APEX INSTITUTE : Class - XII Maths Sample Paper

Instructions

1. Questions 1 to 10 carry 1 mark each.
2. Questions 11 to 22 carry 4 marks each.
3. Questions 23 to 29 carry 6 marks each.

1. Evaluate:

Sol. Let I =

Taking x as first function and sin 3x as second function and integrating by parts, we obtain

2. Find the principal value of tan⁻¹ (−1).

Sol. Let tan⁻¹ (−1) = y. Then, tan y = −1 = −tan(π/4) = tan(π/4)
We know that the range of the principal value branch of tan⁻¹ is

Therefore, the principal value of tan⁻¹ (−1) is (-π)/4.

3. Show that * : R × R → R given by a * b → a + 2b is not associative.

Sol. The operation * is not associative, since
(8 * 5) * 3 = (8 + 10) * 3 = (8 + 10) + 6 = 24,
While 8 * (5 * 3) = 8 * (5 + 6) = 8 * 11 = 8 + 22 = 30.

4. If A and B are symmetric matrices of the same order, prove that AB + BA is symmetric.

Sol. Let P = AB + BA
            P′ = (AB + BA)′
   = (AB)′ + (BA)′
           = B′A′ + A′B′
       = BA + AB [ A′=A,B′ = B]
       = AB + BA
           = P.

5. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector

Sol. It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is

It is known that the line which passes through point A and parallel to

is given by

is a constant.

This is the required equation of the line.

6. Evaluate:

Sol.

As sin⁷(−x) = (sin (−x))⁷ = (−sin x)⁷ = −sin⁷x, therefore, sin⁷x is an odd function.

It is known that, if f(x) is an odd function, then

7. If matrix A = (1,2,3) , write (AA′) , where A′ is the transpose of matrix A.

Sol. AA′ = (1 2 3)

= (1×1 + 2×2 + 3×3)

= (1 + 4 + 9)

= 14.

8. Find the position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).

Sol. The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, – 2) is given by,

9. If

, then for what value of α is A an identity matrix?

Sol.

If A is an identity matrix, then:

Thus, for α = 0°, A is an identity matrix.

10. The line y = mx + 1 is a tangent to the curve y² = 4x .Find the value of m.

Sol. The equation of the tangent to the given curve is y = mx + 1.

Now, substituting y = mx + 1 in y² = 4x, we get:

⇒ (mx + 1)² = 4x
⇒ m²x² + 1 + 2mx – 4x = 0
⇒ m²x² + x (2m – 4) + 1 =0 …(i)

Since a tangent touches the curve at one point, the roots of equation (i) must be equal.

Therefore, we have:
Discriminant = 0

(2m – 4)²- 4(m²)(1) =
⇒ 4m² + 16 - 16m – 4m² = 0
⇒ 16 – 16m = 0
⇒ m = 1

Hence, the required value of m is 1.

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11. Find

Sol. The given relationship is

Differentiating this relationship with respect to x, we obtain

12. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Sol. Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen that (1, 1), (2, 2), (3, 3) ∉R.
∴ R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R
However, (1, 1) ∉ R
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.

13. Solve the following differential equation: (x² − y²) dx + 2 xy dy = 0.

Sol. Given that y = 1 when x = 1

It is a homogeneous differential equation.

Substituting (2) and (3) in (1), we get:

Integrating both sides, we get:

It is given that when x = 1, y = 1
(1)²+ (1)² = C (1)
⇒ C = 2

Thus, the required equation is y² + x² = 2x.

14. Prove

Sol. Let x = sinθ. Then, sin^-1x = θ
We have,
R.H.S. =

= sin^-1 (sin 3θ)
= 3θ
= 3 sin^-1 x
= L.H.S.

15. Evaluate:

Sol. The given integral is

I =

[as sin 2x = 2sin x cos x]

On putting (a + b cos x) = t ⇒ −b sin x dx = dt, we obtain

where C is a constant

16. Let A be a nonsingular square matrix of order 3 × 3. Then find

Sol. We know that,

17. By using properties of determinants, show that:

Sol.

Taking out common factors a, b, and c from R₁, R₂, and R₃respectively, we have:

Applying R₂ → R₂ − R₁and R₃ → R₃ − R₁, we have:

Applying C₁ → aC₁, C₂→ bC_2,and C₃ → cC₃, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

18. Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.

Sol. he area in the first quadrant bounded by y = 4x², x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as

19. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Sol. Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
∴ Marginal Revenue (MR)

= 13(2x) + 26 = 26x + 26

When x = 7,
MR = 26(7) + 26 = 182 + 26 = 208
Hence, the required marginal revenue is Rs 208.

20. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

Sol. The normal vector is,

It is known that the equation of the plane with position vector

is given by,

This is the vector equation of the required plane.

21. On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Sol. The repeated guessing of correct answers from multiple choice questions are Bernoulli trials.

Let X represent the number of correct answers by guessing in the set of multiple choice questions.

Let p(a correct answer) =

Clearly, X has a binomial distribution with n = 5 and

P (guessing more than 4 correct answers)

= P(X ≥ 4)

= P(X = 4) + P(X = 5)

Thus, the required probability is

22. If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

Sol. The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x₁, y₁ z₁) is

where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

23. If a young man rides his motorcycle at 25 km/hour, he had to spend Rs 2 per km on petrol. If he rides at a faster speed of 40 km/hour, the petrol cost increases at Rs 5 per km. He has Rs 100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this as an LPP and solve it graphically.

Sol. Let the young man ride with the speed 25 km/hr for x hours and with the speed 40 km/hr for y hours respectively.

Money spent at speed 25 km/hr = 50x
And Money spent at speed 40 km/hr = 200y

Maximum distance :
Z = 25x + 40y ….(1)

subject to the constraints
x + y ≤ 1 ….(2)
50x + 200y ≤ 100 ….(3)
x , y ≥ 0 ….(4)

Graph the inequalities (2) to (4). The feasible region (shaded) is shown as below:

Corner Point	Z = 25x + 40y
A (0, 1/2)	20
B (2/3, 1/3)	30Maximum
C(1, 0)	25

Thus, the maximum distance he can travel is30 km.

24. Evaluate:

Sol. We have

Dividing numerator and denominator by x²

25. Solve system of linear equations, using matrix method.

x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Sol. The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

26. A family has 2 children. Find the probability that both are boys, if it is known that

(i) at least one of the children in a boy,
(ii) the elder child is a boy.

Sol. Let b stands for boy and g for girl.
The sample space of the experiment is

(i) Let E and F denote the following events:
E: both the children are boys
F: at least one of the children is a boy

Then, E = {(b, b)} and F = {(b, b), (g, b), (b, g)}

(ii) Let P and Q denote the following events:
P: both the children are boys
Q: elder child is a boy

Then, P = {(b, b)} and Q = {(b, b,), (g, b)}

27. Find the equation of the plane passing through the points (1, 2, 3) and (0, −1, 0) and parallel to the line

Sol. Equation of any plane passing through the point (1, 2, 3) is given by:

a (x − 1) + b (y − 2) + c (z − 3) = 0 … (1)

This plane is parallel to the line

∴ a × 2 + b × 3 + c × (−3) = 0
⇒ 2a + 3b −3c = 0 … (2)

Also, plane (1) passes through the point (0, −1, 0). Therefore,

a (0 − 1) + b (−1 − 2) + c (0 − 3) = 0
⇒ −a − 3b − 3c = 0
⇒ a + 3b + 3c = 0 … (3)

Solving equations (2) and (3), we obtain

Hence, the required equation of the plane is given by,

6(x − 1) −3 (y − 2) + 1 (z − 3) = 0
⇒ 6x − 6 −3y + 6 + z − 3 = 0
⇒ 6x −3y + z − 3 = 0

28. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Sol. The volume of a sphere (V) with radius (r) is given by,