ISC Class 11 Maths Specimen Question Paper 2018 With Answers - Free Sample Paper PDF Download

ISC Class 11 Maths 2018 Specimen Question Paper with Solutions are available at BYJU’S, that can be accessed anytime by downloading a pdf. These solutions of ISC Maths Class 11 specimen papers are given by the expert faculty, hence they are accurate and efficient. All the important methods of solving maths problems are covered here for each of these questions.

Class 11 Students can download and practise board papers of ISC Class 11 Maths as a part of their examination preparation. This will help them in facing a variety of problems and methods of solving them so that they can improve their analytical skills to reach the right approach.

SECTION – A

Question 1

(i) Let f : R → R be a function defined by f(x) = (x – m)/ (x – n), where m ≠ n. Then show that f is one-one but not onto.

Solution:

Given,

f(x) = (x – m)/ (x – n), m ≠ n

f'(x) = [(x – n)(1) – (x – m)(1)]/ (x – n)²

= (m – n)/ (x – n)²

Thus, f'(x) < 0 or f'(x) > 0

Therefore, f is one one.

Let y = (x – m)/ (x – n)

xy – ny = x – m

xy – x = ny – m

x(y – 1) = ny – m

x = (ny – m)/ (y – 1)

Thus, y should not equal 1.

y ∈ R – {1}

Therefore, f is one-one but not onto.

Alternative method:

Let x₁, x₂ be the two elements in the domain R such that f(x₁) = f(x₂).

⇒ (x₁ – m)/ (x₁ – n) = (x₂ – m)/ (x₂ – n)

⇒ (x₁ – m)(x₂ – n) = (x₁ – n)(x₂ – m)

⇒ x₁x₂ – nx₁ – mx₂ + mn = x₁x₂ – mx₁ – nx₂ + mn

⇒ mx₁ – nx₁ = mx₂ – nx₂

⇒ (m – n)x₁ = (m – n)x₂

⇒ x₁ = x₂

Therefore, f is one one.

Let f(x) = y

y = (x – m)/ (x – n)

xy – ny = x – m

xy – x = ny – m

x(y – 1) = ny – m

x = (ny – m)/ (y – 1)

Thus, y should not equal 1.

y ∈ R – {1}

Therefore, f is one-one but not onto.

(ii) Find the domain and range of the function f(x) = [sin x].

Solution:

Given,

f(x) = [sin x]

Domain = (-∞, ∞)

or

Domain = {x: x ∈ R}

And

-1 ≤ sin x ≤ 1

Range = [-1, 1]

(iii) Find the square root of the complex number 11 – 60i.

Solution:

Given,

11 – 60i

Comparing with a + ib,

a = 11, b = -60 < 0

= ± [(36)½ – i (25)½]

= ± [6 – i5]

Therefore, the square root of 11 – 6i is ± (6 – i5).

(iv) For what value of k will the equations x² – kx – 21 = 0 and x² – 3kx + 35 = 0 have one common root.

Solution:

Given,

x² – kx – 21 = 0….(i)

x² – 3kx + 35 = 0….(ii)

Let α be the common root of (i) and (ii),

⇒ α² – kα – 21 = 0….(iii)

And

α² – 3kα + 35 = 0….(iv)

Subtracting (iv) from (iii),

α² – kα – 21 – (α² – 3kα + 35) = 0

(3α – α)k – 56 = 0

2αk = 56

α = 56/2k

α = 28/k

Substituting α = 28/k in (iii),

(28/k)² – k(28/k) – 21 = 0

(28/k)² – 28 – 21 = 0

(28/k)² = 49

⇒ 28/k = ±7

⇒ k = ±28/7

⇒ k = ±4

(v) In a ΔABC, show that ∑(b + c) cos A = 2s where, s = (a + b + c)/2

Solution:

∑(b + c) cos A = (b + c) cos A + (c + a) cos B + (a + b) cos C

= b cos A + c cos A + c cos B + a cos B + a cos C + b cos C

= (b cos A + a cos B) + (c cos A + a cos C) + (c cos B + b cos C)

= c + b + a…(i)

We know that,

In a ΔABC, s = (a + b + c)/2

2s = a + b + c….(ii)

From (i) and (ii),

∑(b + c) cos A = 2s

(vi) Find the number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together.

Solution:

Given,

Number of men = 6

Number of women = 5

The number of ways of 6 men can sit at a round table = (n – 1)! = (6 – 1)!= 5!

There will be six places between the 6 men.

Thus, 5 women can sit in 6P₅ ways.

Required number of ways = 5! × 6P₅

= 5! × 6!

(vii) Prove that sin 20° sin 40° sin 80° = √3/8.

Solution:

LHS = sin 20° sin 40° sin 80°

= (1/2) (2 sin 20° sin 40°) sin 80°

= (1/2) [cos (20° – 40°) – cos (20° + 40°)] sin 80°

= (1/2) [cos (-20°) – cos 60°] sin 80°

= (1/2) [cos 20° – (1/2)] sin 80

= (1/2) sin 80° cos 20° – (1/4) sin 80°

= (1/2) sin (90° – 10°) cos 20° – (1/4) sin 80°

= (1/4) [2 cos 10° cos 20°] – (1/4) sin 80°

= (1/4) [cos (10° + 20°) + cos (10° – 20°)] – (1/4) sin 80°]

= (1/4) [cos 30° + cos (-10°)] – (1/4) sin 80°

= (1/4) [cos 30° + cos 10°] – (1/4) sin 80°

= (1/4) cos 30° + (1/4) cos (90° – 80°) – (1/4) sin 80°

= (1/4) (√3/2) + (1/4) sin 80° – (1/4) sin 80°

= √3/8

= RHS

Hence proved.

(viii) If two dice are thrown simultaneously, find the probability of getting a sum of 7 or 11.

Solution:

Given,

Two dice are thrown simultaneously.

Thus, the total number of outcomes = n(S) = 36

Let A be the event of getting a sum of 7.

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

n(A) = 6

P(A) = n(A)/n(S) = 6/36

Let B be the event of getting a sum of 11.

B = {(5, 6), (6, 5)}

n(B) = 2

P(B) = n(B)/n(S) = 2/36

P(A U B) = P(A) + P(B)

= (6/36) + (2/36)

= 8/36

= 2/9

Hence, the probability of getting a sum of 7 or 11 is 2/9.

(ix)

Solution:

(x) Find the point on the curve y² = 4x, the tangent at which is parallel to the straight line y = 2x + 4.

Solution:

Given curve is:

y² = 4x….(i)

Equation of the straight line is y = 2x + 4

This is of the form y = mx + c,

Slope = m = 2

Differentiating the equation of given curve,

2y (dy/dx) = 4

dy/dx = (4/2y)

dy/dx = 2/y

We know that the slope of the tangent at a point is given by the value of the derivative of the equation of the curve at that point.

2/y = 2

⇒ y = 1

Substituting y = 1 in (i),

(1)² = 4x

⇒ x = 1/4

Hence, the required point is (¼, 1).

Question 2

Draw the graph of the function y = |x – 2| + |x – 3|.

Solution:

|x – 2| = x – 2 if x > 2, and |x – 2| = -x + 2 if x < 2

Similarly,

|x – 3| = x – 3 if x > 3, and |x – 3| = -x + 3 if x < 3.

If x < 2, then |x – 2| + |x – 3| = -2x + 5.

If x > 3, then |x – 2| + |x – 3| = 2x – 5.

Now consider another case:

When x is between the values of 2 and 3 then |x – 2| = x – 2 and |x – 3| = -x + 3.

⇒ |x – 2| + |x – 3| = (x – 2) + (-x + 3) = 1

Thus, for x values between 2 and 3, we get the equation y = 1, which is just a horizontal line.

Question 3

Prove that cot A + cot(60 + A) + cot(120 + A) = 3 cot 3A.

Solution:

LHS = cot A + cot (60 + A) + cot (120 + A)

= cot A + cot (60 + A) + cot [180 – (60 – A)]

= cot A + cot (60 + A) – cot (60 – A)

= cot A + [(cot 60 cot A – 1)/ (cot 60 + cot A)] – [(cot 60 cot A + 1)/ (cot A – cot 60)]

= cot A + {[(1/√3)cot A – 1]/ [(1/√3) + cot A]} – {[(1/√3) cot A + 1]/ [cot A – (1/√3)]}

= cot A + [(cot A – √3)/ (√3 cot A + 1)] – [(cot A + √3)/ (√3 cot A – 1)]

= cot A + [(cot A – √3)(√3 cot A – 1) – (cot A + √3)(√3 cot A + 1)]/ [(√3 cot A + 1)(√3 cot A – 1)]

= cot A + [(√3 cot²A – cot A – 3 cot A + √3 – √3 cot²A – cot A – 3 cot A – √3)/ (3 cot²A – 1)

= cot A + [(-8 cot A)/ (3 cot²A – 1)

= (3 cot³A – 9 cot A)/ (3 cot²A – 1)

= 3[(cot³A – 3 cot A)/ (3 cot²A – 1)]

= 3 cot 3A

= RHS

Hence proved.

OR

In a ∆ABC prove that b cos C + c cos B = a.

Solution:

Let ABC be a triangle.

Case 1:

ABC is an acute-angled triangle.

a = BC = BD + CD….(i)

cos B = BD/AB

⇒ BD = AB cos B

⇒ BD = c cos B (AB = c)

And

cos C = CD/AC

⇒ CD = AC cos C

⇒ CD = b cos C (AC = b)

From the above,

a = c cos B + b cos C

Case 2:

ABC is a right-angled triangle.

a = BC

cos B = BC/AB

⇒ BC = AB cos B

⇒ BC = c cos B, [since, AB = c]

a = c cos B

⇒ a = c cos B + 0

b = c cos A + a cos C

⇒ a = c cos B + b cos C (C = 90° ⇒ cos C = cos 90 = 0)

Case 3:

ABC is an obtuse angle triangle.

a = BC = CD – BD

cos C = CD/AC

⇒ CD = AC cos C

⇒ CD = b cos C (since AC = b)

And,

cos (π – B) = BD/AB

⇒ BD = AB cos (π – B)

a = b cos C + c cos B

⇒ BD = -c cos B [since, AB = c and cos (π – θ) = -cos θ]

From the above,

a = b cos C – (-c cos B)

⇒ a = b cos C + c cos B

Question 4

Find the locus of a complex number, Z = x + iy, satisfying the relation |(z – 3i)/ (z + 3i)| ≤ √2. Illustrate the locus of Z in the argand plane.

Solution:

Given,

z = x + iy

Squaring on both sides,

x² + (y – 3)² ≤ 2 [x² + (y + 3)²]

x² + y² + 9 – 6y ≤ 2(x² + y² + 9 + 6y)

x² + y² + 9 – 6y ≤ 2x² + 2y² + 18 + 12y

⇒ 2x² + 2y² + 18 + 12y – x² – y² – 9 + 6y ≥ 9

⇒ x² + y² + 18y + 9 ≥ 0

⇒ x² + (y² + 18y + 81) + 9 ≥ 81

⇒ x² + (y + 9)² ≥ 81 – 9

⇒ x² + [y – (-9)]² ≥ 72

⇒ (x – 0)² + [y – (-9)]² ≥ (6√2)²

Therefore, the locus of z represents a circle with centre (0, -9) and radius 6√2 units.

Question 5

Find the number of words which can be formed by taking four letters at a time from the word “COMBINATION”.

Solution:

Given word is:

COMBINATION

Number of letters = 11

O – 2 times

I – 2 times

N – 2 times

Thus, 8 distinct letters and 3 letters repeated twice.

Case 1: Selecting 4 different letters as one word

In this case, the number of words can be formed = 8C₄ × 4! = 70 × 24 = 1680

Case 2: 2 pairs of repeated letters (for example IIOO)

In this case, the number of words can be formed = 3C₂ × (4!/ 2! 2!) = 3 × 6 = 18

Case 3: one pair of repeated letters and 2 different letters (for example NNCB)

In this case, the number of words can be formed = 3C₁ × 7C₂ × (4!/2!)

= 3 × 21 × 12 = 756

Therefore, the total number of words = 1680 + 18 + 756 = 2454

OR

A committee of 7 members has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls

(ii) at least 3 girls and

(iii) at most three girls.

Solution:

Given,

Number of boys = 9

Number of girls = 4

The number of members should be in the committee = 7

(i) Exactly 3 girls

Total number of ways of forming a committee in which exactly 3 girls included = 9C₄ × 4C₃

= [(9 × 8 × 7 × 6)/ (4 × 3 × 2 × 1)] × 4

= 504

(ii) At least 3 girls

The committee consists of 3 girls + 4 boys and 4 girls + 3 boys

Total number of ways in this case = 4C₃ × 9C₄ + 4C₄ × 9C₃

= 504 + 1 × [(9 × 8 × 7)/ (3 × 2 × 1)]

= 504 + 84

= 588

(iii) At most 3 girls

(No girl + 7 boys) or (1 girl + 6 boys) or (2 girls + 5 boys) or (3 girls + 4 boys)

No girl + 7 boys = 4C₀ × 9C₇ = 1 × 9C₂ = (9 × 8)/ (2 × 1) = 36

1 girls + 6 boys = 4C₁ × 9C₆ = 4 × 9C₃ = 4 × [(9 × 8 × 7)/ (3 × 2 × 1)] = 336

2 girls + 5 boys = 4C₂ × 9C₅ = [(4 × 3)/ (2 × 1)] × 9C₄ = 6 × 126 = 756

3 girls + 4 boys = 4C₃ × 9C₄ = 504

Total number of ways = 36 + 336 + 756 + 504 = 1632

Question 6

Prove by the method of induction.

(1/1.2) + (1/2.3) + (1/3.4) + ………. up to n terms = n/(n + 1) where n ∈ N.

Solution:

Using principle of mathematical induction,

= [k(k + 2) + 1]/ [(k + 1)(k + 2)]

= (k² + 2k + 1)/ (k + 1)(k + 2)

= (k + 1)²/ (k + 1)(k + 2)

= (k + 1)/ (k + 2)

Therefore, the given statement is also true for n = k + 1.

Hence proved that (1/1.2) + (1/2.3) + (1/3.4) + ………. up to n terms = n/(n + 1)

Question 7

Solution:

We know that the general term of (a + b)ⁿ is:

T_{r + 1} = nCr (a)^{n – r} (a)^r

T_{r + 1} = 12C_r (√x/ √3)^{12 – r} . (-√3/ 2x)^r

= 12C_r (x/3) ^{(12 – r)/2} . (-3)^r/2 [1/ 2^r x^r]

= 12C_r (3) ^{-(12 – r)/2} (-3)^r/2 2^-r x^{[(12 – r)/2 – r]}

To get the term which is independent of x, assume the value of power of x as 0.

12 – r – 2r = 0

3r = 12

r = 4

Thus, the fourth term will be the required term.

T_{4 + 1} = 12C₄ (3) ^{-(12 – 4)/2} (-3)^4/2 2^-4

= 12C₄ [3^-4 (9)]/ 16

= 495 × 9/ (81 × 16)

= 55/16

OR

Find the sum of the terms of the binomial expansion to infinity:

Solution:

Question 8

Differentiate from first principle: f(x) = √(3x + 4).

Solution:

Given,

f(x) = √(3x + 4)

f(x + h) = √[3(x + h) + 4]

By the first principle,

f'(x) = lim_{h → 0} [f(x + h) – f(x)]/ h

Question 9

Reduce the equation x + y + √2 = 0 to the normal form (x cos α + y sin α = p) and find the values of p and α.

Solution:

Given,

x + y + √2 = 0

x + y = -√2

√[(coefficient of x)² + (coefficient of y)²] = √(1² + 1²)

= √(1 + 1)

= √2

⇒ (1/√2)(x + y) = -(1/√2) × √2

⇒ (1/√2)x + (1/√2)y = -1

⇒ x cos (π/4) + y sin (π/4) = -1

Comparing with the normal form x cos α + y sin α = p,

α = π/4 = 45° and p = -1

Therefore, α = 45° and p = -1.

Question 10

Write the equation of the circle having radius 5 and tangent as the line 3x – 4y + 5 = 0 at (1, 2).

Solution:

Let the equation of the circle be:

(x – h)² + (y – k)² = r²

(x – h)² + (y – k)² = 5² (given radius = 5)

Also, 3x – 4y + 5 = 0 is the tangent to the circle.

⇒ |3h – 4k + 5|/√(3² + 4²) = 5

⇒ |3h – 4k + 5|/ √(9 + 16) = 5

⇒ |3h – 4k + 5|/ √25 = 5

⇒ |3h – 4k + 5| = 25

⇒ 3h – 4k + 5 = ±25

Now,

3h – 4k + 5 = 25

3h – 4k = 20….(i)

And

3h – 4k + 5 = -25

3h – 4k = -30….(ii)

Consider, 3x – 4y + 5 = 0

4y = 3x + 5

y = (3/4)x + (5/4)

Slope = (3/4)

Slope of the line passes through radius and the point (1, 2) = -1/slope of the given line

(k – 2)/ (h – 1) = -1/(3/4) = -4/3

3(k – 2) = -4(h – 1)

3k – 6 = -4h + 4

4h + 3k = 10….(iii)

By solving (i), (ii), and (iii),

h = 4, k = -2 ad h = -2 and k = 6

Hence, the required equation of circles are:

(x – 4)² + (y + 2)² = 25

(x + 2)² + (y – 6)² = 25

Question 11

In a ΔABC, prove that cot A + cot B + cot C = (a² + b² + c²)/4Δ.

Solution:

Area of triangle = Δ

We know that in a triangle ABC,

Δ = (1/2) bc sin A

bc = 2Δ/sin A….(i)

Δ = (1/2) ac din B

ac = 2Δ/sin B….(ii)

Δ = (1/2) ab sin C

ab = 2Δ/sin C….(iii)

Using cosine rule,

a² = b² + c² – 2bc cos A

b² = a² + c² – 2ac cos B

c² = a² + b² – 2ab cos C

Now,

a² + b² + c² = 2a² + 2b² + 2c² – 2ab cos C – 2ac cos B – 2bc cos A

⇒ a² + b² + c² = 2ab cos C + 2ac cos B + 2bc cos A….(iv)

From (i), (ii), (iii), and (iv),

a² + b² + c² = 2 [(2Δ/sin C) cos C + (2Δ/sin B) cos B + (2Δ/sin A) cos A]

a² + b² + c² = 2 × 2Δ (cot A + cot B + cot C)

⇒ cot A + cot B + cot C = (a² + b² + c²)/4Δ

Hence proved.

Question 12

Find the nth term and deduce the sum to n terms of the series:

4 + 11 + 22 + 37 + 56 + ….

Solution:

Given series is:

4 + 11 + 22 + 37 + 56 + ….

Let S_n = 4 + 11 + 22 + 37 + 56 + … + a_{n – 1} + a_n (n terms)

⇒ 0 = 4 + 7 + 11 + 15 + 19 + … + (a_n – a_{n – 1}) – a_n (n + 1 terms)

⇒ a_n = 4 + 7 + 11 + 15 + 19 + … + (a_n – a_{n – 1})

7 + 11 + 15 + 19 + … + (a_n – a_{n – 1}) is an AP with a = 7 and d = 4

Sum of these n – 1 terms = (n – 1)/2 [2 × 4 + (n – 1 – 1)4]

= (n – 1)/2 [14 + (n – 2)4]

= (n – 1)[7 + (n – 2)2]

a_n = 4 + (n – 1)(7 + 2n – 4)

a_n = 4 + (n – 1)(2n + 3)

= 4 + 2n² + 3n – 2n – 3

= 2n² + n + 1

OR

If (p + q)th term and (p – q)th terms of G.P are a and b respectively, prove that the pth term is √(ab).

Solution:

Let t₁ be the first term and r be the common ratio of a GP.

Given,

(p + q)th term and (p – q)th terms of G.P are a and b respectively.

t₁ × r(p + q – 1) = a….(i)

t₁ × r(p – q – 1) = b….(ii)

Multiplying (i) and (ii),

t₁ × r^{(p + q – 1)} × t₁ × r^{(p – q – 1)} = ab

(t₁)² × r^{(p + q – 1 + p – q – 1)} = ab

(t₁)² × r^{(2p – 2)} = ab

(t₁)² × r^{2(p – 1)} = ab

(t₁ × r^{p – 1})² = ab

t₁ × r^{p – 1} = √ab

Therefore, pth terms = √ab

Question 13

If x is real, prove that the value of the expression [(x – 1)(x + 3)] / [(x – 2)(x + 4)] cannot be between 4/9 and 1.

Solution:

Let [(x – 1)(x + 3)] / [(x – 2)(x + 4)] = y

(x – 1)(x + 3) = y(x – 2)(x + 4)

x² + 3x – x – 3 = y(x² + 4x – 2x – 8)

x² + 2x – 3 – x²y – 2xy + 8y = 0

x²(1 – y) + 2x(1 – y) + (8y – 3) = 0

Given that x is real.

Therefore, discriminant ≥ 0

4(1 + y² – 2y) – 4(8y – 3 – 8y² + 3y) ≥ 0

4 + 4y² – 8y – 44y + 12 + 32y² ≥ 0

36y² – 52y + 16 ≥ 0

4(9y² – 13y + 4) ≥ 0

9y² – 13y + 4 ≥ 0

9y² – 9y – 4y + 4 ≥ 0

9y(y – 1) – 4(y – 1) ≥ 0

(9y – 4)(y – 1) ≥ 0

(y – 4/9)(y – 1) ≥ 0

Therefore, y cannot lie between 4/9 and 1.

OR

Solution:

Given,

General term is:

T_{r + 1} = 2nC_r (x²)^{2n – r} (1/x)^r

= 2nCr x^{4n – 2r} x^(-r)

= 2nC_r x^{(4n – 2r – r)}

= 2nC_r x^{(4n – 3r)}

Let x^p occur in this expansion.

⇒ 4n – 3r = p

⇒ 3r = 4n – p

⇒ r = (4n – p)/3

Coefficient of x^p = 2nC_r

= (2n)!/ [r!(2n – r)!]

= (2n)!/ {[(4n – p)/3]! [2n – (4n – p)/3]!}

= (2n)!/ {[(4n – p)/3]! [(6n – 4n + p)/3]!}

= (2n)!/ {[(4n – p)/3]! [(2n + p)/3]!}

Hence proved.

Question 14

Calculate the standard deviation of the following distribution:

Solution:

Variance = 1/(N – 1) [∑f_ix_i² – (1/N) (∑f_ix_i)²]

= [1/(480 – 1)] [468250 – (14500)²/ 480]

= (1/479) [468250 – 438020.833]

= 30229.167/ 479

= 63.109

Standard deviation = √(63.109) = 7.944

SECTION B

Question 15

(a) Find the focus and directrix of the conic represented by the equation 5x² = -12y.

Solution:

5x² = -12y

x² = (-12/5)y

x² = 4(-3/5)y

(x – 0)² = 4(-3/5) (y – 0)

Comparing with (x – h)² = 4p(y – k)

h = 0, k = 0, p = -3/5

Thus, the given equation represents a parabola with vertex at (h, k) = (0, 0) and focal length |p| = 3/5.

Here, the parabola is symmetric around the y-axis.

Therefore, the focus lies at a distance p from the centre (0, 0) along the y-axis.

i.e. (0, 0 + p)

= (0, 0 + (-3/5))

= (0, -3/5)

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance -p from the centre (0, 0).

Directrix is y = 0 – p

y = 0 – (-3/5)

y = 3/5

Hence, the focus of the given conic is (0, -3/5) and the directrix is y = 3/5.

(b) Construct the truth table (~p ⋀ ~q) ⋁ (p ⋀ ~q)

Solution:

(c) Write the converse, contradiction and contrapositive of the statement.

“If x + 3 = 9, then x = 6.”

Solution:

Given statement is:

If x + 3 = 9, then x = 6.

Converse of the given conditional statement is:

If x = 6, then x + 3 = 9.

Contradiction of the given conditional statement is:

If x + 3 ≠ 9, then x ≠ 6.

Contrapositive statement is:

If x ≠ 6, then x + 3 ≠ 9.

Question 16

Show that the point (1, 2, 3) is common to the lines which join A(4, 8, 12) to B(2, 4, 6) and C(3, 5, 4) to D(5, 8, 5).

Solution:

Given,

A(4, 8, 12), B(2, 4, 6), C(3, 5, 4) and D(5, 8, 5).

Let P = (1, 2, 3)

P is a common point to the line AB and CD if ABP and CDP are collinear.

Consider A(4, 8, 12), B(2, 4, 6) and P(1, 2, 3):

= 4(12 – 12) – 8(6 – 6) + 12(4 – 4)

= 4(0) – 8(0) + 12(0)

= 0

Thus, A, B, and P are collinear.

Consider C(3, 5, 4), D(5, 8, 5) and P(1, 2, 3):

= 3(24 – 10) – 5(15 – 5) + 4(10 – 8)

= 3(14) – 5(10) + 4(2)

= 42 – 50 + 8

= 0

Thus, C, D, and P are collinear.

Therefore, the point (1, 2, 3) is common to the lines which join A(4, 8, 12) to B(2, 4, 6) and C(3, 5, 4) to D(5, 8, 5).

OR

Calculate the Cosine of the angle A of the triangle with vertices A(1, -1, 2) B (6, 11, 2) and C(1, 2, 6).

Solution:

Given,

Vertices of a triangle are A(1, -1, 2) B (6, 11, 2) and C(1, 2, 6).

Using distance formula,

AB = c = 13

BC = a = √122

AC = b = 5

cos A = (b² + c² – a²)/2bc

= (25 + 169 – 122)/ (2 × 5 × 13)

= 72/130

= 36/65

Question 17

Find the equation of the hyperbola whose focus is (1, 1), the corresponding directrix 2x + y – 1 = 0 and e = √3.

Solution:

Given,

Focus = S(1, 1)

Directrix is 2x + y – 1 = 0

Eccentricity (e) = √3

Let P(x, y) be any point on the hyperbola.

We know that,

The perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 = |ax1 + by1 + c|/ √(a² + b²)

Here,

(x₁, y₁) = (x, y)

a = 2, b = 1, c = -1

And

SP = e (perpendicular distance)

SP² = e² (perpendicular distance)²

(x – 1)² + (y – 1)² = (√3)² [|2x + y – 1|/ √(2² + 1²)]²

x² – 2x + 1 + y² – 2y + 1 = (3) [|2x + y – 1|²/ (4 + 1)]

x² + y² – 2x – 2y + 2 = (3/5) (4x² + y² + 1 + 4xy – 2y – 4x)

5x² + 5y² – 10x – 10y + 10 = 12x² + 3y² + 3 + 12xy – 6y – 12x

12x² + 3y² + 3 + 12xy – 6y – 12x – 5x² – 5y² + 10x + 10y – 10 = 0

7x² – 2y² + 12xy – 2x + 4y – 7 = 0

Therefore, the equation of the hyperbola is 7x² – 2y² + 12xy – 2x + 4y – 7 = 0.

OR

Find the equation of tangents to the ellipse 4x² + 5y² = 20 which are perpendicular to the line 3x + 2y – 5 = 0.

Solution:

Given,

4x² + 5y² = 20

(4x²/20) + (5y²/20) = 1

(x²/5) + (y²/4) = 1

This is of the form (x²/a²) + (y²/b²) = 1

a² = 5 and b² = 4

Equation of the line is 3x + 2y – 5 = 0

2y = -3x + 5

y = (-3/2)x + (5/2)

Slope = -3/2

Tangent and the given equation are perpendicular lines.

Thus, the slope of tangent line = m = -1/(-3/2) = ⅔

Equation of tangent is:

y = mx ± √(a²m² + b²)

y = (2/3)x ± √[5(4/9) + 4]

y = (2/3)x ± √[(20 + 36)/9

y = (1/3) [2x ± √56]

3y = 2x ± √56

Question 18

Show that the equation 16x² – 3y² – 32x – 12y – 44 = 0 represents a hyperbola. Find the lengths of axes and eccentricity.

Solution:

Given,

16x² – 3y² – 32x – 12y – 44 = 0

(16x² – 32x) – (3y² + 12y) – 44 = 0

16(x² – 2x) – 3(y² + 4y) – 44 = 0

16(x² – 2x + 1) – 3(y² + 4y + 4) – 44 = 16 – 12

16(x – 1)² – 3(y + 2)² – 44 – 4 = 0

16(x – 1)² – 3(y + 2)² = 48

⇒ [(x – 1)²/ 3] – [(y + 2)²/ 16] = 1

This is the equation of parabola in which a² = 3 and b² = 16.

Centre = (1, -2)

And

b² = a²(e² – 1)

16 = 3(e² – 1)

e² – 1 = 16/3

e² = (16/3) + 1

e² = 19/3

e = √(19/3)

SECTION C

Question 19

(i) Two sample sizes of 50 and 100 are given.The mean of these samples respectively are 56 and 50. Find the mean of size 150 by combining the two samples.

Solution:

Let n₁ and n₂ be the sizes of two samples.

X₁ and X₂ are the means of two samples.

According to the given,

n₁ = 50 and n₂ = 100

X₁ = 56 and X₂ = 50

New sample size = n₁ + n₂ = 50 + 100 = 150

New mean = (n₁X₁ + n₂X₂)/ (n₁ + n₂)

= [50(56) + 100(50)]/ 150

= (2800 + 5000)/ 150

= 7800/150

= 52

Therefore, the mean of size 150 by combining the two samples is 52.

(ii) Calculate P95, for the following data:

Solution:

N = 60

P at 95 = p = (95/100) × 60 = 57

Cumulative frequency greater than and nearest to 57 is 60 which lies in the interval 50 – 60.

Percentile class = 50 – 60

Lower limit of the percentile class = l = 50

Frequency of the percentile class = f = 4

Cumulative frequency of the class preceding the percentile class = cf = 56

Class height = h = 10

P₉₅ = l + [(p – cf)/ f] × h

= 50 + [(57 – 56)/ 4] × 10

= 50 + (10/4)

= 50 + 2.5

= 52.5

OR

Calculate mode for the following data.

Solution:

Let us arrange the data as shown below:

From the given data,

Maximum frequency = 16

Modal class is 7.5 – 10.5

Frequency of the modal class = f₁ = 16

Frequency of the class preceding the modal class = f₀ = 11

Frequency of the class succeeding the modal class = f₂ = 8

Lower limit of the modal class = l = 7.5

Class height = h = 3

Mode = l + [(f₁ – f₀)/ (2f₁ – f₀ – f₂)] × h

= 7.5 + [(16 – 11)/ (2 × 16 – 11 – 8)] × 3

= 7.5 + [5/ (32 – 19)] × 3

= 7.5 + (15/ 13)

= 7.5 + 1.154

= 8.654

Question 20

(i) Find the covariance between X and Y when N = 10, ∑X = 50, ∑Y = -30, and ∑XY = 115.

Solution:

Given,

N = 10,

∑X = 50

∑Y = -30

∑XY = 115

Covariance between X and Y = (1/N) ∑XY – (1/N2) ∑X ∑Y

= (1/10) × 115 – (1/100) × 50 × (-30)

= 11.5 + 15

= 26.5

(ii) Calculate Spearman’s Rank Correlation for the following data and interpret the result:

Solution:

n = 10

∑d² = 27

The spearman’s rank correlation coefficient = 1 – [(6 × ∑d²)/ n(n² – 1)]

= 1 – [(6 × 27)/ 10(100 – 1)]

= 1 – [162/ 10 × 99]

= 1 – (162/990)

= (990 – 162)/ 990

= 828/990

= 0.836

OR

Find Karl Pearson’s Correlation Coefficient from the given data:

Solution:

Mean of x = ∑x/n = 302/10 = 30.2

Mean of y = ∑y/n = 1292/10 = 129.2

Karl Pearson’s Correlation Coefficient = (∑d_x d_y)/ (√∑d²_x ∑d²_y)

= 405.6 / √(405.6 × 405.6)

= 405.6/405.6

= 1

Question 21

Find the consumer price index for 2007 on the basis of 2005 from the following data using weighted average of price relative method:

Solution:

Index number of weighted average of price relatives = ∑Iw/ ∑w

= 49600/ 80

= 620

Question 22

Using the following data. Find out the trend using Quarterly moving average and plot them on graph:

Solution:

Trend: