Plus Two Maths Model Question Papers Paper 3 is part of Plus Two Maths Previous Year Question Papers and Answers. Here we have given Plus Two Maths Model Question Papers Paper 3.

## Plus Two Maths Model Question Papers Paper 3

Board |
SCERT |

Class |
Plus Two |

Subject |
Maths |

Category |
Plus Two Previous Year Question Papers |

**Time : 2 1/2 Hours**

**Cool off time : 15 Minutes**

**Maximum : 80 Score**

**General Instructions to Candidates :**

- There is a ‘Cool off time’ of 15 minutes in addition to the writing time.
- Use the ‘Cool off time’ to get familiar with questions and to plan your answers.
- Read questions carefully before you answering.
- Read the instructions careully.
- When you select a question, all the sub-questions must be answered from the same question itself.
- Calculations, figures and graphs should be shown in the answer sheet itself.
- Malayalam version of the questions is also provided.
- Give equations wherever necessary.
- Electronic devices except non programmable calculators are not allowed in the Examination Hall.

**QUESTIONS**

**Question 1 to 7 carry 3 scores each. Answer any six questions only
**

Question 1.

a. Let f : R → R be a function defined by f (x) = x^{3} + 5. Then f^{1} (x) is

i. (x+5)^{1/3}

ii. (x-5)^{1/3}

iii. (5-x)^{1/3}

iv. 5-x

b. Let * be a binary operation defined on Q

a*b = a-b + ab. Check whether

i. It is commutative?

ii. Is * associative ?

Question 2.

Question 3.

Question 4.

Question 5.

Prove that the function f given by f (x)= log sin x is strictly increasing on ( 0,\(\frac { \pi }{ 2 } \))

Question 6.

Question 7.

**Question 8 to 17 carry 4 scores each. Answer any eight questions only
**

Question 8.

a. Show that the relation R in set of real numbers defined as R = {(a,b): a < b^{2}} is neither reflexive nor symmetric not transitive.

b. Show that the operation * on Q, defined by a*b = a+b-ab is commutative, and ex-its and identity elements find it.

Question 9.

a. The principal value of the expression cos-1 cos (680) is …………..

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

a. Find the distance between the planes x-y + z-5 = 0 and 2x-2y + 2z = 0.

b. Write the vector equation corresponding to Cartesian equation of a line

Question 16.

Find the shortest distance between the lines

Question 17.

**Question 18 to 25 carry 6 scores each. Answer any 5 questions only
**

Question 18.

Question 19.

a. Use differential to approximate (0.999)^{1/10}

b. A window is in the form of rectangle sur-mounted by a semicircular opening. The total perimeter of the windows is 1 Om. Find the dimensions of the window to admit maximum light through the whole opening.

Question 20.

Find the area lying above x axis and included between the circle x^{2} + y^{2} = 8x and inside of the parabola y^{2} = 4x. Also draw a neat diagram.

Question 21.

Evaluate:

Question 22.

Minimize and maximize Z = 5x + 10 y subject to x + 2y < 120, x + y > 60, x – 2y > 0, x, y > 0

a. Draw the feasible region

b. Find the comer points

c. Find the maximum and minimum profit.

Question 23.

a. Find the distance of the point (-1, -5, -10) from the point of intersection of the line

Question 24.

a. Two numbers are selected at random (with-out replacement) from the Pt six positive integers.

Let X denote the larger of the two numbers obtained. Find E(X) and Var(X)

b. A card from a pack of 52 cards is lost from the remaining X cards of the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade.

**ANSWERS**

Answer 1.

a. ii

b. a * b = a – b + ab

b*a = b- a + ba = b – a + ab

∴ * is not commutative.

(a,b) * c = d * c = d- c + dc

= a- b + ab – c + ac – bc + abc

= a- b + ab – c + ac – bc + abc

= a – b – c + ab – bc + ca + abc

a * (b * c) = a * d = a- d + ad

= a – (b – c + be) + a (b – c + be)

= a- b + c – bc + ab – ac + abc

= a – b + c ab – be + ca + abc

∴ * is not associative.

Answer 2.

Answer 3.

We are giving that

Answer 4.

Answer 5.

Answer 6.

Answer 7.

Answer 8.

a. R = {(a,b): a < b^{2}}

Relation R is defined in the set of real numbers.

i. Reflexive

Consider a ∈ R

If a ∈ R ⇒ a < a^{2} which is false (a, a) ∈ R

R is not reflexive.

ii. Symmetric

Let a,b ∈ R and

(a,b) ∈ R ⇒ a < b^{2} and b < a^{2},

which is false ⇒ (a,b) ∈ R, but (b,a) ∈ R

∴ R is not symmetric.

iii. Transitive

Let a, b, c ∈ R

Answer 9.

Answer 10.

Answer 11.

Answer 12.

Answer 13.

Answer 14.

Answer 15.

Answer 16.

Answer 17.

Answer 18.

Answer 19.

Answer 20.

The given equation of the circle x^{2} + y^{2} = 8x can be expressed as (x – 4)^{2} + y^{2} = 16. Thus, the centre of the circle is (4,0) and radius is 4. Its intersection with the parabola y^{2} = 4x gives

x^{2} + 4x = 8x

or

x^{2} – 4x = 0

or

x(x-4) = 0

or

x = 0, x = 4

Thus, the point of intersection of these two curves are 0 (0,0) and P (4,4) above the x-axis.

From the above the required area of the region OPQCO included between these two curves above x-axis is

= (area of the region OCPO)

+ (area of the region PCQP)

Answer 21.

Answer 22.

a. The feasible region determine! by the constraints,

x + 2y < 120, x + y > 60, x – 2y > 0,

x > 0 and y > 0

is as follows.

b. The comer points of the feasible are region are A(60,0), C(60,30) and D (40,20).

The values of Z at these comer points are as follows.

Corner point | Z=5x + 10y | |

A (60,0) | 300 | → Minimum |

B (120,0) | 600 | → Maximum |

C (60,30) | 600 | → Maximum |

D (40,20) | 600 |

a. The minimum value of Z is 300 at (60,0) and the maximum value of Z is 600 at all the points on the line segment joining (120,0) and (60,30)

Answer 23.

Answer 24.

a. The two positive integers can be sele-cted from the fist six positive integers without replacement in 6 x 5 = 30 ways.

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2,3,4,5 or 6.

For X=2, the possible observations are (1,1)and(2,1)

∴ P (x = 2) = \(\frac { 2 }{ 30 } \) = \(\frac { 1 }{ 15 } \)

For X = 3 the possible observations are (1,3), (2,3), (3,1) and (3,2).

∴ p (x = 3) = \(\frac { 4 }{ 30 } \) = \(\frac { 2 }{ 15 } \)

For x = 4 the possible observations are

(1,4), (2,4), (3,4), (4,3), (4,2) and (4,1).

∴ p (x = 4) = \(\frac { 6 }{ 30 } \) = \(\frac { 1 }{ 5 } \)

For X = 5, the possible observations are (1.5) , (2,5), (3,5), (4,5), (5,4), (5,3), (5,2) and (5,1).

∴ p (x = 5) = \(\frac { 8 }{ 30 } \) = \(\frac { 4 }{ 15 } \)

For X = 6, the possible observations are (1.6), (2,6), (3,6), (4,6), (5,6), (6,4), (6,3), (6,2) and (6,1)

∴ p (x = 6) = \(\frac { 10 }{ 30 } \) = \(\frac { 1 }{ 3 } \)

Therefore, the required probability distribution is as follows.

b. Let E and E, be the respective events of choosing a spade card and a card which is not spade. Out of 52 cards, 13 cards are spade and 39 cards are not spade.

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