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.

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

 

isc class 11 maths specimen question paper with solutions 2018 01
isc class 11 maths specimen question paper with solutions 2018 02
isc class 11 maths specimen question paper with solutions 2018 03
isc class 11 maths specimen question paper with solutions 2018 04
isc class 11 maths specimen question paper with solutions 2018 05
isc class 11 maths specimen question paper with solutions 2018 06
isc class 11 maths specimen question paper with solutions 2018 07
isc class 11 maths specimen question paper with solutions 2018 08
isc class 11 maths specimen question paper with solutions 2018 09
isc class 11 maths specimen question paper with solutions 2018 10
isc class 11 maths specimen question paper with solutions 2018 11
isc class 11 maths specimen question paper with solutions 2018 12
isc class 11 maths specimen question paper with solutions 2018 13
isc class 11 maths specimen question paper with solutions 2018 14
isc class 11 maths specimen question paper with solutions 2018 15
isc class 11 maths specimen question paper with solutions 2018 16
isc class 11 maths specimen question paper with solutions 2018 17
isc class 11 maths specimen question paper with solutions 2018 18
isc class 11 maths specimen question paper with solutions 2018 19
isc class 11 maths specimen question paper with solutions 2018 20
isc class 11 maths specimen question paper with solutions 2018 21
isc class 11 maths specimen question paper with solutions 2018 22
isc class 11 maths specimen question paper with solutions 2018 23
isc class 11 maths specimen question paper with solutions 2018 24
isc class 11 maths specimen question paper with solutions 2018 25
isc class 11 maths specimen question paper with solutions 2018 26
isc class 11 maths specimen question paper with solutions 2018 27
isc class 11 maths specimen question paper with solutions 2018 28
isc class 11 maths specimen question paper with solutions 2018 29

 

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]

ISC Class 11 maths 2018 Sp Q 1 (ii) sol

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

ISC class 11 maths 2018 Sp Q 1(iii) sol

= ± [(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)

ISC Class 11 maths 2018 SP Q 1(ix)

Solution:

ISC Class 11 maths 2018 SP Q 1(ix) sol

(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.

ISC class 11 maths 2018 SP Q 2 sol

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.

ISC Class 11 maths 2018 SP Q 3 (b) case 1

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.

ISC Class 11 maths 2018 SP Q 3 (b) case 2

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.

ISC Class 11 maths 2018 SP Q 3 (b) case 3

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

ISC Class 11 maths 2018 SP Q 4 sol

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.

ISC Class 11 maths 2018 SP Q 4 sol graph

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,

ISC Class 11 maths 2018 SP Q 6 sol

= [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

ISC Class 11 maths 2018 SP Q 7(a)

Solution:

[(√x/ √3) – (√3/ 2x)]12

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

Tr + 1 = nCr (a)n – r (a)r

Tr + 1 = 12Cr (√x/ √3)12 – r . (-√3/ 2x)r

= 12Cr (x/3) (12 – r)/2 . (-3)r/2 [1/ 2r xr]

= 12Cr (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] = 0

12 – r – 2r = 0

3r = 12

r = 4

Thus, the fourth term will be the required term.

T4 + 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:

ISC Class 11 maths 2018 SP Q 7(b)

Solution:

ISC Class 11 maths 2018 SP Q 7(b) sol

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) = limh → 0 [f(x + h) – f(x)]/ h

ISC class 11 maths 2018 SP Q 8 sol

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

ISC class 11 maths 2018 SP Q 11 sol

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 Sn = 4 + 11 + 22 + 37 + 56 + … + an – 1 + an (n terms)

⇒ 0 = 4 + 7 + 11 + 15 + 19 + … + (an – an – 1) – an (n + 1 terms)

⇒ an = 4 + 7 + 11 + 15 + 19 + … + (an – an – 1)

7 + 11 + 15 + 19 + … + (an – an – 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]

an = 4 + (n – 1)(7 + 2n – 4)

an = 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 × r2(p – 1) = ab

(t1 × rp – 1)2 = ab

t1 × rp – 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

[2(1 – y)]2 – 4(1 – y)(8y – 3) ≥ 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

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

[y – (4/9)](y – 1) is positive only when y < 4/9 or y > 1.

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

OR

ISC Class 11 maths 2018 SP Q 13(b)

Solution:

Given,

[x2 + (1/x)]2n

General term is:

Tr + 1 = 2nCr (x2)2n – r (1/x)r

= 2nCr x4n – 2r x(-r)

= 2nCr x(4n – 2r – r)

= 2nCr x(4n – 3r)

Let xp occur in this expansion.

⇒ 4n – 3r = p

⇒ 3r = 4n – p

⇒ r = (4n – p)/3

Coefficient of xp = 2nCr

= (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 (fi)

Class mark (xi)

fixi

fi(xi)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

∑fi = 480

∑fixi = 14500

∑fi(xi)2 = 468250

Variance = 1/(N – 1) [∑fixi2 – (1/N) (∑fixi)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:

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):

ISC Class 11 maths 2018 SP Q 16 sol (a) i

= 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):

ISC Class 11 maths 2018 SP Q 16 sol (a) ii

= 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,

ISC Class 11 maths 2018 SP Q 16 (b) sol

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:

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

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.

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 = 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:

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 (R1)

Statistics

Rank (R2)

d = R1 – R2

d2

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

∑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:

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)

dx

d2x

y

(y – mean)

(y – 129.2)

dy

d2y

dx dy

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

∑dx = 0

∑d2x = 405.6

∑y = 1292

∑dy = 0

∑d2y = 405.6

∑dx dy = 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 = (∑dx dy)/ (√∑d2x ∑d2y)

= 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 (p0)

Year 2007 (p1)

Price relative

I = (p1/p0) × 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:

ISC Class 11 maths 2018 SP Q 22 sol

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class