# Class 12 Mathematics Sample Paper With Solutions Set J

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

SECTION – A

1. If A and B are square matrices, each of order 2 such that |A| = 3 and |B| = −2 , then write the value of |3AB|.
Sol. |3AB| = 32 AB| = 9|A||B| = 9×3×(−2) = −54 .

2. Write the derivative of |x – 5| at x = 2.
Sol.

3. Find : ∫ax .exdx .
Sol.

4. Find the cosine of the angle which the vector √2î + ĵ+ k̂ makes with y-axis.
Sol.

SECTION – B

5.

Sol.

6. Evaluate :

Sol.

7. Find the inverse (A–1) of the matrix

using elementary operations.
Sol. Using elementary row operations, we have : A = IA

8. If the curves y = 2ex and y = ae−x intersect orthogonally, find the value of a.
Sol. We have y = 2ex and y = ae−x           ∴ dy/dx = 2ex and dy/dx = ae −x
As these curves cut orthogonally, so (2ex )×(−ae−x ) = −1       ∴  a =1/2 .

9. Find the points on the curve y =12x − x3 , where the tangent drawn is parallel to x-axis.
Sol. Given curves is y =12x − x3
Let the point be P(a, b). Clearly we have b =12a − a3…(i)
Now dy/dx = 12 − 3x2   ⇒ dy/dx]at P    = 12a − 3a= 0    (As any line parallel to x-axis has slope as 0.)
∴  a = 2,−2. So, by (i), b =16,−16 .
Hence the required points are (2, 16) and (–2, –16).

10. Find :

Sol.

11. Find a vector perpendicular to the plane of ABC, where A, B and C are points (3, –1, 2), (1, –1, –3) and (4, –3, 1) respectively.
Sol. Let a = A̅B̅ = (î − ĵ− 3k̂) − (3î − ĵ+ 2k̂) = −2î −5k̂ and,
b = A̅C̅ = (4î − 3ĵ+ k̂) − (3î − ĵ+ 2k̂) = î − 2ĵ− k̂

12. Eight cards numbered 1 to 8 (one number on one card) are placed in a box, mixed up
thoroughly and then a card is drawn randomly. If it is known that the number on the drawn card is more than 2, then find the probability that it is an odd number.
Sol. Let E : the number on card is odd, F : the number on the card is more than 2.
∴ S = {1, 2,…,8}, E = {1,3,5,7}, F ={3, 4,5,6,7,8}    ∴ E ∩ F ={3,5,7}

SECTION – C

13. Using properties of determinants, prove that :

Sol.

OR

Sol.

By equality of matrices, we get :
a + 4x = −7, 2a + 5x = −8, 3a + 6x = −9, c + 4d = 2, 2c + 5d = 4, 3c + 6d = 6 .
On solving these equations, we get : a =1, x = −2, c = 2, d = 0 .

14. Find the value of k for which the given function f (x) is continuous at x = 0 ,

Sol. Here f (0) = k…(i)

∴ By (i), (ii) and (iii), we get : k = 4.

OR

If xy = ex−y  show that dy/dx = log x/(1 + log x)2
Sol. We have xy = ex−y                          (Taking logarithm on both sides)
⇒ log xy = log ex−y       ⇒ ylog x = (x − y) log e = x − y      ⇒ y(1+ log x) = x           (∵log e =1

15. If x = a(cos t + t sin t) and y = a(sin t − t cos t) , find d2y/dx.
Sol. Given x = a (cos t + t sin t) , y = a (sin t − t cos t )

16. Find :

Sol.

OR

Find :

Sol.

17. Evaluate :

Sol.

18. Find the general solution of the following differential equation :
ydx + x log (y/x) dy − 2xdy = 0
Sol.

19. For the differential equation xy dy/dx = (x + 2)(y + 2), find the solution curve passing through the point (1, –1).
Sol.

⇒ y − 2log |y + 2| = x + 2log |x| + c…(i)
As (i) passes through (1, –1) so, −1− 2log |−1+ 2| =1+ 2log |1| + c i.e., c = −2 .
Hence the required curve is y − 2log |y + 2| = x + 2log |x| − 2 .

20. If the scalar product of î + ĵ+ k̂ with a unit vector along the sum of vectors 2î + 4ĵ−5k̂ and λî + 2ĵ+3k̂ is equal to 1, find the value of λ .
Sol. Let a̅ = 2î + 4ĵ−5k̂ + λî + 2ĵ+3k̂ = (2 + λ)î + 6ĵ− 2k̂

⇒ 36 + λ2 +12λ = 4 + λ2 + 4λ + 40            ⇒ 8λ = 8                  ∴ λ =1.

21. Find the coordinates of the foot of perpendicular drawn from the point P(1, 8, 4) to the line joining the points A(0, –1, 3) and B(2, –3, –1). Also find the length of this perpendicular.
Sol.

∴ Coordinates of random point on this line is Q(2λ,−2λ −1,−4λ + 3) .
The d.r.’s of PQ are 2λ −1,−2λ −9,−4λ −1.
As PQ ⊥ AB so, 2(2λ −1) − 2(−2λ −9) − 4(−4λ −1) = 0

22. Find the probability distribution of number of doublets in two throws of a pair of dice. Hence find the mean of the distribution.
Sol. Let X : number of doublets in two throws of a pair of dice.           ∴ X = 0,1, 2.

23. A and B throw a pair of dice alternatively till one of them gets a sum of 5, of the numbers on the two dice and wins the game. Find their respective probabilities of winning, if A starts the game.
Sol. Sum of 5 is obtained when we get following outcomes on the throw of pair of dice :
(1, 4), (4, 1), (2, 3), (3, 2)

SECTION – D

24. Let R be the relation defined in the set A ={1, 2,3, 4,5,6,7,8,9} by R = {(x, y) : x, y ∈ A, x and y are either both odd or both even}. Show that R is an equivalence relation. Write all the equivalence classes of set A.
Sol. We’ve R = {(x, y) : x, y ∈ A, x and y are either both odd or both even} and,
A ={1, 2,3, 4,5,6,7,8,9}.
Reflexivity : Let any element a ∈ A, both a and a must be either odd or even, so that (a, a) ∈ R.
Symmetry : Let (a, b) ∈ R ⇒ both a and b must be either odd or even ⇒ (b, a) ∈ R. So, R is symmetric.
Transitivity : Let (a, b) ∈ R and (b, c) ∈ R ⇒ all elements a, b, c, must be either even or odd
simultaneously ⇒ (a, c) ∈ R. Hence, R is a transitive relation.
Since the relation R is reflexive, symmetric and transitive so, it is an equivalence relation.
Now let (1, x) ∈ R , clearly x will be odd. Hence  = {1, 3, 5, 7, 9}.
Similarly  =  =  =  = {1, 3, 5, 7, 9}.
Also let (2, y) ∈ R , clearly y will be even. Hence  = {2, 4, 6, 8}.
Similarly  =  =  =  = {2, 4, 6, 8}.

OR

Let A be the set of all real numbers except –1 and let * be a binary operation on A
defined by a * b = a + b + ab, ∀ a, b ∈ A. Prove that (i) * is commutative and associative, and (ii) number 0 is its identity element.
Sol. We’ve a * b = a + b + ab, ∀ a, b ∈ A, where A is the set of real numbers except –1.
(i) Commutativity : Let a,b ∈ A . As a *b = a + b + ab = b + a + ba = b*a , so * is commutative.
Associativity : Let Let a, b,c ∈ A.
Now (a *b)*c = (a + b + ab)*c = a + b + c + ab + bc + ca + abc…(A)
Also, a *(b*c) = a *(b + c + bc) = a + b + c + ab + bc + ca + abc…(B)
By (A) and (B), (a *b)*c = a *(b*c) . Hence * is associative.
(ii) Let e be the identity element.
Then, a *e = a = e*a for all a ∈ A = R −{−1}    ⇒ a *e = a and e*a = a
⇒ a + e + ae = a and e + a + ea = a   ⇒ e (1+ a) = 0 and e (1+ a) = 0
∴ e = 0              [∵ a ∈ R −{−1} ⇒ a ≠ −1 ⇒ 1+ a ≠ 0]
Hence 0 is the identity element for * defined in set A = R–{–1}.

25. Solve the following system of equations, using matrix method :
5x − y + z = 4, 3x + 2y −5z = 2, x + 3y − 2z = 5.
Sol. We have 5x − y + z = 4, 3x + 2y −5z = 2, x + 3y − 2z = 5

By equality of matrices, we get : x =1, y = 2, z =1.

26. Find the intervals in which the function f given by f (x) = sin 3x − cos3x, 0 < x < π , is (a) strictly increasing and, (b) strictly decreasing.
Sol.
Given function is f (x) = sin 3x − cos3x, 0 < x < π
⇒ f ‘(x) = 3cos3x + 3sin 3x = 3(cos3x + sin 3x)
For critical points, f ‘(x) = 3(cos3x + sin 3x) = 0              ⇒ tan 3x = −1

OR

Prove that the height of a solid cylinder of given surface and greatest volume is equal to
the diameter of its base.
Sol.

Now we know that S = 6πr2 (from above steps)
By (i), 2πr h + 2πr2 = 6πr2          ⇒2πr h = 4πr      ⇒2h = 4r          ∴h = 2r
Therefore, the height of a solid cylinder of given surface area and greatest volume is equal to the
diameter of its base.

27. If the area between the curve x = y2 and the line x = 4 is divided into two equal parts by the line x = a , find the value of a.
Sol.

OR

By the method of limit of sum, find the value of the following definite integral :

Sol.

28.

are coplanar. Hence find the equation of the plane containing these lines.
Sol.

Rewriting these lines in vector form, we get :
r̅ = î + 3ĵ+ λ(2î + 4ĵ− k̂) and r̅ = 4î + ĵ+ k̂ + λ(3î − 2ĵ+ k̂)
Here a̅1 = î + 3ĵ, b̅1 = 2î + 4ĵ− k̂, a̅2 = 4î + ĵ+ k̂, b̅2 = 3î − 2ĵ+ k̂.
As a̅2 − a̅1 = 3î − 2ĵ+ k̂ .

∴ Eq. of plane : r̅.(2î −5ĵ−16k̂) = (î + 3ĵ).(2î −5ĵ−16k̂) i.e., r̅.(2î − 5ĵ−16k̂) = −13
Therefore, 2x −5y −16z +13 = 0 is the required equation of plane.

29. If a class XII student aged 17 years, rides his motor cycle at 40 km/hr, the petrol cost is Rs 2 per km. If he rides at a speed of 70 km/hr, the petrol cost increases to Rs 7 per km. He has Rs 100 to spend on petrol and wishes to cover the maximum distance within one hour.
(i) Express the above as as an LPP.
(ii) What are the benefits of driving a vehicle at a slow speed?
(iii) Should a child below 18 years be allowed to drive a motorcycle? Give reasons.
Sol. (i) Let x and y represent the distance travelled by the student at speed of 40 km/hr and 70 km/hr
respectively.
To Maximize : Z = x + y (in km).
Subject to constraints : x/40 + y/70 ≤ 1, 2x + 7y ≤ 100, x≥0, y≥0
(ii) It saves petrol and hence saves money too.
(iii) No, because according to the law driving license is issued when a person is above the 18
years of age.