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ISC Class 11 Maths Mock Sample Question Paper 2 with Solutions
SECTION – A
Question 1 [10×2]
(i) Let f : R → R be defined as f(x) = x^{2} + 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) = x^{2} + 1
f^{1}(7) = {x ∈ R ; f(x) = 7}
Now,
x^{2} + 1 = 7
x^{2} = 7 – 1
x^{2} = 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 + 3i^{2} + i^{3} – (1 – 3i + 3i^{2} – i^{3})
= 1 + 3i + 3i^{2} + i3 – 1 + 3i – 3i^{2} + i^{3}
= 6i + 2i^{3}
= 6i + 2i(i^{2})
= 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 (sin^{2}B – sin^{2}C)
= 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 px^{2} – qx + r = 0, find α^{3} + β^{3}.
Solution:
Given,
α and β are the roots of the equation px^{2} – qx + r = 0.
Sum of the roots = α + β = (q)/p – q/p
Product of the roots = αβ = r/p
α^{3} + β^{3} = (α + β) (α^{2} + β^{2} – αβ)
= (q/p) [α^{2} + β^{2} + 2αβ – 3αβ]
= (q/p) [(α + β)^{2} – 3αβ]
= (q/p) [(q/p)^{2} – 3(r/p)]
= (q/p) [(q^{2}/p^{2}) – (3r/p)]
= (q/p) [(q^{2} – 3pr)/ p^{2}]
= (q^{3} – 3pqr)/ p^{3}
(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) = (x^{2} + 1)/ (x + 1)
Solution:
Given,
f(x) = (x^{2} + 1)/ (x + 1)
Let u = x^{2} + 1 and v = x + 1
⇒ du/dx = d/dx (x^{2} + 1) = 2x
And
dv/dx = d/dx (x + 1) = 1
We know that,
d/dx (u/v) = [v(du/dx) – u(dv/dx)] / v^{2}
= [(x + 1)(2x) – (x^{2} + 1)(1)]/ (x + 1)^{2}
= [2x^{2} + 2x – (x^{2} + 1)]/ (x + 1)^{2}
= (2x^{2} + 2x – x^{2} – 1)/ (x + 1)^{2}
= (x^{2} + 2x – 1)/ (x + 1)^{2}
Therefore, the derivative of a given f(x) is (x^{2} + 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) = y^{2}
⇒ 2x – 3 = 1/y^{2}
⇒ 2x = (1/y^{2}) + 3
⇒ x = (1 + 3y^{2})/ 2y^{2}
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
= 3^{4} – 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)} . 3^{2} – 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
⇒ √[x^{2} + (y + 2)^{2}] < √[(2x)^{2} + (2y + 1)^{2}]
Squaring on both sides,
⇒ x^{2} + y^{2} + 4y + 4 < 4x^{2} + 4y^{2} + 1 + 4y
⇒ y^{2} + 4y + 4 < 3x^{2} + 4y^{2} + 4y + 1
⇒ 3x^{2} + 4y^{2} – y^{2} > 4 – 1
⇒ 3x^{2} + 3y^{2} > 3
⇒ x^{2} + y^{2} > 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 = 16C_{11}
= 16C_{5}
= (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 14C_{9} ways.
= 14C_{5}
= (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 14C_{11} ways.
= 14C_{3}
= (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:
[2x^{2} – (3/x^{3})]^{25}Solution:
Given,
[2x^{2} – (3/x^{3})]^{25}Let (r + 1)th be the term in the given expression which is independent of x.
T_{r+1} = nC_{r} x^{nr} a^{r}
= 25C_{r} (2x^{2})^{25r} (3/x^{3})^{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,
T_{11} = T_{10 + 1}
= (1)^{10} 25C_{10} × 2^{(25 – 10)} × 3^{10}
= 25C_{10} (2^{15} × 3^{10})
Hence, the term independent of x is 25C_{10} (2^{15} × 3^{10}).
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,
a_{1} = 6, b_{1} = 8, c_{1} = 5
a_{2} = 7, b_{2} = 24, c_{2} = 8
a_{1}a_{2} + b_{1}b_{2} = 6(7) + (8)(24)
= 42 – 192
= 150 < 0
Thus, the obtuse angle bisector is given by:
(a_{1}x + b_{1}y + c_{1})/ √(a_{1}^{2} + b_{1}^{2}) = (a_{2}x + b_{2}y + c_{2})/ √(a_{2}^{2} + b_{2}^{2})
6x – 8y + 5/ √[(6)^{2} + (8)^{2}] = (7x + 24y – 8)/ √(7^{2} + 24^{2})
(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 x^{2} + y^{2} – 8x + 4y + 7 = 0.
Solution:
Given equation of circle is:
x^{2} + y^{2} – 8x + 4y + 7 = 0
Comparing with x^{2} + y^{2} + 2gx + 2fy + c = 0
2g = 8, 2f = 4
Centre = (g, f) = (4, 2)
Radius = √(g^{2} + f^{2} – 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} = r^{2}….(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),
⇒ h^{2} + k^{2} + 1 – 2h = r^{2}….(ii)
Substituting (2, 7) in (i),
⇒ h^{2} + k^{2} + 53 – 4h + 14k = r^{2}….(iii)
Substituting (8, 1) in (i),
⇒ (8 – h)^{2} + (1 – k)^{2} = r^{2}
h^{2} + k^{2} + 65 – 16h – 2k = r^{2}….(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),
r^{2} = 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 sin^{2}θ + cos^{2}θ = 1,
[√(1 – sin^{2}A)]/ a = [√(1 – sin^{2}B)]/ bSquaring on both sides,
(1 – sin^{2}A)/a^{2} = (1 – sin^{2}B)/b^{2}
⇒ (1/a^{2}) – (sin^{2}A/a^{2}) = (1/b^{2}) – (sin^{2}B/b^{2})
From (i),
sin^{2}A/a^{2} = sin^{2}B/b^{2}
⇒ 1/a^{2} = 1/b^{2}
⇒ a^{2} = b^{2}
⇒ 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 = (x^{2} – 3x + 4)/ (x^{2} + 3x + 4).
Solution:
Given,
y = (x^{2} – 3x + 4)/ (x^{2} + 3x + 4)
Let dy/dx = 0
dy/dx = [(x^{2} + 3x + 4) (2x – 3) – (x^{2} – 3x + 4)(2x + 3)]/ (x^{2} + 3x + 4)2 = 0
2x^{3} – 3x^{2} + 6x^{2} – 9x + 8x – 12 = 2x^{3} + 3x^{2} – 6x^{2} – 9x + 8x + 12
6x^{2} = 24
x^{2} = 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 kx^{2} +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 kx^{2} +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)
= (k^{2} + m^{2} + 2km)/ km
Therefore, the quadratic equation is:
x^{2} – (sum of the roots)x + (product of roots) = 0
x^{2} – [(lm + kl)/ km]x + [(k^{2} + m^{2} + 2km)/ km] = 0
kmx^{2} + (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,
a_{1} = a
a_{2} = a + d
a_{3} = a + 2d
According to the given,
a_{1} + a_{2} + a_{3} = 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)
a^{2} + d^{2} + 1 + 2ad – 2d – 2a = a^{2} + 2ad + a
(7 – a)^{2} – 3a + 1 – 2(7 – a) = 0
49 + a^{2} – 14a – 3a + 1 – 14 + 2a = 0
a^{2} – 15a + 36 = 0
a^{2} – 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:
[1/(1 × 6)] + [1/(6 × 11)] + [1/(11 × 16)] + …. + [1/(5n – 4)(5n + 1)]Tr = 1/ (5r – 4)(5r + 1)
= (1/5) [1/(5r – 4) – 1/(5r + 1)]
T_{1} = (1/5)[1 – (1/6)]
T_{2} = (1/5) [(1/6) – (1/11)]
….
S_{n} = T_{1} + T_{2} + T_{3} + …. + 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:
Class 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
70 – 80 
80 – 90 
Frequency 
8 
10 
15 
25 
20 
18 
9 
5 
Solution:
Class 
Frequency (f_{i}) 
Class mark (x_{i}) 
f_{i}x_{i} 
f_{i}(x_{i})^{2} 
10 – 20 
8 
15 
120 
1800 
20 – 30 
10 
25 
250 
6250 
30 – 40 
15 
35 
525 
18375 
40 – 50 
25 
45 
1125 
50625 
50 – 60 
20 
55 
1100 
60500 
60 – 70 
18 
65 
1170 
76050 
70 – 80 
9 
75 
675 
50625 
80 – 90 
5 
85 
425 
36125 
Total 
∑f_{i} = 110 
∑f_{i}x_{i} = 5390 
∑f_{i}(x_{i})^{2} = 300350 
Variance = 1/(N – 1) [∑f_{i}x_{i}^{2} – (1/N) (∑f_{i}x_{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 x^{2} = 9y is the abscissa three times that of ordinate?
Solution:
Given equation of parabola is:
x^{2} = 9y
Given that abscissa is three times that of ordinate.
Thus, (3y_{1}, y_{1}) be the point of the given parabola.
Substituting this point in the given equation,
(3y_{1})^{2} = 9(y_{1})
9y_{1}^{2} = 9y_{1}
y_{1}^{2} – y_{1} = 0
y_{1}(y_{1} – 1) = 0
y_{1} = 0 or y_{1} – 1 = 0
y_{1} = 0 or y_{1} = 1
3y_{1} = 3(1) = 3
Hence, the point is (3, 1).
(b) Construct a truth table for (p v q) ⋀ (~p v ~q).
Solution:
p 
q 
~p 
~q 
p v q 
~p v ~q 
(p v q) ⋀ (~p v ~q) 
T 
T 
T 
F 
T 
T 
T 
T 
F 
F 
T 
T 
T 
T 
F 
T 
T 
F 
T 
T 
T 
F 
F 
T 
T 
F 
T 
F 
(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 x^{2}/a^{2} + y^{2}/b^{2} = 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 semimajor axis = 20/2 = 10 units
⇒ a = 10
Now,
c^{2} = a^{2} – b^{2}
(5√3)^{2} = (10)^{2} – b^{2}
b^{2} = 100 – 75 = 25
Therefore, the equation of an ellipse is:
x^{2}/100 + y^{2}/25 = 1
OR
(b) Find the eccentricity, coordinates of the foci, equations of directrices, and length of the latusrectum of the hyperbola 4x^{2} – 3y^{2} = 36.
Solution:
Given equation of hyperbola is:
4x^{2} – 3y^{2} = 36
(4x^{2}/36) – (3y^{2}/36) =1
(x^{2}/9) – (y^{2}/12) = 1
This is of the form (x^{2}/a^{2}) – (y^{2}/b^{2}) = 1,
a^{2} = 9, b^{2} = 12
Now,
b^{2} = a^{2}(e^{2} – 1)
12 = 9(e^{2} – 1)
e^{2} – 1 = 12/9
e^{2} = (4/3) + 1
e^{2} = (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 latusrectum = 2b^{2}/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 yzplane.
Solution:
We know that the xcoordinate is 0 in the yzplane.
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) = (x_{1}, y_{1}, z_{1})
(3, 5, 4) = (x_{2}, y_{2}, z_{2})
m : n = k : 1
Using the section formula,
P(x, y, z) = [(kx_{2} + x_{1})/ (k + 1), (ky_{2} + y_{1})/ (k + 1), (kz_{2} + z_{1})/ (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) = (x_{1}, y_{1}, z_{1})
(4, 3b, 10) = (x_{2}, y_{2}, z_{2})
(8, 14, 2c) = (x_{3}, y_{3}, z_{3})
Centroid = [(x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3, (z_{1} + z_{2} + z_{3})/3]
= [(2a – 4 + 8)/ 3, (2 + 3b + 14)/3, (6 – 10 + 2c)/3]
According to the given,
[(4 + 2a)/3, (16 + 3b)/3, (4 + 2c)/3] = (0, 0, 0)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)/ √(4^{2} + 3^{2})
Squaring on both sides,
x^{2} + 1 + 2x + y^{2} + 1 – 2y = (4x + 3y – 24)^{2}/ (16 + 9)
25(x^{2} + y^{2} + 2x – 2y + 2) = 16x^{2} + 9y^{2} + 576 + 24xy – 144y – 192x
25x^{2} + 25y^{2} + 50x – 50y + 50 – 16x^{2} – 9y^{2} – 24xy + 144y + 192x – 576 = 0
9x^{2} + 16y^{2} – 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]
CI 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
Frequency 
3 
10 
17 
7 
6 
4 
3 
Solution:
Marks (C.I) 
Frequency 
Cumulative frequency 
0 – 10 
3 
3 
10 – 20 
10 
13 
20 – 30 
17 
30 
30 – 40 
7 
37 
40 – 50 
6 
43 
50 – 60 
4 
47 
60 – 70 
3 
50 
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
P_{65} = 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.
Class 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
Frequency 
11 
29 
18 
4 
5 
3 
Solution:
C.I 
Frequency 
0 – 10 
11 
10 – 20 
29 
20 – 30 
18 
30 – 40 
4 
40 – 50 
5 
50 – 60 
3 
From the given data,
Maximum frequency = 29
Modal class is 10 – 20
Frequency of the modal class = f_{1} = 29
Frequency of the class preceding the modal class = f_{0} = 11
Frequency of the class succeeding the modal class = f_{2} = 18
Lower limit of the modal class = l = 10
Class height = h = 10
Mode = l + [(f_{1} – f_{0})/ (2f_{1} – f_{0} – f_{2})] × 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 ∑d^{2} = 22 and rank correlation r = 31/ 42 then find the value of n. [2]
Solution:
Given,
∑d^{2} = 22
Rank correlation = r = 31/42
We know that,
Rank correlation = 1 – [(6 × ∑d^{2})/ (n(n^{2} – 1)
31/42 = 1 – [(6 × 22)/ (n(n^{2} – 1)]
⇒ 132/ n(n^{2} – 1) = 1 – (31/42)
⇒ 132/ n(n^{2} – 1) = (42 – 31)/ 42
⇒ n(n^{2} – 1) = (42/11) × 132
⇒ n(n^{2} – 1) = 504
⇒ n(n^{2} – 1) = 8 × 63
⇒ n(n^{2} – 1) = 8 × (64 – 1)
⇒ n(n^{2} – 1) = 8 × [(8)^{2} – 1]
⇒ n = 8
(b) Find the correlation coefficient r(x, y) if:
n = 7, ∑x = 19, ∑y = 565, ∑x^{2} = 75, ∑y^{2} = 46775, ∑xy = 1380 [4]
Solution:
Given,
n = 7
∑x = 19
∑y = 565
∑x^{2} = 75
∑y^{2} = 46775
∑xy = 1380
Therefore, correlation coefficient = 1075/ 1159.66 = 0.93
OR
Find Spearman’s rank correlation coefficient for the below data.
X 
50 
175 
270 
375 
425 
580 
710 
790 
890 
980 
Y 
180 
120 
200 
100 
100 
120 
80 
60 
100 
85 
Solution:
X 
Rank (R_{1}) 
Y 
Rank (R_{2}) 
d = R_{1} – R_{2} 
d^{2} 
50 
10 
180 
2 
8 
64 
175 
9 
120 
3.5 
5.5 
30.25 
270 
8 
200 
1 
7 
49 
375 
7 
100 
6 
1 
1 
425 
6 
100 
6 
0 
0 
580 
5 
120 
3.5 
1.5 
2.25 
710 
4 
80 
9 
5 
25 
790 
3 
60 
10 
7 
49 
890 
2 
100 
6 
4 
16 
980 
1 
85 
8 
7 
49 
n = 10
∑d^{2} = 285.5
The spearman’s rank correlation coefficient = 1 – [(6 × ∑d^{2})/ n(n^{2} – 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.
Commodity 
Price in 1995 (in Rs.) 
Price in 1990 (in Rs.) 
Weight 
A 
5.20 
4.25 
30 
B 
3.75 
2.95 
40 
C 
1.95 
2.15 
15 
D 
8.10 
8.85 
15 
Solution:
Commodity 
Weights (w) 
Price in Year 1990 (p0) 
Price in Year 1995 (p1) 
Price relative I = (p1/p0) × 100 
Iw 
A 
30 
4.25 
5.20 
122.35 
3670.5 
B 
40 
2.95 
3.75 
127.12 
5084.8 
C 
15 
2.15 
1.95 
90.7 
1360.5 
D 
15 
8.85 
8.10 
91.53 
1372.95 
Total 
∑w = 100 
∑Iw = 11488.75 
Index number of weighted average of price relatives = ∑Iw/ ∑w
= 11488.75/100
= 114.88
Question 22 [4]
Calculate the 5year moving average for the following data.
Year 
1995 
1996 
1997 
1998 
1999 
2000 
2001 
2002 
2003 
2004 
No. of students 
332 
317 
357 
392 
402 
405 
410 
427 
405 
438 
Solution:
Year 
No. of students 
5year moving total 
5year moving average 
1995 
332 
_ 

1996 
317 
_ 

1997 
357 
1800 
360 
1998 
392 
1873 
374.6 
1999 
402 
1966 
393.2 
2000 
405 
2036 
407.2 
2001 
410 
2049 
409.8 
2002 
427 
2085 
417 
2003 
405 
_ 

2004 
438 
_ 
Graph: