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.
Download ISC Class 11 Maths Sample Paper 2018
Download ISC Class 11 Maths Specimen Question Paper 2018 Answers
ISC Class 11 Maths Specimen Question Paper 2018 with Solutions
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 oneone 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 oneone but not onto.
Alternative method:
Let x_{1}, x_{2} be the two elements in the domain R such that f(x_{1}) = f(x_{2}).
⇒ (x_{1} – m)/ (x_{1} – n) = (x_{2} – m)/ (x_{2} – n)
⇒ (x_{1} – m)(x_{2} – n) = (x_{1} – n)(x_{2} – m)
⇒ x_{1}x_{2} – nx_{1} – mx_{2} + mn = x_{1}x_{2} – mx_{1} – nx_{2} + mn
⇒ mx_{1} – nx_{1} = mx_{2} – nx_{2}
⇒ (m – n)x_{1} = (m – n)x_{2}
⇒ x_{1} = x_{2}
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 oneone 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 x^{2} – kx – 21 = 0 and x^{2} – 3kx + 35 = 0 have one common root.
Solution:
Given,
x^{2} – kx – 21 = 0….(i)
x^{2} – 3kx + 35 = 0….(ii)
Let α be the common root of (i) and (ii),
⇒ α^{2} – kα – 21 = 0….(iii)
And
α^{2} – 3kα + 35 = 0….(iv)
Subtracting (iv) from (iii),
α^{2} – kα – 21 – (α^{2} – 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 6P_{5} ways.
Required number of ways = 5! × 6P_{5}
= 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 y^{2} = 4x, the tangent at which is parallel to the straight line y = 2x + 4.
Solution:
Given curve is:
y^{2} = 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 cot^{2}A – cot A – 3 cot A + √3 – √3 cot^{2}A – cot A – 3 cot A – √3)/ (3 cot^{2}A – 1)
= cot A + [(8 cot A)/ (3 cot^{2}A – 1)
= (3 cot^{3}A – 9 cot A)/ (3 cot^{2}A – 1)
= 3[(cot^{3}A – 3 cot A)/ (3 cot^{2}A – 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 acuteangled 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 rightangled 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,
x^{2} + (y – 3)^{2} ≤ 2 [x^{2} + (y + 3)^{2}]
x^{2} + y^{2} + 9 – 6y ≤ 2(x^{2} + y^{2} + 9 + 6y)
x^{2} + y^{2} + 9 – 6y ≤ 2x^{2} + 2y^{2} + 18 + 12y
⇒ 2x^{2} + 2y^{2} + 18 + 12y – x^{2} – y^{2} – 9 + 6y ≥ 9
⇒ x^{2} + y^{2} + 18y + 9 ≥ 0
⇒ x^{2} + (y^{2} + 18y + 81) + 9 ≥ 81
⇒ x^{2} + (y + 9)^{2} ≥ 81 – 9
⇒ x^{2} + [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 = 8C_{4} × 4! = 70 × 24 = 1680
Case 2: 2 pairs of repeated letters (for example IIOO)
In this case, the number of words can be formed = 3C_{2} × (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 = 3C_{1} × 7C_{2} × (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 = 9C_{4} × 4C_{3}
= [(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 = 4C_{3} × 9C_{4} + 4C_{4} × 9C_{3}
= 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 = 4C_{0} × 9C_{7} = 1 × 9C_{2} = (9 × 8)/ (2 × 1) = 36
1 girls + 6 boys = 4C_{1} × 9C_{6} = 4 × 9C_{3} = 4 × [(9 × 8 × 7)/ (3 × 2 × 1)] = 336
2 girls + 5 boys = 4C_{2} × 9C_{5} = [(4 × 3)/ (2 × 1)] × 9C_{4} = 6 × 126 = 756
3 girls + 4 boys = 4C_{3} × 9C_{4} = 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)]
= (k^{2} + 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:
[(√x/ √3) – (√3/ 2x)]^{12}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)/2 – r] = 012 – r – 2r = 0
3r = 12
r = 4
Thus, the fourth term will be the required term.
T_{4 + 1} = 12C_{4} (3)^{ (12 – 4)/2 }(3)^{4/2 }2^{4}
= 12C_{4} [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}] = √(1^{2} + 1^{2})
= √(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} = r^{2}
(x – h)^{2} + (y – k)^{2} = 5^{2} (given radius = 5)
Also, 3x – 4y + 5 = 0 is the tangent to the circle.
⇒ 3h – 4k + 5/√(3^{2} + 4^{2}) = 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 = (a^{2} + b^{2} + c^{2})/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,
a^{2} = b^{2} + c^{2} – 2bc cos A
b^{2} = a^{2} + c^{2} – 2ac cos B
c^{2} = a^{2} + b^{2} – 2ab cos C
Now,
a^{2} + b^{2} + c^{2} = 2a^{2} + 2b^{2} + 2c^{2} – 2ab cos C – 2ac cos B – 2bc cos A
⇒ a^{2} + b^{2} + c^{2} = 2ab cos C + 2ac cos B + 2bc cos A….(iv)
From (i), (ii), (iii), and (iv),
a^{2} + b^{2} + c^{2} = 2 [(2Δ/sin C) cos C + (2Δ/sin B) cos B + (2Δ/sin A) cos A]
a^{2} + b^{2} + c^{2} = 2 × 2Δ (cot A + cot B + cot C)
⇒ cot A + cot B + cot C = (a^{2} + b^{2} + c^{2})/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 + 2n^{2} + 3n – 2n – 3
= 2n^{2} + 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 t_{1} 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.
t_{1} × r(p + q – 1) = a….(i)
t_{1} × r(p – q – 1) = b….(ii)
Multiplying (i) and (ii),
t_{1} × r^{(p + q – 1)} × t_{1} × r^{(p – q – 1)} = ab
(t_{1})^{2} × r^{(p + q – 1 + p – q – 1)} = ab
(t_{1})^{2} × r^{(2p – 2)} = ab
(t_{1})^{2} × r^{2(p – 1)} = ab
(t_{1} × r^{p – 1})^{2} = ab
t_{1} × 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)
x^{2} + 3x – x – 3 = y(x^{2} + 4x – 2x – 8)
x^{2} + 2x – 3 – x^{2}y – 2xy + 8y = 0
x^{2}(1 – y) + 2x(1 – y) + (8y – 3) = 0
Given that x is real.
Therefore, discriminant ≥ 0
[2(1 – y)]^{2} – 4(1 – y)(8y – 3) ≥ 04(1 + y^{2} – 2y) – 4(8y – 3 – 8y^{2} + 3y) ≥ 0
4 + 4y^{2} – 8y – 44y + 12 + 32y^{2} ≥ 0
36y^{2} – 52y + 16 ≥ 0
4(9y^{2} – 13y + 4) ≥ 0
9y^{2} – 13y + 4 ≥ 0
9y^{2} – 9y – 4y + 4 ≥ 0
9y(y – 1) – 4(y – 1) ≥ 0
(9y – 4)(y – 1) ≥ 0
(y – 4/9)(y – 1) ≥ 0
Factor 
y < 4/9 
y ∈ (4/9, 1) 
y > 1 
y – (4/9) 
ve 
+ve 
+ve 
y – 1 
ve 
ve 
+ve 
[y – (4/9)] (y – 1) 
+ve 
ve 
+ve 
Therefore, y cannot lie between 4/9 and 1.
OR
Solution:
Given,
[x^{2} + (1/x)]^{2n}General term is:
T_{r + 1} = 2nC_{r} (x^{2})^{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:
Age 
20 – 25 
25 – 30 
30 – 35 
35 – 40 
40 – 45 
45 – 50 
No. of persons 
170 
110 
80 
45 
40 
35 
Solution:
Class 
Frequency (f_{i}) 
Class mark (x_{i}) 
f_{i}x_{i} 
f_{i}(x_{i})^{2} 
20 – 25 
170 
22.5 
3825 
86062.5 
25 – 30 
110 
27.5 
3025 
83187.5 
30 – 35 
80 
32.5 
2600 
84500 
35 – 40 
45 
37.5 
1687.5 
63281.25 
40 – 45 
40 
42.5 
1700 
72250 
45 – 50 
35 
47.5 
1662.5 
78968.75 
Total 
∑f_{i} = 480 
∑f_{i}x_{i} = 14500 
∑f_{i}(x_{i})^{2} = 468250 
Variance = 1/(N – 1) [∑f_{i}x_{i}^{2} – (1/N) (∑f_{i}x_{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 5x^{2} = 12y.
Solution:
5x^{2} = 12y
x^{2} = (12/5)y
x^{2} = 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 yaxis.
Therefore, the focus lies at a distance p from the centre (0, 0) along the yaxis.
i.e. (0, 0 + p)
= (0, 0 + (3/5))
= (0, 3/5)
Since the parabola is symmetric around the yaxis, the directrix is a line parallel to the xaxis 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:
p 
q 
~p 
~q 
~p ⋀ ~q 
p ⋀ ~q 
(~p ⋀ ~q) ⋁ (p ⋀ ~q) 
T 
T 
F 
F 
F 
F 
F 
T 
F 
F 
T 
F 
T 
T 
F 
T 
T 
F 
F 
F 
F 
F 
F 
T 
T 
T 
F 
T 
(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 = (b^{2} + c^{2} – a^{2})/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 (x_{1}, y_{1}) to the line ax + by + c = 0 = ax1 + by1 + c/ √(a^{2} + b^{2})
Here,
(x_{1}, y_{1}) = (x, y)
a = 2, b = 1, c = 1
And
SP = e (perpendicular distance)
SP^{2} = e^{2} (perpendicular distance)^{2}
(x – 1)^{2} + (y – 1)^{2} = (√3)^{2} [2x + y – 1/ √(2^{2} + 1^{2})]^{2}
x^{2} – 2x + 1 + y^{2} – 2y + 1 = (3) [2x + y – 1^{2}/ (4 + 1)]
x^{2} + y^{2} – 2x – 2y + 2 = (3/5) (4x^{2} + y^{2} + 1 + 4xy – 2y – 4x)
5x^{2} + 5y^{2} – 10x – 10y + 10 = 12x^{2} + 3y^{2} + 3 + 12xy – 6y – 12x
12x^{2} + 3y^{2} + 3 + 12xy – 6y – 12x – 5x^{2} – 5y^{2} + 10x + 10y – 10 = 0
7x^{2} – 2y^{2} + 12xy – 2x + 4y – 7 = 0
Therefore, the equation of the hyperbola is 7x^{2} – 2y^{2} + 12xy – 2x + 4y – 7 = 0.
OR
Find the equation of tangents to the ellipse 4x^{2} + 5y^{2} = 20 which are perpendicular to the line 3x + 2y – 5 = 0.
Solution:
Given,
4x^{2} + 5y^{2} = 20
(4x^{2}/20) + (5y^{2}/20) = 1
(x^{2}/5) + (y^{2}/4) = 1
This is of the form (x^{2}/a^{2}) + (y^{2}/b^{2}) = 1
a^{2} = 5 and b^{2} = 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 ± √(a^{2}m^{2} + b^{2})
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 16x^{2} – 3y^{2} – 32x – 12y – 44 = 0 represents a hyperbola. Find the lengths of axes and eccentricity.
Solution:
Given,
16x^{2} – 3y^{2} – 32x – 12y – 44 = 0
(16x^{2} – 32x) – (3y^{2} + 12y) – 44 = 0
16(x^{2} – 2x) – 3(y^{2} + 4y) – 44 = 0
16(x^{2} – 2x + 1) – 3(y^{2} + 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 a^{2} = 3 and b^{2} = 16.
Centre = (1, 2)
And
b^{2} = a^{2}(e^{2} – 1)
16 = 3(e^{2} – 1)
e^{2} – 1 = 16/3
e^{2} = (16/3) + 1
e^{2} = 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 n_{1} and n_{2} be the sizes of two samples.
X_{1} and X_{2} are the means of two samples.
According to the given,
n_{1} = 50 and n_{2} = 100
X_{1} = 56 and X_{2} = 50
New sample size = n_{1} + n_{2} = 50 + 100 = 150
New mean = (n_{1}X_{1} + n_{2}X_{2})/ (n_{1} + n_{2})
= [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:
Marks 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
Frequency 
3 
7 
11 
12 
23 
4 
Solution:
Marks (C.I) 
Frequency 
Cumulative frequency 
0 – 10 
3 
3 
10 – 20 
7 
10 
20 – 30 
11 
21 
30 – 40 
12 
33 
40 – 50 
23 
56 
50 – 60 
4 
60 
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
P_{95} = 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.
C.I. 
17 – 19 
14 – 16 
11 – 13 
8 – 10 
5 – 7 
2 – 4 
Frequency 
12 
4 
8 
16 
11 
4 
Solution:
Let us arrange the data as shown below:
C.I 
Frequency 
1.5 – 4.5 
4 
4.5 – 7.5 
11 
7.5 – 10.5 
16 
10.5 – 13.5 
8 
13.5 – 16.5 
4 
16.5 – 19.5 
12 
From the given data,
Maximum frequency = 16
Modal class is 7.5 – 10.5
Frequency of the modal class = f_{1} = 16
Frequency of the class preceding the modal class = f_{0} = 11
Frequency of the class succeeding the modal class = f_{2} = 8
Lower limit of the modal class = l = 7.5
Class height = h = 3
Mode = l + [(f_{1} – f_{0})/ (2f_{1} – f_{0} – f_{2})] × 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:
Marks in Mathematics 
36 
48 
27 
36 
29 
30 
36 
39 
42 
48 
Marks in Statistics 
27 
45 
24 
27 
31 
33 
35 
45 
41 
45 
Solution:
Mathematics 
Rank (R_{1}) 
Statistics 
Rank (R_{2}) 
d = R_{1} – R_{2} 
d^{2} 
36 
6 
27 
8.5 
2.5 
6.25 
48 
1.5 
45 
2 
0.5 
0.25 
27 
10 
24 
10 
0 
0 
36 
6 
27 
8.5 
2.5 
6.25 
29 
9 
31 
7 
2 
4 
30 
8 
33 
6 
2 
4 
36 
6 
35 
5 
1 
1 
39 
4 
45 
2 
2 
4 
42 
3 
41 
4 
1 
1 
48 
1.5 
45 
2 
0.5 
0.25 
n = 10
∑d^{2} = 27
The spearman’s rank correlation coefficient = 1 – [(6 × ∑d^{2})/ n(n^{2} – 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:
x 
21 
24 
26 
29 
32 
43 
25 
30 
35 
37 
y 
120 
123 
125 
128 
131 
142 
124 
129 
134 
136 
Solution:
x 
(x – mean) (x – 30.2) d_{x} 
d^{2}_{x} 
y 
(y – mean) (y – 129.2) d_{y} 
d^{2}_{y} 
d_{x} d_{y} 
21 
9.2 
84.64 
120 
9.2 
84.64 
84.64 
24 
6.2 
38.44 
123 
6.2 
38.44 
38.44 
26 
4.2 
17.64 
125 
4.2 
17.64 
17.64 
29 
1.2 
1.44 
128 
1.2 
1.44 
1.44 
32 
1.8 
3.24 
131 
1.8 
3.24 
3.24 
43 
12.8 
163.84 
142 
12.8 
163.84 
163.84 
25 
5.2 
27.04 
124 
5.2 
27.04 
27.04 
30 
0.2 
0.04 
129 
0.2 
0.04 
0.04 
35 
4.8 
23.04 
134 
4.8 
23.04 
23.04 
37 
6.8 
46.24 
136 
6.8 
46.24 
46.24 
∑x = 302 
∑d_{x} = 0 
∑d^{2}_{x} = 405.6 
∑y = 1292 
∑d_{y} = 0 
∑d^{2}_{y} = 405.6 
∑d_{x} d_{y} = 405.6 
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})/ (√∑d^{2}_{x} ∑d^{2}_{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:
Items 
Food 
Rent 
Cloth 
Fuel 
Prive in 2005 (Rs.) 
20 
100 
150 
50 
Price in 2007 (Rs.) 
280 
200 
120 
100 
Weighted 
30 
20 
20 
10 
Solution:
Items 
Weights (w) 
Year 2005 (p_{0}) 
Year 2007 (p_{1}) 
Price relative I = (p_{1}/p_{0}) × 100 
Iw 
Food 
30 
20 
280 
1400 
42000 
Rent 
20 
100 
200 
200 
4000 
Cloth 
20 
150 
120 
80 
1600 
Fuel 
10 
50 
100 
200 
2000 
Total 
∑w = 80 
∑Iw = 49600 
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:
Year/ Quarter 
Q1 
Q2 
Q3 
Q4 
1994 
29 
37 
43 
34 
1995 
90 
42 
55 
43 
1996 
47 
51 
63 
53 
Solution:
Year 
Quarter 
Quarterly moving total 
Quarterly moving average 
Centered moving average (trend values) 

1994 
I 
29 

1994 
II 
37 

143 
35.75 

1994 
III 
43 
43.375 

204 
51 

1994 
IV 
34 
51.625 

209 
52.25 

1995 
I 
90 
53.75 

221 
55.25 

1995 
II 
42 
56.375 

230 
57.5 

1995 
III 
55 
52.125 

187 
46.75 

1995 
IV 
43 
47.875 

196 
49 

1996 
I 
47 
50 

204 
51 

1996 
II 
51 
52.25 

214 
53.5 

1996 
III 
63 

1996 
IV 
53 
Trend: