ISC Class 11 Maths Mock Sample Question Paper 2 With Answers - Free PDF Download

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SECTION – A

Question 1 [10×2]

(i) Let f : R → R be defined as f(x) = x² + 1, find f^-1(-7) = x.

Solution:

We know that,

If f : A → B such that y ∈ B, then f-(y) = {x ∈ A : f(x) = y}

Given,

f(x) = x² + 1

f^-1(-7) = {x ∈ R ; f(x) = -7}

Now,

x² + 1 = -7

x² = -7 – 1

x² = -8

This is not possible.

Therefore, f^-1(-7) = Φ

(ii) If a ∈ {2, 4, 6, 9} and b ∈ {4, 6, 18, 27}, then form the set of all ordered pairs (a, b) such that a divides b and a < b.

Solution:

Given,

a ∈ {2, 4, 6, 9} and b ∈{4, 6, 18, 27}

Now,

2 divides 4, 6, 18 and is also 2 < 4, 6, 8

4 divides 4 but both are equal

6 divides 6, 18 and 6 is less than 18

9 divides 18, 27 and is less than 18 and 27

Let us form ordered pairs (a, b) such that a divides b and a < b.

Hence, the required ordered pairs of (a, b) are {(2, 4), (2, 6), (2, 18), (6, 18), (9, 18) and (9, 27)}

(iii) Prove that cos x + cos (120° – x) + cos (120° + x) = 0.

Solution:

LHS = cos x + cos (120° – x) + cos (120° + x)

= cos x + (cos 120° cos x + sin 120° sin x) + (cos 120° cos x – sin 120° sin x)

= cos x + 2 cos 120° cos x

= cos x + 2 cos x cos (90° + 30°)

= cos x + 2 cos x (- sin 30°)

= cos x + 2 × (1/2) cos x

= cos x – cos x

= 0

= RHS

Hence proved.

(iv) Express (1 + i)³ – (1 – i)³ in the form of a + ib, find the value of a and b.

Solution:

(1 + i)³ – (1 – i)³

= (1)³ + 3(1)²(i) + 3(1)(i)² + (i)³ – [(1)³ – 3(1)²(i) + 3(1)(i)² – (i)³]

= 1 + 3i + 3i² + i³ – (1 – 3i + 3i² – i³)

= 1 + 3i + 3i² + i3 – 1 + 3i – 3i² + i³

= 6i + 2i³

= 6i + 2i(i²)

= 6i + 2i(-1)

= 6i – 2i

= 4i

= 0 + 4i

This is of the form a + ib.

Therefore, a = 0 and b = 4.

(v) Prove that b sin B – c sin C = a sin (B – c).

Solution:

By sine rule in a triangle,

a/sin A = b/sin B = c/sin C = k

⇒ a = k sin A, b = k sin B, c = k sin C…..(i)

LHS = b sin B – c sin C

k sin B sin B – k sin C sin C [From (i)]

= k (sin²B – sin²C)

= k [sin (B + c) sin (B – C)]….(ii)

We know that,

A + B + C = π (by the angle sum property of a triangle)

B + C = π – A….(iii)

From (ii) and (iii),

= k [sin (π – A) sin (B – C)]

= k [sin A sin (B – C)]

= (k sin A) sin (B – C)

= a sin (B – C) [again from (i)]

= RHS

Hence proved.

(vi) If α and β are the roots of the equation px² – qx + r = 0, find α³ + β³.

Solution:

Given,

α and β are the roots of the equation px² – qx + r = 0.

Sum of the roots = α + β = -(-q)/p – q/p

Product of the roots = αβ = r/p

α³ + β³ = (α + β) (α² + β² – αβ)

= (q/p) [α² + β² + 2αβ – 3αβ]

= (q/p) [(α + β)² – 3αβ]

= (q/p) [(q/p)² – 3(r/p)]

= (q/p) [(q²/p²) – (3r/p)]

= (q/p) [(q² – 3pr)/ p²]

= (q³ – 3pqr)/ p³

(vii) In how many ways can 10 members of a committee sit around a round table so that the secretary and the joint secretary are always as the neighbors of the present?

Solution:

Let us assume the President, Secretary, and Joint secretary be 1 member.

Remaining = 10 – 3 = 7

Now we consider the total members as 8 (i.e. 7 members and 1 member that represents three specified people).

The number of ways of arranging these 8 members around a table = (8 – 1)! = 7!

And, the number of ways of arranging three specified people such that Secretary and the Joint secretary be the neighbors of the President = 2

Hence, the total number of arrangements = 7! × 2

= 5040 × 2

= 10080

(viii) In a single throw of three dice, find the probability of getting a total of 17 or 18.

Solution:

Given that three dice are thrown simultaneously.

Total number of possible outcomes = 63 = 216

i.e. n (S) = 216

Let E be the event of getting a total of 17 or 18.

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

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 216

= 1/54

(ix) Evaluate:

Solution:

(x) Differentiate: f(x) = (x² + 1)/ (x + 1)

Solution:

Given,

f(x) = (x² + 1)/ (x + 1)

Let u = x² + 1 and v = x + 1

⇒ du/dx = d/dx (x² + 1) = 2x

And

dv/dx = d/dx (x + 1) = 1

We know that,

d/dx (u/v) = [v(du/dx) – u(dv/dx)] / v²

= [(x + 1)(2x) – (x² + 1)(1)]/ (x + 1)²

= [2x² + 2x – (x² + 1)]/ (x + 1)²

= (2x² + 2x – x² – 1)/ (x + 1)²

= (x² + 2x – 1)/ (x + 1)²

Therefore, the derivative of a given f(x) is (x² + 2x – 1)/ (x + 1)².

Question 2 [4]

Find the domain and range of the function f(x) = 1/ √(2x – 3).

Solution:

f(x) = 1/ √(2x – 3)

Let 1/ √(2x – 3) = y….(i)

The given function will be defined only if 2x – 3 > 0

⇒ 2x > 3

⇒ x > 3/2

Thus, the function is defined for all the real values greater than 3/2.

From (i),

1/√(2x – 3) = y

Squaring on both sides,

1/(2x – 3) = y²

⇒ 2x – 3 = 1/y²

⇒ 2x = (1/y²) + 3

⇒ x = (1 + 3y²)/ 2y²

This can be defined when y > 0.

Therefore, the domain of f(x) is (3/2, ∞) and the range is (0, ∞).

Question 3 [4]

(a) Solve for x: cos x + sin x = cos 2x + sin 2x

Solution:

cos x + sin x = cos 2x + sin 2x

cox – cos 2x = sin 2x – sin x

-2 [sin (2x + x)/2] [sin (2x – x)/2] = 2 [cos (2x + x)/2] [sin (2x – x)/2]

2 sin (3x/2) sin (x/2) = 2 cos (3x/2) sin (x/2)

sin (3x/2) sin (x/2) – cos (3x/2) sin (x/2) = 0

sin (x/2) [sin (3x/2) – cos (3x/2)] = 0

sin (x/2) = 0 or sin (3x/2) = cos (3x/2)

sin (x/2) = sin mπ or sin (3x/2) / cos (3x/2) = 0

sin (x/2) = sin mπ or tan (3x/2) = 1

sin (x/2) = sin mπ or tan (3x/2) = tan π/4

x/2 = mπ or 3x/2 = nπ + π/4

x = 2mπ or x = 2nπ/3 + π/6

Hence, the general solution is

x = 2mπ or 2nπ/3 + π/6, where m, n ϵ Z.

OR

(b) If A + B + C = π, prove that tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C.

Solution:

Given,

A + B + C = π

A + B = π – C

Multiplying by 2 on both sides,

2A + 2B = 2π – 2C

Taking “tangent” on both sides,

tan (2A + 2B) = tan (2π – 2C)

(tan 2A + tan 2B)/ (1 – tan 2A tan 2B) = – tan 2C

tan 2A + tan 2B = -tan 2C (1 – tan 2A tan 2B)

tan 2A + tan 2B = -tan 2C + tan 2A tan 2B tan 2C

tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

Hence proved.

Question 4 [4]

Prove that 3^{2n + 2} – 8n – 9 is divisible by 8 for all n ϵ N.

Solution:

Let P(n) = 3^{2n + 2} – 8n – 9 is divisible by 8

Let us check for n = 1,

P(1) = 3^{(2.1 + 2)} – 8(1) – 9

= 3⁴ – 8 – 9

= 81 – 17

= 64

= 8 (8), which is divisible by 8.

P(n) is true for n = 1.

Now, let us check for P(n) is true for n = k, and have to prove that P(k + 1) is also true.

P(k): 3^{(2k + 2)} – 8k – 9 is divisible by 8

= 3^{(2k + 2)} – 8k – 9 = 8M … (i)

We have to prove,

3^{(2k + 4)} – 8(k + 1) – 9 is divisible by 8.

3^{[2(k + 1) + 2]} – 8(k + 1) – 9

= 3^{(2k + 2)} . 3² – 8k – 8 – 9

= (8M + 8k + 9)9 – 8k – 17 [From (i)]

= 72M + 72k + 81 – 8k – 17 using equation (1)

= 72M + 64k + 64

= 8(9M + 8k + 8)

= 8P

P(n) is also true for n = k + 1.

Hence, P(n) is true for all n ∈ N.

Question 5 [4]

Find the locus of z satisfying the inequality |(z + 2i)/ (2z + 1)| < 1, where z = x + iy. Represent it in the argand plane.

Solution:

Given,

z = x + iy

|(z + 2i)/ (2z + 1)| < 1

|(z + 2i)/ (2z + i)| < 1

⇒ |z + 2i|/ |2z + i| < 1

⇒ |z + 2i| < |2z + i|

⇒ |x + iy + 2i| < |2(x + iy) + i|

⇒ |x + (y + 2)i| < |2x + (2y + 1)i|

⇒ √[x² + (y + 2)²] < √[(2x)² + (2y + 1)²]

Squaring on both sides,

⇒ x² + y² + 4y + 4 < 4x² + 4y² + 1 + 4y

⇒ y² + 4y + 4 < 3x² + 4y² + 4y + 1

⇒ 3x² + 4y² – y² > 4 – 1

⇒ 3x² + 3y² > 3

⇒ x² + y² > 1

Thus, the locus of z represents the circle with centre (0, 0) and radius 1 unit.

Question 6 [4]

(a) In how many ways can a football team of 11 players be selected from 16 players? How many of these will:

(i) Include 2 particular players

(ii) Exclude 2 particular players?

Solution:

Total number of players = 16

We have to select a team of 11 players among the given.

So, the number of ways = 16C₁₁

= 16C₅

= (16 × 15 × 14 × 13 × 12)/ (5 × 4 × 3 × 2 × 1)

= 4368

(i) If two particular players are included, then 9 players can be selected from the remaining 14 players in 14C₉ ways.

= 14C₅

= (14 × 13 × 12 × 11 × 10)/ (5 × 4 × 3 × 2 × 1)

= 2002

(ii) If two particular players are excluded, then all 11 players can be selected from the remaining 14 players in 14C₁₁ ways.

= 14C₃

= (14 × 13 × 12)/ (3 × 2 × 1)

= 364

OR

(b) How many permutations can be formed by the letters of the word “VOWELS”, when

(i) there is no restriction on letters

(ii) each word begins with E

(iii) each word begins with O and ends with L

(iv) all vowels come together?

Solution:

Given word is “VOWELS”.

Number of letters = 6

(i) There is no restriction on letters:

The permutation of letters of the given word = 6! = 720

(ii) Each word begins with E.

Thus, the position of E is fixed.

Remaining number of letters = 6 – 1 = 5

Therefore, the permutation of letters in this case = 5! = 120

(iii) Each word begins with O and ends with L:

Thus, the positions of O and L are fixed.

Remaining number of letters = 6 – 2 = 4

Therefore, the permutation of letters in this case = 4! = 24

(iv) All vowels come together:

Vowels = {O, E}

These 2 vowels can be taken as one.

Total = 4 letters + 1 (2 vowels) = 5

Therefore, the permutation of letters in this case = 5! × 2!

= 120 × 2

= 240

Question 7 [4]

Find the term independent of x in the expansion of the following:

Solution:

Given,

Let (r + 1)th be the term in the given expression which is independent of x.

T_r+1 = nC_r x^n-r a^r

= 25C_r (2x²)^25-r (-3/x³)^r

= (-1)^r 25C_r × 2^{(25 – r)} x^{(50 – 2r)} 3^r (x)^-2r

= (-1)^r 25C_r × 2^{(25 – r)} 3^r x^{(50 – 2r – 3r)}

For this term to be independent of x, the power of x must be 0.

50 – 2r – 3r = 0

50 – 5r = 0

5r = 50

r = 50/5

= 10

So, the required term is 11th term.

We have,

T₁₁ = T_{10 + 1}

= (-1)¹⁰ 25C₁₀ × 2^{(25 – 10)} × 3¹⁰

= 25C₁₀ (2¹⁵ × 3¹⁰)

Hence, the term independent of x is 25C₁₀ (2¹⁵ × 3¹⁰).

Question 8 [4]

Find the equation of obtuse angle bisector of lines:

6x – 8y + 5 = 0 and 7x + 24y – 8 = 0

Solution:

Given,

6x – 8y + 5 = 0 and 7x + 24y – 8 = 0

Comparing with the standard form,

a₁ = 6, b₁ = -8, c₁ = 5

a₂ = 7, b₂ = 24, c₂ = 8

a₁a₂ + b₁b₂ = 6(7) + (-8)(24)

= 42 – 192

= -150 < 0

Thus, the obtuse angle bisector is given by:

(a₁x + b₁y + c₁)/ √(a₁² + b₁²) = -(a₂x + b₂y + c₂)/ √(a₂² + b₂²)

|6x – 8y + 5|/ √[(6)² + (-8)²] = -(7x + 24y – 8)/ √(7² + 24²)

(6x – 8y + 5)/ √(36 + 64) = -(7x + 24y – 8)/ √(49 + 576)

(6x – 8y + 5)/ √100 = -(7x + 24y – 8)/ √625

25(6x – 8y + 5) = -10(7x + 24y – 8)

150x – 200y + 125 = -70x – 240y + 80

150x – 220y + 125 + 70x + 240y – 80 = 0

220x + 20y + 45 = 0

⇒ 44x + 4y + 9 = 0

Hence, the equation of obtuse angle bisector of given lines is 44x + 4y + 9 = 0.

Question 9 [4]

(a) Prove that the line 2x – 3y – 27 = 0 is a tangent to the circle x² + y² – 8x + 4y + 7 = 0.

Solution:

Given equation of circle is:

x² + y² – 8x + 4y + 7 = 0

Comparing with x² + y² + 2gx + 2fy + c = 0

2g = -8, 2f = 4

Centre = (-g, -f) = (4, -2)

Radius = √(g² + f² – c)

= √[(4)2 + (-2)2 – 7]

= √(16 + 4 – 7)

= √13

Now, the perpendicular distance between centre (4, -2) and the line 2x – 3y – 27 = 0 is given by:

d = |2(4) – 3(-2) – 27|/ √[(2)² + (-3)²]

= |8 + 6 – 27| √(4 + 9)

= |-13|/ √13

= 13/ √13

= (√13 × √13)/ √13

= √13 = Radius

Therefore, the line 2x – 3y – 27 = 0 is the tangent to the given circle.

OR

(b) Show that the points A(1, 0), B(2, -7), C(8, 1) and D(9, -6) all lie on the same circle. Find the equation of this circle, its centre, and radius.

Solution:

The general form of the equation of a circle is:

(x – h)² + (y – k)² = r²….(i)

where (h, k) is the centre and r is the radius.

Consider the points (1, 0), (2, -7) and (8, 1) lie on the circle.

Substituting (1, 0) in (i),

⇒ h² + k² + 1 – 2h = r²….(ii)

Substituting (2, -7) in (i),

⇒ h² + k² + 53 – 4h + 14k = r²….(iii)

Substituting (8, 1) in (i),

⇒ (8 – h)² + (1 – k)² = r²

h² + k² + 65 – 16h – 2k = r²….(iv)

Subtracting (ii) from (iii),

h – 7k – 26 = 0….(v)

Subtracting (ii) from (iv),

7h + k – 32 = 0….(vi)

Solving (v) and (vi),

h = 5 and k = -3

Substituting h = 5 and k = -3 in (iv),

r² = 25

Centre = (5, -3)

Radius = 25

Thus, the equation of circle is:

(x – 5)² + (y + 3)² = 25

Check for (9, -6):

Substituting (9, -6) in the above equation of circle.

(9 – 5)² + (-6 + 3)² = 25

16 + 9 = 25

25 = 25

Hence, the points A(1, 0), B(2, -7), C(8, 1) and D(9, -6) all lie on the same circle.

Question 10 [4]

Differentiate the function cos (3x – 2) by the first principle of differentiation.

Solution:

Let f(x) = cos (3x – 2)

By the first principle of differentiation,

We know that,

lim x → 0 (sin x)/x = 1

= [-2 sin (3x – 2 + 0)] × 1 × (3/2)

= -3 sin (3x – 2)

Question 11 [6]

In triangle ABC, if cos A/a = cos B/b, show that the triangle is isosceles.

Solution:

In any triangle ABC,

a/sin A = b/sin B = c/sin C = 2R….(i)

Given,

cos A/a = cos B/b

Using the identity sin²θ + cos²θ = 1,

Squaring on both sides,

(1 – sin²A)/a² = (1 – sin²B)/b²

⇒ (1/a²) – (sin²A/a²) = (1/b²) – (sin²B/b²)

From (i),

sin²A/a² = sin²B/b²

⇒ 1/a² = 1/b²

⇒ a² = b²

⇒ a = b

Thus, the two sides of a triangle are equal.

Therefore, triangle ABC is an isosceles triangle.

Hence proved.

Question 12 [6]

(a) If x is real, then find the maximum and minimum values of y = (x² – 3x + 4)/ (x² + 3x + 4).

Solution:

Given,

y = (x² – 3x + 4)/ (x² + 3x + 4)

Let dy/dx = 0

dy/dx = [(x² + 3x + 4) (2x – 3) – (x² – 3x + 4)(2x + 3)]/ (x² + 3x + 4)2 = 0

2x³ – 3x² + 6x² – 9x + 8x – 12 = 2x³ + 3x² – 6x² – 9x + 8x + 12

6x² = 24

x² = 4

x = ±2

When x = 2,

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

= (4 – 6 + 4)/ (4 + 6 + 4)

= 2/14

= 1/7

When x = -2,

y = [(-2)² – 3(-2) + 4]/ [(-2)² + 3(-2) + 4]

= (4 + 6 + 4)/ (4 – 6 + 4)

= 14/2

= 7

The minimum value of y is 1/7 when x = 2 and the maximum value of y is 7 when x = -2.

OR

(b) If α, β are the roots kx² +lx + m = 0, then form an equation whose roots are α + (1/β) and β + (1/α).

Solution:

Given,

α, β are the roots of the quadratic equation kx² +lx + m = 0.

Sum of the roots = α + β = -l/k

Product of the roots = αβ = m/k

Let α + (1/β) and β + (1/α) be the roots.

Sum of roots = α + (1/β) + β + (1/α)

= (α + β) + [(1/α) + (1/β)]

= (α + β) + [(α + β)/ αβ]

= (-l/k) + [(-l/k)/ (m/k)]

= (-l/k) – (l/m)

= -(lm + kl)/km

Product of roots = [α + (1/β)] [β + (1/α)]

= αβ + 1 + 1 + (1/αβ)

= (m/k) + 2 + (k/m)

= (k² + m² + 2km)/ km

Therefore, the quadratic equation is:

x² – (sum of the roots)x + (product of roots) = 0

x² – [-(lm + kl)/ km]x + [(k² + m² + 2km)/ km] = 0

kmx² + (lm + kl)x + (k + m)² = 0

Question 13 [6]

(a) The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

Solution:

Let a be the first term and d be the common difference of an AP.

Thus,

a₁ = a

a₂ = a + d

a₃ = a + 2d

According to the given,

a₁ + a₂ + a₃ = 21

a + a + d + a + 2d = 21

3a + 3d = 21

a + d = 7

d = 7 – a…..(i)

Given that when the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P.

i.e. a, a + d – 1 and a + 2d + 1

Thus, (a + d – 1)/a = (a + 2d + 1)/ (a + d – 1)

(a + d – 1)² = a(a + 2d + 1)

a² + d² + 1 + 2ad – 2d – 2a = a² + 2ad + a

(7 – a)² – 3a + 1 – 2(7 – a) = 0

49 + a² – 14a – 3a + 1 – 14 + 2a = 0

a² – 15a + 36 = 0

a² – 12a – 3a + 36 = 0

a(a – 12) – 3(a – 12) = 0

(a – 3)(a – 12) = 0

a = 3 or a = 12

d = 7 – a

d = 7 – 3 or d = 7 – 12

d = 4 or -5

When a = 3 and d = 4, then A.P is 3, 7, 11

When a = 12 and d = -5, then A.P is 12, 7, 2

Therefore, the numbers are 3, 7, 11 or 12, 7, 2.

OR

(b) Find the sum of the series:

Solution:

Tr = 1/ (5r – 4)(5r + 1)

= (1/5) [1/(5r – 4) – 1/(5r + 1)]

T₁ = (1/5)[1 – (1/6)]

T₂ = (1/5) [(1/6) – (1/11)]

….

S_n = T₁ + T₂ + T₃ + …. + T_n

S_n = (1/6) – (1/30) + (1/30) – (1/55) + (1/55) +….+ (1/5) [ 1/(5n – 4) – 1/(5n + 1)]

= (1/5) – (1/5) [1/(5n + 1)]

= (1/5) [(5n + 1 – 1)/ (5n + 1)]

= (1/5) [5n/ (5n + 1)]

= n/(5n + 1)

Question 14 [6]

Find the standard deviation for the following data:

Solution:

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

= [1/(110 – 1)] [300350 – (5390)²/ 110]

= (1/109) [300350 – 264110]

= 36240/ 109

= 332.477

Standard deviation = √(332.477) = 18.234

SECTION B (20 Marks)

Question 15 [3×2]

(a) At what point of the parabola x² = 9y is the abscissa three times that of ordinate?

Solution:

Given equation of parabola is:

x² = 9y

Given that abscissa is three times that of ordinate.

Thus, (3y₁, y₁) be the point of the given parabola.

Substituting this point in the given equation,

(3y₁)² = 9(y₁)

9y₁² = 9y₁

y₁² – y₁ = 0

y₁(y₁ – 1) = 0

y₁ = 0 or y₁ – 1 = 0

y₁ = 0 or y₁ = 1

3y₁ = 3(1) = 3

Hence, the point is (3, 1).

(b) Construct a truth table for (p v q) ⋀ (~p v ~q).

Solution:

(c) Write Converse and inverse of the given conditional statement:

I go to the beach whenever it is a sunny day.

Solution:

Given conditional statement is:

I go to a beach whenever it is a sunny day.

Converse: If I go to a beach, then it is a sunny day.

Inverse: I do not go to a beach, if it is not a sunny day.

Question 16 [4]

(a) The length of the major axis of an ellipse is 20 units and its foci are (±5√3, 0) Find the equation of the ellipse.

Solution:

We know that,

The general form of equation of an ellipse is x²/a² + y²/b² = 1

Where, vertices = (±a, 0) and foci = (±c, 0)

Given,

Foci = (±5√3, 0)

i.e. c = 5√3

Length of the major axis of an ellipse = 20 units

Thus, the length of the semi-major axis = 20/2 = 10 units

⇒ a = 10

Now,

c² = a² – b²

(5√3)² = (10)² – b²

b² = 100 – 75 = 25

Therefore, the equation of an ellipse is:

x²/100 + y²/25 = 1

OR

(b) Find the eccentricity, coordinates of the foci, equations of directrices, and length of the latus-rectum of the hyperbola 4x² – 3y² = 36.

Solution:

Given equation of hyperbola is:

4x² – 3y² = 36

(4x²/36) – (3y²/36) =1

(x²/9) – (y²/12) = 1

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

a² = 9, b² = 12

Now,

b² = a²(e² – 1)

12 = 9(e² – 1)

e² – 1 = 12/9

e² = (4/3) + 1

e² = (4 + 3)/3

e = √(7/3)

ae = ±3 × (√(7/3)

= ±(√3 × √3) × (√7/ √3)

= ±√21

Therefore, the coordinates of foci are (±ae, 0) = (±√21, 0)

Equation of directrices:

x = ±a/e

= ±3 × (√3/ √7)

√7x = ±3√3

The length of latus-rectum = 2b²/a

= (2 × 12)/3

= 8

Question 17 [4]

(a) Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Solution:

We know that the x-coordinate is 0 in the yz-plane.

Let P(0, y, z) divide the line segment joining the points A(2, 4, 5) and B(3, 5, 4) in the ratio k : 1.

Here,

(2, 4, 5) = (x₁, y₁, z₁)

(3, 5, 4) = (x₂, y₂, z₂)

m : n = k : 1

Using the section formula,

P(x, y, z) = [(kx₂ + x₁)/ (k + 1), (ky₂ + y₁)/ (k + 1), (kz₂ + z₁)/ (k + 1)]

(0, y, z) = [(3k + 2)/ (k + 1), (5k + 4)/ (k + 1), (4k + 5)/ (k + 1)]

Now equating the corresponding elements,

(3k + 2)/ (k + 1) = 0

3k + 2 = 0

3k = -2

k = -2/3

Therefore, P divides the given points externally in the ratio 2 : 3.

OR

(b) If the origin is the centroid of a triangle with vertices (2a, 2, 6), (-4, 3b,-10) and (8, 14, 2c), then find the values of a, b, and c.​

Solution:

Let the given vertices of a triangle be:

(2a, 2, 6) = (x₁, y₁, z₁)

(-4, 3b, -10) = (x₂, y₂, z₂)

(8, 14, 2c) = (x₃, y₃, z₃)

Centroid = [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3]

= [(2a – 4 + 8)/ 3, (2 + 3b + 14)/3, (6 – 10 + 2c)/3]

According to the given,

By equating the corresponding coordinates,

(4 + 2a)/3 = 0

4 + 2a = 0

2a = -4

a = -4/2 = -2

Now,

(16 + 3b)/3 = 0

16 + 3b = 0

3b = -16

b = -16/3

And

(-4 + 2c)/3 = 0

-4 + 2c = 0

2c = 4

c = 4/2 = 2

Therefore, a = -2, b = -16/3, and c = 2.

Question 18 [6]

Find the equation of the parabola whose focus is (-1, 1) and directrix is 4x + 3y – 24 = 0.

Solution:

Given,

Focus = S(-1, 1)

Equation of directrix (M) is 4x + 3y – 24 = 0

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

Distance from focus to the point = Distance from the directrix to the point

SP = PM

√[(x + 1)² + (y – 1)²] = (4x + 3y – 24)/ √(4² + 3²)

Squaring on both sides,

x² + 1 + 2x + y² + 1 – 2y = (4x + 3y – 24)²/ (16 + 9)

25(x² + y² + 2x – 2y + 2) = 16x² + 9y² + 576 + 24xy – 144y – 192x

25x² + 25y² + 50x – 50y + 50 – 16x² – 9y² – 24xy + 144y + 192x – 576 = 0

9x² + 16y² – 24xy + 242x + 94y – 526 = 0

This is the required equation of parabola.

SECTION C (20 Marks)

Question 19

(a) The mean weight of 100 students in a class is 46 kg. The mean weight of girls is 50 and of the boys is 40 kg. Find the number of girls and boys in the class. [2]

Solution:

Given,

Mean weight of 100 students = 46 kg

Mean weight of girls = 50 kg

Mean weight of boys = 40 kg

Let x and y be the number of girls and boys in the class respectively.

Thus, x + y = 100

x = 100 – y….(i)

According to the given,

46 = (50x + 40y)/ (x + y)

50x + 40y = 46 × 100

50(100 – y) + 40y = 4600 [From (i)]

5000 – 50y + 40y = 4600

5000 – 4600 = 50y – 40y

10y = 400

y = 400/10

y = 40

Substituting y = 40 in (i),

x = 100 – 40 = 60

Hence, the number of girls are 60 and the number of boys are 40.

(b) Calculate P65 for the following distribution. [4]

Solution:

N = 60

P at 65 = p = (65/100) × 50 = 32.5

Cumulative frequency greater than and nearest to 32.5 is 37 which lies in the interval 30 – 40.

Percentile class = 30 – 40

Lower limit of the percentile class = l = 30

Frequency of the percentile class = f = 7

Cumulative frequency of the class preceding the percentile class = cf = 30

Class height = h = 10

P₆₅ = l + [(p – cf)/ f] × h

= 30 + [(32.5 – 30)/ 7] × 10

= 30 + (25/7)

= 30 + 3.57

= 33.57

OR

Find mode from the following distribution.

Solution:

From the given data,

Maximum frequency = 29

Modal class is 10 – 20

Frequency of the modal class = f₁ = 29

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

Frequency of the class succeeding the modal class = f₂ = 18

Lower limit of the modal class = l = 10

Class height = h = 10

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

= 10 + [(29 – 11)/ (2 × 29 – 11 – 18)] × 10

= 10 + [18/ (58 – 29)] × 10

= 10 + (180/ 29)

= 10 + 6.207

= 16.207

Question 20

(a) In a sample of ‘n’ observations given that ∑d² = 22 and rank correlation r = 31/ 42 then find the value of n. [2]

Solution:

Given,

∑d² = 22

Rank correlation = r = 31/42

We know that,

Rank correlation = 1 – [(6 × ∑d²)/ (n(n² – 1)

31/42 = 1 – [(6 × 22)/ (n(n² – 1)]

⇒ 132/ n(n² – 1) = 1 – (31/42)

⇒ 132/ n(n² – 1) = (42 – 31)/ 42

⇒ n(n² – 1) = (42/11) × 132

⇒ n(n² – 1) = 504

⇒ n(n² – 1) = 8 × 63

⇒ n(n² – 1) = 8 × (64 – 1)

⇒ n(n² – 1) = 8 × [(8)² – 1]

⇒ n = 8

(b) Find the correlation coefficient r(x, y) if:

n = 7, ∑x = 19, ∑y = 565, ∑x² = 75, ∑y² = 46775, ∑xy = 1380 [4]

Solution:

Given,

n = 7

∑x = 19

∑y = 565

∑x² = 75

∑y² = 46775

∑xy = 1380

Therefore, correlation coefficient = -1075/ 1159.66 = -0.93

OR

Find Spearman’s rank correlation coefficient for the below data.

Solution:

n = 10

∑d² = 285.5

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

= 1 – [(6 × 285.5)/ 10(100 – 1)]

= 1 – [1713/ 10 × 99]

= 1 – (1713/990)

= (990 – 1713)/ 990

= -723/990

= -0.73

Question 21 [4]

Calculate the Index Number for the following data using the Weighted average of price relatives Method for the year 1995 with respect to 1990 as the base.

Solution:

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

= 11488.75/100

= 114.88

Question 22 [4]

Calculate the 5-year moving average for the following data.

Solution:

Graph: