordinates of a point of the parabola y = x2 + 7x + 2 which is closest to the straight line y = 3x – 3.
Solution: Given line is y = 3x – 3 i.e., 3x – y – 3 = 0…(i)
Also let the required point on the parabola y = x2 + 7x + 2 be P(h, k) ∴ k = h2 + 7h + 2…(i)
Distance of P from line 3x – y – 3 = 0 is, s = |3h – k – 3|/√32 + (-1)2
By (i), k = (-2)2 + 7(-2) + 2 = – 8
Hence the coordinates of the required points on the given parabola are P(–2, –8)
