# Class 12 Mathematics Sample Paper With Solutions Set K

Please refer to Class 12 Mathematics Sample Paper With Solutions Set K 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 K

SECTION – A

1. If for any 2 x 2 square matrix A, A(adj.A) =

then write the value of |A|.
Solution:

2. Determine the value of ‘k’ for which the following function is continuous at x = 3 :

Solution:

3. Find :

Solution:

4. Find the distance between the planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20 .
Solution: Note that the d.r.’s of the normal to both the planes are proportional (i.e., 2/5 = -1/-2.5 = 2/5) which implies that the planes are parallel.
Now π1 : 2x – y + 2z – 5 = 0 and π2 :5x – 2.5y + 5z – 20 = 0 can be rewritten as
π1 : 5x – 2.5y + 5z -12.5 = 0 and π2 : 5x – 2.5y + 5z – 20 = 0
Now the distance between the parallel planes is |-12.5 + 20|/√25 + 6.25 + 25 = 75/√5625 = 1 unit .

SECTION – B

5. If A is a skew-symmetric matrix of order 3, then prove that det A = 0.
Solution: As A is skew-symmetric of order 3. So, A = -AT

6. Find the value of c in Rolle’s Theorem for the function f (x) = x3 – 3x in [- √3, 0].
Solution: Since the polynomial function f (x) = x3 – 3x is everywhere continuous and differentiable, so
i) f (x) is continuous on [- √3, 0] and,
ii) f (x) is differentiable on (- √3, 0) .
iii) Also f (-√3) = -3√3 + 3 √3 = 0 and f (0) = 0  ⇒ f (- √3) = f (0) .
∴ all the conditions of Rolle’s theorem are satisfied. So there must exist one point c ∈ (- √3,0)
such that f ‘(c) = 0. Now f ‘(x) = 3x2 – 3
For f ‘(c) = 0, 3c2 – 3 = 0  ∴ c = -1 ∈(- √3, 0) .

7. The volume of a cube is increasing at the rate of 9 cm3/s. How fast is its surface area increasing when the length of an edge is 10 cm?
Solution: As volume of cube, V = a3  ⇒  dV/dt = 3a2  da/dt = 9cm3/s  ⇒  da/dt = 3/a2 cm/s
Now Surface area of cube, S = 6a2  ⇒  dS/dt = 12a  da/dt = 12a x 3/a2 = 36/a cm2s-1

8. Show that the function f (x) = x3 – 3x2 + 6x – 100 is increasing on R.
Solution: Here f (x) = x3 – 3x2 + 6x -100   ⇒  f ‘(x) = 3x2 – 6x + 6 = 3(x2 – 2x + 1) + 3
⇒ f ‘(x) = 3(x -1)2 + 3 > 0 for all x ∈ R  ⇒ f (x) is increasing on R.

9. The x coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, –2) is 4. Find its z-coordinate.
Solution: Equation of line PQ : x-2/3 = y-2/-1 = z-1/-3 = λ
Coordinates of any random point on the line PQ is A(3λ + 2,-λ + 2,-3λ +1) .
As x coordinate of a point on the line PQ is 4 so, 3λ + 2 = 4 ⇒ λ = 2/3 .
So, z coordinate is -3λ + 1= -3(2/3) + 1= -1.

10. A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find if A and B are independent events.
Solution: Here A = {2, 4,6}, B = {1, 2,3}, A∩B = {2}
So, P(A) = 3/6 = 1/2 , P(B) = 3/6 = 1/2 , P(A ∩ B) = 1/6 ≠ P(A)P(B) [∴ P(A)P(B) = 1/4]
Hence the events A and B are not independent events.

11. Two tailors, A and B, earn `300 and `400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP.
Solution: Let the number of days for which the tailors A and B work be x and y respectively.
To minimize: Z = ₹ (300x + 400y)
Subject to constraints: 6x + 10y ≥ 60, 4x + 4y ≥ 32; x, y ≥ 0 or, 3x + 5y ≥ 30, x + y ≥ 8; x, y ≥ 0 .

12. Find ∫ dx/5 – 8x – x2 .
Solution:

SECTION – C

13. If tan-1 x-3/x-4 + tan-1 x+3/x+4 = π/4 , then find the value of x.
Solution:

14. Using properties of determinants, prove that

Solution:

OR

Find matrix A such that

Solution:

15. If xy + yx = ab , the find dy/dx .
Solution:

OR

If ey (x +1) =1, then show that d2y/dx2 = (dy/dx)2 .
Solution:

16. Find

Solution:

17. Evaluate :

Solution:

OR

Evaluate :

Solution:

18. Solve the differential equation (tan-1 x – y)dx = (1+ x2 )dy .
Solution: Given differential equation can be rewritten as dy/dx + (1/1+x2) y = tan-1/1+ x2 .
This is linear differential equation of the form dy/dx + P(x)y = Q(x) where P(x) = 1/1+x2

19. Show that the points A, B, C with position vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂, and 3î – 4ĵ – 4k̂respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
Solution:

20. Find the value of λ , if four points with position vectors  3î + 6ĵ + 9k̂, î + 2ĵ + 3k̂, 2î + 3ĵ + k̂ and 4î + 6ĵ + λk̂ are coplanar.
Solution:

21. There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.
Solution: Clearly X can take values 4, 6, 8, 10. 12.
Here S = {(1,3),(1,5),(1,7),(3,1),(3,5),(3,7),(5,1),(5,3),(5,7),(7,1),(7,3),(7,5)}

22. Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at random from the school and he was found to have an A grade.
What is the probability that the student has 100% attendance? Is regularity required only in school? Justify your answer.
Solution: Let E : the student has grade A, E1 : the student has 100% attendance and E2 : the student is irregular.
Clearly P(E1) = 30%, P(E2 ) = 70%, P(E | E1 ) = 70% and P(E | E2 ) = 10%.
By Bayes’ Theorem, P(E1 | E) = P(E | E1 )P(E1 )/P(E | E1 )P(E1 ) + P(E | E2 )P(E2) = 70 x 30/70 x 30 + 10 x 70 = 3/4 = 75% .
The value of Regularity helps in all aspects of life. Regularity in school as well as professional life brings success for us.

23. Maximize : Z = x + 2y
Subject to the constraints : x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200, x ≥ 0, y ≥ 0 .
Solve the L.P.P. graphically
Solution: To Maximize : Z = x + 2y
Subject to : x + 2y ≥ 100, 2x – y ≤ 0,
2x + y ≤ 200, x ≥ 0, y ≥ 0.

Clearly, Maximum value of Z = 400 at A(0,200) .

24. Determine the product

and use it to solve the system of equations x – y + z = 4, x – 2y – 2z = 9 , 2x + y + 3z =1.
Solution:

25. Consider f : R -{-4/3} → R – {4/3} given by f (x) = 4X+3/3X+4 . Show that f is bijective. Find th inverse of f and hence find f -1(0) and x such that f -1(x) = 2 .
Solution:

OR

Let A = QxQ and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d)∈ A . Determine, whether * is commutative and associative. Then, with respect to * on A find the (i) identity element in A (ii) invertible elements of A.
Solution: Given A = Q x Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b)*(c,d) = (ac, b + ad) for (a, b),(c,d) ∈ A.
Commutativity : As (c,d)*(a, b) = (ca,d + cb) . As (ca,d + cb) ≠ (ac,b + ad) so, (a, b)*(c,d) ≠ (c,d)*(a, b) . Hence, * isn’t commutative.
Associativity : Let (a, b),(c,d),(x, y)∈ A.
[(a, b)*(c,d)]*(x, y) = [(ac,b + ad)]*(x, y) = (acx,b + ad + acy)…(i)
Also, (a, b)*[(c,d)*(x, y)] = (a, b)*(cx,d + cy) = (acx, b + ad + acy)…(ii)
By (i) and (ii), we can say that * is associative.
(i) Let (e,e’) be the identity element of * in A. Then (a, b)*(e,e’) = (a,b) = (e,e’)*(a, b)

26. Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.
Solution: Let the length and width of the cuboid of square base be x and its height be y.
Therefore volume of the cuboid, V = xx xx y

Since length of cuboid = width of cuboid = height of cuboid therefore, it is a cube.

27. Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C(8, 4).
Solution: Here A(4, 1), B(6, 6) and C(8, 4).
Eq. of AB, BC and CA are respectively :

OR

Find the area enclosed between the parabola 4y = 3x2 and the line 3x – 2y + 12 = 0.
Solution: We have y = 3x2/4 …(i) and, 3x – 2y + 12 = 0 ..(ii)
Solving these curves simultaneously, we’ve :
3x-2 x 3x2/4 + 12 = 0  ⇒ x2 – 2x – 8 = 0
⇒ (x-4)(x+2) = 0  ∴ x = 4 , -2
So, points of intersections are (4,12) and, (–2, 3).

28. Find the particular solution of the differential equation (x-y) dy/dx = (x+2y) , given that y = 0 when x =1.
Solution: Given differential equation can be rewritten as dy/dx = x+2y/x-y

Q29. Find the coordinates of the point where the line through the points (3, –4, –5) and (2, –3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2, –3) and (0, 4, 3).
Solution: Equation of plane determined by the points (1, 2, 3), (4, 2, –3) and (0, 4, 3) :

On expanding along R1, we get : 2x + y + z = 7…(i)
Now equation of line through the points (3, –4, –5) and (2, –3, 1) is : x-3/-1 = y+4/1 = z+5/6 = λ
Coordinates of random point on line : P(-λ + 3, λ – 4, 6λ – 5) .
For the point of intersection of line and plane, point P must satisfy (i), so we get : λ = 2.
Hence the required point of intersection is : P(1, – 2, 7) .

OR

A variable plane which remains at a constant distance 3p from the origin cuts the axes at A, B, C. Show that the locus of the centroid of triangle ABC is 1/x2 + 1/y2 + 1/z2 = 1/p2 .
Solution: Let A(a, 0, 0), B(0, b, 0) and C(0, 0, c). So, equation of plane : x/a + y/b + z/c = 1 ..(i)