Class 12 Maths Important Questions

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Short Answer Type-I Questions (2 marks each)

Ques: Show that the relation R on IR defined as R = {(a, b) : (a ≤ b)}, is reflexive and transitive but not symmetric. 

Ans: Given a relation R = {{a, b): a ≤ b} on IR (the set of real numbers).

Reflexivity:

Since a ≤ a is true for all values of a ∈ IR.

(a, a) ∈ R ∀ a ∈ IR

Hence, the given relation is reflexive.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R be any arbitrary elements.

Then, we have a ≤ b and b ≤ c

⇒ a ≤ b ≤ c

⇒ a ≤ c

⇒ (a,c) ∈ R

Hence, the given relation is transitive.

Symmetricity:

Note - (2,3) ∈ R as 2 < 3

but (3, 2) ∉ R as 3 ≮ 2

Hence, the given relation is not symmetric.

Hence proved.

Ques: Write the value of tan^-1 (√3) – cot^-1(- √3)

Ans: We have, tan^-1 (√3) – cot^-1(- √3)

= tan^-1 (√3) –{ π-cot^-1( √3)} [∵ cot^-1(- x) = π - cot^-1(x) x ∈ R]

= tan^-1 (√3) – π+cot^-1( √3)

= {tan^-1 (√3-cot^-1( √3)}- π

= π² – π = – π² [∵ tan^-1 (x) – cot^-1(x) =π² x ∈ R]

Which is the required principal value.

Ques: What are the points of discontinuity of the function f(x) = \(\frac{(x-3) + (x+4)}{(x-5)(x-9)}\)

Ans: From the given function, it can be noticed that the function is not defined at the values x = 5,9

Being a polynomial, we can say that the function is continuous for the points x = 3 and x = -4. Hence, the points of discontinuity will be situated at the points x = 5 and x = 9. 

Ques: Solve the initial value problem x\(\frac{dy}{dx}\) + 1 = 0; y(-1) = 0

Ans: We have x\(\frac{dy}{dx}\) + 1 = 0

\(\frac{dy}{dx}\) = \(\frac{-1}{x}\)

dy = \(\frac{-1}{x}\) dx [separating variables]

Integrating both sides we get,

∫dy =∫ \(\frac{-1}{x}\)dx

y = -log|x| + C …(i)

It is given that y(-1) = 0 

So, 0 = -log|-1| +C 

C = 0

Substituting the value of C in (i) ,we get 

Y = -log|x|

Hence, y = -log|x| is the solution to the given.

Ques: A fair coin and an unbiased die are tossed. Let Abe the event head appear on the coin and B be the event 3 on the die. Check whether A and B are independent events or not.

Ans: Let A be the sample space of a given experiment.

S = {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

A: Head appear on the coin

B: 3 appear on the dice

A = {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}

B = {(H,3),(T,3)}

A B = {( H, 3)}

P(A) = \(\frac{6}{12}= \frac{1}{2}\), P(B) = \(\frac{2}{12}= \frac{1}{6}\)

P(A B) = \(\frac{1}{12}\)

P(A).P(B) = \( \frac{1}{2}\).\( \frac{1}{6}\) = \(\frac{1}{12}\) = P(A B)

Hence A and B are independent events.

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Short Answer Type-II Questions (3 marks each)

Ques: Write the set of points of continuity of g(x)= |x-1| +| x+1| . 

Ans: Observing the breakpoints of the piecewise function,

It is observed that the function is a polynomial at all the breakpoints, polynomials are

continuous for their entire domain and hence the points of continuity of g(x) is the whole real no set.

Ques: Evaluate ∫ e^2xsin3xdx

Ans: Let I = ∫ e^2xsin3xdx

Using integration by parts, 

I = e^2x(- \(\frac{cos 3x}{3}\)) -∫2e^2x (-\(\frac{cos 3x}{3}\)) dx

I = - \(\frac{1}{3}\)e^2xcos3x + \(\frac{2}{3}\)∫2e^2xcos 3x dx

I = - \(\frac{1}{3}\)e^2xcos3x + \(\frac{2}{3}\){e^2x \(\frac{sin 3x}{3}\) - ∫2e^2x\(\frac{sin 3x}{3}\)dx}

I = - \(\frac{1}{3}\)e^2xcos3x + \(\frac{2}{9}\)e^2xsin 3x - \(\frac{4}{9}\)∫e^2xsin 3x dx

I = - \(\frac{1}{3}\)e^2xcos3x + \(\frac{2}{9}\)e^2xsin 3x - \(\frac{4}{9}\)I

I + \(\frac{4}{9}\)I = \(\frac{1}{9}\)e^2x (2 sin 3x - 3 cos 3x)

\(\frac{13}{9}\)I = \(\frac{1}{9}\)e^2x (2sin 3x - 3 cos 3x)

I = \(\frac{1}{13}\)e^2x(2sin 3x - 3 cos 3x) + C

Ques: Minimize Z= 13x - 15y, subject to the constraints:

x+y≤7, 2x - 3y +6 ≥ 0, x ≥ 0, y ≥ 0

Ans: Consider x +y = 7

When x = 0, then y = 7 and

when y = 0, then x = 7

So, A(O, 7) and B(7, 0) are the points on line, x + y = 7

Consider 2x -3y + 6 = 0

When x = 0, then y = 2 and when y = 0, then x = - 3, So C(O, 2) and D(-3, 0) are the

points on line 2x -3y + 6 = 0

Also, we have x > 0 and v > 0.

The feasible region OBEC is bounded, so, minimum value will obtain at a comer I

point of this feasible region.

Corner points are 0(0, 0), B(7, 0), E(3, 4) and C(O, 2)

Z = 13x - 15y

At 0(0, 0), Z = 0

At B(7, 0), Z = 13(7) - 15(0) = 91

At E(3 , 4), Z = 13(3) - 15(4) = - 21

At C(O, 2), Z = 13(0) - 15(2)

= -30 (minimum)

Hence , the minimum value is -30 at the point (0, 2).

Ques: Show that the lines r = 3i + 2j - 4k + λ ( i+ 2j + 2k) and r = 5i - 2j + μ (3i + 2j +6k) are intersecting. Hence, find their point of intersection.

Ans: The position vectors of two arbitrary points on the given lines are 3i + 2j - 4k + λ ( i+ 2j + 2k) = (3 + λ )i + (2 + 2λ)j + ( 2λ - 4)k and 5i - 2j + μ (3i + 2j +6k) = (5 + 3μ)i + ( -2 + 2μ)j + 6μk

If the lines intersect, they have a common point. So, for some values of μ and λ we must have, 

3+λ = 5 + 3μ …..(i)
2+2λ = -2+ 2μ….(ii)

2λ – 4 = 6μ…..(iii)

Solving (i) and (ii) , we get 

λ = -4 

μ = -2 

Substituting the values λ = -4 

μ = -2 in (iii), we get LHS = 2λ - 4 

μ =2(-4) - 4

μ = -12

RHS = 6μ = 6(-2) = -12

LHS = RHS

Since λ = -4 and μ = -2 satisfy (iii), the lines intersect μ = -2 in the second line , we get 

r = 5i - 2j - 6i -4j - 12k = -i -6j- 12k as the position vector of the point of intersection. 

Thus, the coordinates of the point of intersection are (-1, -6, -12).

Ques: Determine the value of the constant ‘k’ so that the function f(x)= { \(\frac{kx}{|x|}\), if x < 0 and 3 if x ≥ 0 is continuous at x = 0. 

Ans: Let, f(x)= { \(\frac{kx}{|x|}\), if x < 0 and 3 if x≥0

is continuous at x = 0. 

Then lim x -> 0⁺f(x)= x -> 0^-f(x)= f(0)

lim h -> 0f(0 +h)= lim h -> 0f(0 - h) = f(0)

3 = lim h -> 0\( \frac{k(-h)}{|-h|}\) = 3

lim h -> 0\(\frac{-kh}{h}\) = 3

lim h ->0 (-k) = 3 

k = -3

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Long Answer Type Questions (4 marks each)

Ques: Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation. 

Ans: The given relation is R = {(a, b): a, b ∈ Z and a – b is divisible by 5}.To prove R is an equivalence relation, we have to prove R is reflexive, symmetric and transitive.

Reflexive:

As for any x ∈ Z, we have x – x = 0, which is divisible by 5. ⇒ (x – x) is divisible by 5.

⇒ (x, x) ∈ R, V x ∈ Z Therefore, R is reflexive.

Symmetric:

Let (x, y) ∈ R, where x, y ∈ Z. ⇒ (x -y) is divisible by 5. [by definition of R]

⇒ x – y = 5B for some B Z. ⇒ y-x = 5(-B) ⇒ (y- x) is also divisible by 5.

⇒ (y, x) ∈ R

Therefore, R is symmetric.

Transitive:

Let (x, y) ∈ R, where x, y ∈ Z.

⇒ (x – y) is divisible by 5.

⇒ x – y = 5B for some B ∈ Z Again, let (y, z) ∈ R, where y, z ∈ Z.

⇒ (y – 1) is divisible by 5.

⇒ y – z = 5C for some C ∈ Z.

Now, (x – y) + (y – 2) = 5B + 5C

⇒ x – z = 5(B + C)

⇒ (x – z) is divisible by 5 for some (B + C) ∈ Z

⇒ (x, z) ∈ R

Therefore, R is transitive.

Thus, R is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

Ques: Solve the system of linear equations by Cramer’s rule:

2y - 3z = 0, x + 3y = -4, 3x + 4y = 3

Ans: Given, 2y - 3z = 0, x + 3y = -4, 3x + 4y = 3 Then the determinant will be 

D= |0 2 -3 

1 3 0

3 4 0|

Expanding along R1, = 0(0) - 2(0) - 3(-5) = 15

Also D1= |0 2 -3 

-4 3 0

3 4 0|

Expanding along R1, = 0(0)- 2(0) - 3(-25) = 75

Again D2= |0 0 -3 

1 -4 0

3 3 0|

Expanding along R1, = 0(0)- 0(0) - 3(155) = -45

Again D3= |0 2 0 

1 3 -4

3 4 3|

Expanding along R1, = 0(25)- 2(15) + 0(-5) = -30

Now,x = \(\frac{D1}{D}\)= \(\frac{75}{15}\)=5

y = \(\frac{D2}{D}\) = – \(\frac{45}{15}\)=-3

z = \(\frac{D3}{D}\) = – \(\frac{30}{15}\)=- 2

Hence x = 5, y = -3, z = -2

Ques: To hire a marketing manager , it’s important to find a way to properly assess candidates who can bring radical changes and has leadership experience.Ajay, Ramesh and Ravi attend the interview for the post of a marketing manager.For Ajay, Ramesh and Ravi the chances of being selected as the manager of a firm are in the ratio 4 : 1 : 2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are, 0.3, 0.8 and 0.5. If the change does take place,

(i) Find the probability that it is due to the appointment of Ajay(A).

(ii) Find the probability that it is due to the appointment of Ramesh(B).

Ans: (i) Let E1: Ajay(A) is selected , E2: Ramesh (B) is selected, E3: Ravi(C) is selected .

Let A be the event of making a change

P(E1) = \(\frac{4}{7}\), P(E2) = \(\frac{1}{7}\),P(E3)= \(\frac{2}{7}\)

P(A/E1)= 0.3, P(A/E2) = 0.8, P(A/E3) = 0.5

P(E1/A) = \(\frac{P(E1).P(A/E1)}{P(E1).P(A/E1) +P(E2).P(A/E2) + P(E3).P(A/E3)}\)

 =\(\frac{\frac{4}{7}(0.3)}{ \frac{4}{7}(0.3) + \frac{1}{7}(0.8) + \frac{2}{7}(0.5)}\)

= \(\frac{\frac{1.2}{7}}{ \frac{1.2}{7} +\frac{ 0.8}{7} + \frac{1}{7}}\)

= \(\frac{\frac{1.2}{7}}{\frac{3}{7}}\)

= \(\frac{1.2}{ 3} = \frac{12}{ 30}= \frac{2}{5}\)

(ii) Let E1: Ajay(A) is selected , E2: Ramesh (B) is selected, E3: Ravi(C) is selected .

Let A be the event of making a change

P(E1) = \(\frac{4}{7}\), P(E2) = \(\frac{1}{7}\),P(E3)= \(\frac{2}{7}\)

P(A/E1)= 0.3, P(A/E2) = 0.8, P(A/E3) = 0.5

P(E2/A) = \(\frac{P(E2).P(A/E2)}{P(E1).P(A/E1) +P(E2).P(A/E2) + P(E3).P(A/E3)}\)

 =\(\frac{\frac{1}{7}(0.8)}{ \frac{4}{7}(0.3) + \frac{1}{7}(0.8) +\frac{2}{7}(0.5)}\)

= \(\frac{\frac{0.8}{7}}{\frac{1.2}{7} + \frac{0.8}{7} + \frac{1}{7}}\)

= \(\frac{\frac{0.8}{7}}{\frac{ 3}{7}} = \frac{0.8}{3} = \frac{8}{30}= \frac{4}{ 15}\)

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