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HIGHER MATH

HIGHER MATHS CLASS - 1 TO 3 INSTRUCTOR’S NOTE HIGHER MATHS CLASS - 1 ANSWER KEY 1-[4]

2-[1]

3-[4]

4-[4]

5-[3]

6-[3]

7-[4]

8-[1]

9-[1]

10-[3]

11-[1]

12-[3]

13-[4]

14-[1]

15-[4]

16-[1]

17-[1]

18-[2]

19-[1]

20-[1]

21-[1]

22-[4]

23-[3]

24-[4]

25-[3]

26-[4]

27-[3]

28-[4]

29-[3]

30-[3]

31-[3]

32-[2]

33-[1]

34-[2]

35-[1]

36-[1]

37-[2]

38-[3]

HIGHER MATHS CLASS - 2 1.

I. a, b, c are in A.P. \ –a, –b, –c are in A.P. adding a + b + c to this A.P. \ b + c, a + c, a + b are also in A.P. Thus, I is true. II. a, b, c are in A.P.

II. a, b, c are in A.P. then b – a = c – b. ( b –

Þ

a b c , , are in A.P. (Dividing by abc.) abc abc abc

a)=( c +

ab + bc + ac ab + bc + ac ab + bc + ac – 1, – 1, – bc ca ab 1 are in A.P. .... (subtracting 1 from each term.)

=

=

b+ c ( b + c )( c + a ) c+ a ( b + a )( c + a )

\

ab + ac ab + bc bc + ac , , .... are in A.P. bc ac ab

Þ

a ( b + c) b ( a + c) c ( a + b ) \ bc , ac , ab ..... are in A.P. Thus, II is true. Hence, [3].

1 c+ a 1 b+ c

,

b)

b+ a c+ a- a- b b+ a

Divide both sides by ( c +

\

b )( c –

c- b

=

b + c

b+ c

ab + bc + ac ab + bc + ac ab + bc + ac , , are in A.P. bc ca ab (Multiplying by (ab + bc + ac.)

2.

b - a

b+ c- c- a

1 1 1 , , are in A.P. bc ac ab

\

a )( b +

1 b+ c 1 c+ a

a) c+ a

( b + c )( c + a ) a+ b

( b + a )( c + a )

=

and

1 b+ a 1 b+ a

1 c+ a

are in A.P.

i.e., b + c , c + a and b + a are in H.P. Thus, II is also true. Hence, [3].

I. To prove that bc, ca and ab in H.P. 3.

1 1 1 , , to be in A.P. ab bc ca Since a, b and c are in A.P.

One has to prove

a b c , , are in A.P. abc abc abc

\

1 1 1 , and are in A.P. bc ca ab

Þ

bc, ca and ab are in H.P. Thus, I is true.

Amount paid in the first month = Rs.1000/Every month the amount is increased by Rs.50/\ Total amount paid in 30 months = Sum of 30 terms of A. P. with first term is 1000 and Common difference. 50 30 [1000 + 1000 + (29 × 50)] 2 = 15[2000 + 1450] = 15 × 3450 = Rs.51750. Hence, [3].

=

1

IMS-34-UG-AL-HM-Class-1-2-3

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HIGHER MATH

4.

(1 - r 8 ) a S8 1- r = (1 - r 4 ) S4 a 1- r

1 n + 2 – 2 n æ = 2(n + 2) = 2(n + 2) = 2 ç1 + 2 ö÷ nø è

(1 - r 4 )(1 + r 4 ) 97 = (1 - r 4 ) 81

1 2 Hence, [2]. \ S¥

97 = 1 + r4 81

5.

8.

a = 20; d = 4; S n = 1064 n [2a + (n - 1) d] 2

r4 =

97 97 – 1 = – 1 81 81

Sn =

r4 =

16 2 => r = + . Hence, [2]. 81 3

1064 =

a, b and c are in A.P. \ 2b = a + c x is the GM of a and b \ x2 = ab y is the GM of b and c \ y2 = bc x2 + y2 = ab + bc = b(a + c) = b(2b) = 2b2 \ b2 is the AM between x2 and y2. Hence, [1].

7.

Let the first n terms of the series be T1, T2 ... and T n and S n be the sum of the first n terms. 1 1 T n = (n + 1) (n + 2) = (n + 1)

9.

10.

Let the four numbers be (a – 3d ), (a – d), (a + d), (a + 3d) (a – 3d ) + (a – d) + (a + d) + (a + 3d) = 40 \ 4a = 40 \ a = 10 (a – 3d) (a + 3d) = 64 a2 – 9d 2 = 64 9d2 = 36 \ d2 = 4 \ d = ±2 \ The numbers are 4, 8, 12 and 16. Hence, [3]. Alternatively, look at the options and solve.

11.

The reciprocals of the terms are 2 13 9 23 , , , - - - - - - etc , 5 20 10 20

1 – (n + 2)

These terms form an Arithmetic progression \ of which a = \ Tn =

1 n

1 n +1

=

1 1 Tn = – n +1 n + 2 Sn =

n [2a + (n - 1) d] = n [2 ´ 1 + (n - 1) 2] . 2 2

= n + n(n – 1) = n 2 Hence, [2].

1 1 – 3 4

Similarly, T n–1 =

The sequence is 1, 3, 5, 7, — a = 1, d = 2 Sn =

1 1 T1 = – 2 3 T2 =

n [2 ´ 20 + (n - 1) 4] = 20n + 2n 2 – 2n 2

2n 2 + 18n – 1064 = 0 n 2 + 9n – 532 = 0 (n + 28) (n – 19) = 0 \ n = – 28 or n = + 19 n = – 28 is not possible ; \ n = 19. Hence, [4]

Let a1 be the first term and d be the common difference of the A.P. \ a5 = a1 + 4d and a10 = a1 + 9d \ 5a5 = 10a10 \ 5(a1 + 4d) = 10(a1 + 9d) \ 5a1 + 20d = 10a1 + 90d \ 5a1 = –70d \ a1 = –14d \ a1 + 14d = 0 But a1 + 14d = a15 i.e., a 15 = 0. Hence, [3].

6.

=

1 1 – 2 n + 2

IMS-34-UG-AL-HM-Class-1-2-3

2

2 and d = 13 - 2 = 5 = 1 5 20 5 20 4

1 1 = 2 1 a + (n - 1) d + (n - 1) 5 4

1 3 n + 20 4

=

20 . Hence, [2]. 3 + 5n

®

HIGHER MATH 12.

Case 2: P = 3x x + 2x + 3x + (3x) = 360 o Þ x = 40 o Angles are 40 o, 80 o, 120 o, 120 o; Reject it.

a = 10, d = 5, n = 120 n [2a + (n - 1) d ] Sn = 2 S120 =

13.

120 [2 ´ 10 + 119 ´ 5] 2

Case 3: P = 2x \ x + 2x + 3x + (2x) = 8x = 360 o Þ x = 45 o \ angles are 45 o , 90 o, 135 o, 90 o. Thus, the smallest angle = 45 o. Hence, [2].

= 60 [20 + 595] = 36900 meters = 36.9 km. Hence, [3]. First the ball falls through a distance of 24 m. Height to which it rebound =

2 (24) . 3

It falls through the same distance i.e.,

16.

2 (24) . 3

2

(24) é2

= 24 + 2 ´ ê

êë 3

Þ 5

and so on \ Total distance travelled

(24) +

æ2ö ç ÷ è3ø

2

(24) +

ù - - - - - ¥ú úû

5

= log 5x (By definition of logarithm.)

2

(ii) log 327 + log 3x = 4 Þ log327x = 4

Þ Þ

17.

é2 ù ´ 24 ´ 3ú = 24 + 2(48) = 120 m. ë3 û

= 24 + 2 ê

27x = 3 4 = 81 x = 3. Hence, [1].

log 73 54 × log 52 493 =

Hence,[3].

.15.

(log 5 x) = 2

= x i.e. x = 5 5 = 3125 Hence, [3].

é2 ù a ù é ê ´ 24 ú = 24 + 2 ê 3 ú ..... êS¥ = 1 - r ú 2 ë û ê1ú 3 ûú ëê

14.

5

Þ ( 5)2

Again, the ball rebound to and falls back through æ2ö 2´ç ÷ è3ø

log

(i)

4 6 log 7 5 × log 5 7 3 2

æ ö a ççQ log bβ a α = log b a ÷÷ b è ø

There are two possibilities, either the smallest angle is 60 o or the middle one is 60 o. If the largest is 60 o then the others will obviously be 60 o so that the sum is 180o. Now, if the smallest is 60o then 60 + 60r + 60r 2 = 180. Solving this equation we get r = 1 or –2. But r = –2 is not possible. Hence, r = 1 and the triangle is equilateral. If middle angle is of 60 o then A + C = 120, AC = 3600. Again, A = C = B = 60o. Hence, the triangle is equilateral triangle. Hence, [1].

4 6 × × log 75 × log57 3 2 = 4 ( Q logab × log ba = 1). Hence, [4].

=

18.

Let the 3 angles (in AP) be x, x + a, x + 2a Given that (x + 2a) – (x + a) = x i.e. 2a – a = x i.e. a = x Hence the angles are of the form x, 2x, 3x let the 4 th angle be P. Three cases exist when P is equal to one of x, 2x and 3x respectively.

75 5 32 – 2log + log 16 9 243 = log 75 – log 16 – 2(log 5 – log 9) + log 32 – log 243 = log 75 – log 16 – log 25 + log 81 + log 32 – log 243 = log 25 + log 3 – log 16 – log 25 + log 81 + log 2 + log 16 – log 81 – log 3 = log 2 Alternatively,

log

log

Case 1: P = x x + 2x + 3x + (x) = 360 o Þ 7x = 360 o Þ x = 51.4 Hence the angles are 51.4, 51.4, (51.4) × 2, (51.4) × 3 Since no angle is 90 o, we reject this possibility.

75 5 32 – 2log + log 16 9 243

= log

75 25 32 – log + log 16 81 243

81 32 ö æ 75 ´ ´ ÷ = log ç 16 25 243 è ø = log2. Hence, [1].

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IMS-34-UG-AL-HM-Class-1-2-3

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HIGHER MATH

19.

log 2(9 – 2 x ) = 10 log

10

( 3 – x)

= 3 – x (Q blog n = n) b

25.

Þ 2 3 – x = 9 – 2 x (By definition of logarithm.) Þ 23 – x + 2x = 9 Þ x = 0 ( Q RHS is an odd number and the only power of 2 which gives an odd number is 0) and for log 10(3 – x) to be meaningful, 3 – x > 0 i.e., x < 3. Hence, [1]. Alternatively, Since x < 3 and the only option satisfying this is [1].

21.

26.

log

=

2 log

( 8)

2

8

=

8 4 log

8

8

32

5 4 ´ log

52

4 4 log 5 ´ log 3 3 5 2 2

We know that log ab = k Þ b = a k

Þ log 2 (log 3 (log x)) = ( 2 )2 = 2

x

(70 2

log

Þ log 3 (log x) = ( 2 )2 = 2

- 33x ) = 1

Þ log x = ( 3 )2 = 3

\ x = e3. Hence, [4].

= 70 - 33x

3

( 3 )4

é êlog 5 ë

9 = log

5

5

( 5)

= log 10(1) = 0. Hence, [3].

x

29.

=

3

33 = 5 + x 5 + x = 27 x = 27 – 5 = 22. Hence, [1].

1 81

\ 3x = – 4

30.

2 log

10

log log

10 10

\ x = - 4 . Hence, [1]. 3

6 = 0.7782

10

8

log (5 + x ) = 3

33x = 3-4

log

3

3

= 625 5 . Hence, [1].

1ö ÷ = x è 81 ø

(27 )x

log (5 + x ) + log 8 = 2 2 log (5 + x ) + 1 = 4

æ Let log 27 ç

\

log10 {log 95 × log 59] é log 5 log 9 ù = log 10 ê log 9 ´ log 5 ú ë û

x 9

\ x =

28.

ù xú = 4 û

= log

8 = 2. 4

34

2 (1 + 2 + 3 + … + n) = n(n+1) = n 2 + n. Hence, [4]. log

=

= 2( log3 5 ´ log5 3) = 2 × 1 = 2. Hence, [4].

x 2 + 33x – 70 = 0 (x + 35) (x – 2) = 0 \ x = –35 or x = 2 Hence, [1].

24.

8´1

=

\ log 2 [log 2 (log 3 (log x))] = 2

2 1

23.

8

8

log 3 3 + log 3 9 + log 3 27 = 2 + 4 + 6 + 8 + … =

(x )

22.

8

2 log

Hence, [2].

27. 20.

8 log 8

6 = 2 × 0.7782

6 2 = 2 × 0.7782 36 = 1.5564. Hence, [3].

IMS-34-UG-AL-HM-Class-1-2-3

4

log(x + 3) + log(x + 5) = log35 log[(x + 3) (x + 5)] = log35 (x + 3) (x + 5) = 35 x 2 + 8x + 15 = 35 x 2 + 8x – 20 = 0 (x + 10) (x – 2) = 0 \ x = –10 or x = 2 But x cannot be –10 as log(x + 3) will be log(–7) and log of negative numbers is not defined \ x = 2 Hence, [1].

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HIGHER MATH

HIGHER MATHS CLASS - 3 1.

2.

3.

4.

5.

f[g(x)] = f[2x + 4] = (2x + 4) + 5 = 2x + 9 g[f(x)] = g[x + 5] = 2(x + 5)+ 4 = 2x + 10 + 4 = 2x + 14 \ f[g(x)] – g[f(x)] = (2x + 9) – (2x + 14) = 2x + 9 – 2x – 14 = –5. Hence, [3]. f(x) = ax 2 + bx + 2 f(1) = a(1) 2 + b(1) + 2 = 3 a + b + 2 = 3 a + b = 1 …….. (i) f(3) = a ´ (3) 2 + b(3) + 2 = –7 9a + 3b + 2 = –7 9a + 3b = –9 3a + b = –3 …….. (ii) Solving (i) and (ii), a = –2 and b = 3. Hence, [2]. f(x + 1) 1 + (x + 1 + x + = 1 + x 2 + x – –x + 3x 2x = –2

= f(x+ 2) 1) – (x + 1)2 = 1 + (x + 2) – (x + 2) 2 1 – (x 2 + 2x + 1) + 2 – (x 2 + 4x + 4) x 2 – 2x – 1 = 3 + x – x 2 – 4x – 4 = –1 –1 \ x = –1. Hence, [3].

f(x) = (x – 1) (x – 2) (x + 3) for –2 < x < 4 \ f(x) = 0 when x = 1, or x = 2 x = –3 is not considered as it does not satisfy –2 < x < 4. f(x) = (5 – x) for all other values of x. \ f(x) = 0 for x = 5. Hence, [4]. f(1) = (1 – 1) (1 – 2) (1 + 3) = 0 g(1) = 13 – 1 = 0 f(6) = (5 – 6) = –1 g(–1) =

1 = –1 \ f(6) = f(–1) 1

7.

Option [1] : f(1) = 1 + 1 + 1 = 3 while g(–15) = 2(–15) + 30 = 0 Option [2] : f(–2305) = g(–2305) {as –2305 < 0} Option [3] : f(0) = g(0) ¹ g(1) Hence, [2].

8.

g(1) = 2(1) + 30 = 32 while f(1) = 1 + 1 + 1 = 3 also g(2) = 2(2) + 30 = 34 and f(2) = 2 + 22 + 2 3 = 14 Hence g(2) > f(2). Hence, [2].

9.

f(1) = 3 while f(2) = 14 while f(3) = 39 while f(4) = 84 while Thus we see that Hence, [3].

10.

f(1, –1)

= = Now, g(16) = = = Hence, [2].

11.

f(x) + g(x) = x 2 + 1 + x 2 – 1 = 2x 2

\

(i) is true

\

(ii)

true. f(0) = (–1) (–2) (3) = 6 g(0) = 0 – 1 = 1 \

f(x)  g(x) . 2

It can be seen that other options are not correct. Hence, [4].

\ (iii) is true. Hence, [3].

12. 6.

4(1) 2 – 3(1)(–1) + 3(–1) 2 + 6 4 + 3 + 3 + 6 = 16 (16) 2 – 3(16) + 20 256 – 48 + 20 228.

f(x)  g(x)  x2 = x = h(x) 2

Thus, h(x) =

f (0) 6 = = -6 g ( 0) -1

= 32 = 34 = 36 = 38 and g(3) are the closest.

f(x)  g(x) 2x2   x2 2 2

\

is

g(1) g(2) g(3) g(4) f(3)

 1   f(2) + f(–2) + f   2

f(x): at x = 0, f(x) = a

g(x): at x = 0, g(x) = –a

a a , f(x) = 2 2 at x = a, f(x) = 0

a a , g(x) = – 2 2 at x = –a, g(x) = 0

at x =

 1   = (2 + 2 2 + 2 3) + g(–2) + g   2

at x = –

\ g(x) = –f(–x). Hence, [4]. Alternatively, The graph of g(x) can be obtained by reflecting the graph of f(x) about the x-axis and y-axis one after the other. \ g(x) = –f(–x)

 1   = 14 + [2(–2) + 30] + [2   + 30] 2 = 14 + 26 + 29 = 69. Hence, [1].

5

IMS-34-UG-AL-HM-Class-1-2-3

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HIGHER MATH 13.

14.

15.

f(x) = ax 2 – bx + 7 f(2) = a(2)2 – b(2) + 7 = 5 Þ 4a – 2b + 7 = 5 \ 4a – 2b = –2 —— (i) f(4) = a(4)2 – b(4) + 7 = 11 Þ 16a – 4b + 7 = 11 \ 16a – 4b = 4 ——— (ii) Solving (i) and (ii) a = 1 and b = 3 \ a + b = 1 + 3 = 4. Hence, [4].

= –

18.

19.

3

\ f (x) = –1

=

f(–x) =

20.

38

p

=

(x)

p

px  px

=

= –f(x)

5

f(x) = –3x 3 +

x2

f(–x) = –3(–x) 3 +

3

= 3x +

=

5 (x)2

+ 5x 2 – 5

3

+ 5(–x) 2 –

(x)2

3

2

+ 5x +

3

1 1 f æç ö÷ = –3 æç ö÷ +

èxø

cot x

cos x

ò sin x

=

sin x

=

t

ò

3 t 2

Hence, [1].

sin x

dx

cos x dx = dt =

dt

-2 t

=

-2 sin x

1

ò sin x + cos x

x3

èxø

æ1ö ç ÷ èxø

IMS-34-UG-AL-HM-Class-1-2-3

2

1

dx ö 1 2 æ 1 çç sin x + cos x ÷÷ 2 è 2 ø

ò 1 2

1 2 1 2

ò

dx pö æ sin ç x + ÷ 4ø è

3

21.

f(x) =

(x)3

.

1 + 5 æç ö÷ – èxø

pö ÷ dx 4ø

æ

ò cosec çè x +

pö æx log tan ç + ÷ . 8ø è2

4x , 4 + 2 x

f(1 – x) =

2

5

dx

Hence, [1].

x3

Thus, f(x) ¹ f(–x). II.

x sec 2 x - (sec 2 x - 1) dx

dt

I =

= I.

2

ò

I =

px  px

\ [1] is an odd function. The rest are neither odd nor even functions. Hence, [1]. 17.

x (sec 2 x - 1) dx

1 tan 3 x - tan x + x . 3

=

px  px =  x p  px

2

=

= 2.

px  p(x) x

ò tan

ò tan

x  3 .

Hence, [2].

16.

x dx =

Hence, [2].

= 5, f (5)  3 =

4

=

òt

–1

At x 35

3

ò tan

Þ Let, sin x = t

x  3

3

– 3x 3 = f(x)

x2

Hence, [2].

f(x) = x – 3 i.e., y = x – 3 To find f –1 (x), we have to find y such that f(y) = x i.e., x = y 3 – 3 \ y3 = x + 3

\ y =

5

5x 2 +

+

æ1ö \ f ç x ÷ = f(x). Thus only II is true. è ø

As seen from the graph, the left half rises more rapidly than that on the right side. On close observation the slope m of line y = mx is half (in magnitude) of the slope of the line y = –nx, in fact, here, m = 1, n = –2. So [1] is the best fit. Hence, [1]. 3

3 x3

=

3 æ1ö ç ÷ èxø

3

=

6

4 4x

4 + 2 4x

41 1- x

x

4

+ 2

=

2 4x + 2

f(x) + f(1 – x) = 1. Hence, [2].

.

®

HIGHER MATH

22.

23. 24.

25.

26.

(4 x - 1) (2 x - 1) . x ® 0 x x Hence, [1]. lim

28.

I =

x =

2

– x + 3

– x + 3

)

2

x dx 1 - x2

1 2

Þ cos t =

1 2

Þ t =

x = 0 Þ cos t = 0 Þ t = p/2

\ I= ò

p/4

Answer options [2], [3] and [4] have no meaning. Hence, [1]. 2

ò

0

a × b = 42i + 14j – 21k = 7(6i + 2j – 3k) = 7c a × (b × c) = (a × b) × c = 7c × c = 7(c × c) = 0. Hence, [1].

Let y = e(5 x

1/

Substituting x = cos t, we get, dx = –sintdt

Property of cross product of vectors. Hence, [1].

dy = e (5 x dx

p 4

p 2

p/2 cos t. sin t dt = ò cot( t ) dt = [log(sin(t ))]pp // 24 2 sin t p/4

= log(1) – log

)

1 2

= log 2 . Hence, [3].

Alternatively, d (1 – x 2 ) = –2x dx

d (5x 2 – x + 3) dx

= (10x – 1) e (5 x Hence, [2].

27.

= log4. log2 = 2(log 2)2.

2

– x + 3

)

1 2 1/ \ I = - 2 ò 1 - x 2 dx = - [log(1 - x )]0 2 0

1

=

(1 + cos x ) - 1 cos x dx = ò dx ò 1 + cos x 1 + cos x

1/

2

- 2x

2

1 log 2 = log 2 . 2

1 æ ö ÷ dx = ò ç1 1 + cos x ø è

=

æ

1 - cos x ö ÷ 2 ÷ dx = xø

ò ççè1 - 1 - cos

æ

ò ççè1 -

1 - cos x ö ÷÷dx sin 2 x ø

= ò (1 - cosec 2 x + cot x cosec x ) dx = x + cotx – cosecx + c cos x 1 1 - cos x + c = x – + c sin x sin x sin x Hence, [2].

= x +

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IMS-34-UG-AL-HM-Class-1-2-3

Higher Math_Class_1_3_Exp_Ans

ANSWER KEY 1 + , ac 1 + and ab 1 + are in A.P. 1-[4] 2-[1] 3-[4] 4-[4] 5-[3] 6-[3] 7-[4] 8-[1] 9-[1] 10-[3] 11-[1] 12-[3] 13-[4] 14-[1] 15-[...