## Monday, May 19, 2008

Q) Find the sum to 2n terms of the series whose every even term is 'a' times the term before it and every odd term is 'c' times the term before it, the first term being unity.

A) Let T indicate a term of the progression.

T1 T2 T3.......Tn......T2n

T1 = 1

T2 = a

T3 = ca

T4 = c.a^2

T5 = c^2.a^2

Tk if k is even = a^(k/2). c^(k/2 - 1)

T2n = a^(2n/2).c^(2n/2 -1)

T2n = a^n. c^(n-1)

S 2n = 1 + a + ca + c.a^2 + c^2.a^2 + c^2.a^3 .....a^n. c^(n-1)

= 1 + [ a + ca^2 + c^2.a^3 ....+ a^n.c^(n-1) ] + [ ca + c^2.a^2 + c^3.a^3..... + a^(n-1). c^(n-1) ]

= 1 + [ a.(a^n.c^n - 1) / (ac - 1) ] + [ ac( a^(n-1).c^(n-1) - 1) / (ac - 1) ]

..

..

.. solving further..

---> S 2n = (a^n.c^n - 1)(a + 1) / (ac - 1)

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A) Let T indicate a term of the progression.

T1 T2 T3.......Tn......T2n

T1 = 1

T2 = a

T3 = ca

T4 = c.a^2

T5 = c^2.a^2

Tk if k is even = a^(k/2). c^(k/2 - 1)

T2n = a^(2n/2).c^(2n/2 -1)

T2n = a^n. c^(n-1)

S 2n = 1 + a + ca + c.a^2 + c^2.a^2 + c^2.a^3 .....a^n. c^(n-1)

= 1 + [ a + ca^2 + c^2.a^3 ....+ a^n.c^(n-1) ] + [ ca + c^2.a^2 + c^3.a^3..... + a^(n-1). c^(n-1) ]

= 1 + [ a.(a^n.c^n - 1) / (ac - 1) ] + [ ac( a^(n-1).c^(n-1) - 1) / (ac - 1) ]

..

..

.. solving further..

---> S 2n = (a^n.c^n - 1)(a + 1) / (ac - 1)

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Q) Let x, y and z be three positive numbers such that x + y + z = 12. Then the maximum value of x

A) x+y+z = 12

x/2 + x/2 + y/3 + y/3 + y/3 + z = 12

Use AM > GM inequality:

(x + y + z) / 6 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6

2 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6

2^6 > (x^2 / 4)(y^3 / 27)(z)

2^6 * 4*27 > x^2.y^3.z

So x

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^{2}y^{3}z is ?A) x+y+z = 12

x/2 + x/2 + y/3 + y/3 + y/3 + z = 12

Use AM > GM inequality:

(x + y + z) / 6 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6

2 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6

2^6 > (x^2 / 4)(y^3 / 27)(z)

2^6 * 4*27 > x^2.y^3.z

So x

^{2}y^{3}z < 6912________________________________________________________

Q) If log a, log b ,log c are in Arithmetic Progression and also loga - log2b ,log2b - log3c , log3c - loga are in Arithmetic Progression then :

1)a, b, c are in AP

2)a, 2b, 3c are in AP

3)a, b, c are the sides of the triangle

4)none of the above

A)

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1)a, b, c are in AP

2)a, 2b, 3c are in AP

3)a, b, c are the sides of the triangle

4)none of the above

A)

log a, log b ,log c are in AP loga + logc = 2logb logac = logb^2 ac = b^2 .....................so a,b,c are in GP. log a-log 2b ,log2b-log3c , log3c-loga are in AP loga - log2b + log3c - loga = 2 (log2b - log3c) log(3c/2b) = 2 (log (2b/3c)) 3c/2b = (2b/3c)^2 THUS, 3c/2b = 1 3c = 2b b = 3c/2 ac= b^2 = 9c^2/4 a = 9c/4 b = 3c/2 c = c since this is a pythagorean triplet (sum of any 2 is greater than 3rd) So they are the sides of a triangle. |

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Q) Find domain of 2

A) 2^x + 2^y = 2

2^x = 2 - 2^y

x log2 = log (2 - 2^y)

x = log(2 - 2^y) / log 2

x = log (2 - 2^y) to base 2

2 - 2^y is always <= 2...... as 2^y least value will tend to 0.

log (2 - 2^y) to base 2 will lie in (-infinity, 1)

So domain is (- infinity, 1)

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^{x}+ 2^{y}= 2 .A) 2^x + 2^y = 2

2^x = 2 - 2^y

x log2 = log (2 - 2^y)

x = log(2 - 2^y) / log 2

x = log (2 - 2^y) to base 2

2 - 2^y is always <= 2...... as 2^y least value will tend to 0.

log (2 - 2^y) to base 2 will lie in (-infinity, 1)

So domain is (- infinity, 1)

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Q) A piece of hair shows the

A) Activity is proportional to amount of nuclei.

If activity falls by 75% it means nuclei decreased by 75%

means 25% left

means 2 half lives......(because after first half life, 50%remains, and after next half life half of that)

= 2 x 5760

= 11520 years

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^{14}C activity falling by 75% of the initial activity. Given that^{14}C has half life of 5760 yrs, the age of the hair is ?A) Activity is proportional to amount of nuclei.

If activity falls by 75% it means nuclei decreased by 75%

means 25% left

means 2 half lives......(because after first half life, 50%remains, and after next half life half of that)

= 2 x 5760

= 11520 years

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Q) The quadratic equation whose roots are twice those of 3x^2+5x-2=0 is ?

A) 3x^2+5x-2 = 0

= x^2 - (sum of roots)x + (product of roots)

If roots are multiplied by two, it should become:

x^2 - 2(sum of roots)x + (2*2*product of roots)

So answer is

3x^2 + 2*5x - 2*4 = 0

=> 3x^2 + 10x - 8 = 0

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A) 3x^2+5x-2 = 0

= x^2 - (sum of roots)x + (product of roots)

If roots are multiplied by two, it should become:

x^2 - 2(sum of roots)x + (2*2*product of roots)

So answer is

3x^2 + 2*5x - 2*4 = 0

=> 3x^2 + 10x - 8 = 0

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Q) Let z be a complex number ,then the minimum value of |z-2|+|z-3|+|2z-9| is ?

A) |z-2|+|z-3|+|2z-9|

Let z = a + ib

|z-2|+|z-3|+|2z-9|

= sqrt [ (a-2)^2 + b^2 ] + sqrt [ (a-3)^2 + b^2 ] + sqrt [ (2a - 9)^2 + 4b^2 ]

for minimising, b^2 will be 0.

= sqrt [ (a-2)^2 ] + sqrt [ (a-3)^2 ] + sqrt [ (2a - 9)^2 ]

= |a-2| + |a-3| + |2a - 9|

if seeing first two terms, a - 3> 0 then a > 3

then |2a - 9| will be 9 - 2a

= a - 2 + a - 3 + 9 - 2a

= 4

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A) |z-2|+|z-3|+|2z-9|

Let z = a + ib

|z-2|+|z-3|+|2z-9|

= sqrt [ (a-2)^2 + b^2 ] + sqrt [ (a-3)^2 + b^2 ] + sqrt [ (2a - 9)^2 + 4b^2 ]

for minimising, b^2 will be 0.

= sqrt [ (a-2)^2 ] + sqrt [ (a-3)^2 ] + sqrt [ (2a - 9)^2 ]

= |a-2| + |a-3| + |2a - 9|

if seeing first two terms, a - 3> 0 then a > 3

then |2a - 9| will be 9 - 2a

= a - 2 + a - 3 + 9 - 2a

= 4

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Q) Find domain of

.

A)

=

.

A)

=

=

x^{8} + 1 will always be positive.

x^{4} - x will be least at x = (1/4)^{1/3} ...............found by diff. and equating to 0.

But at this value, the expression under the root is greater than 0.

**Hence domain is ALL REAL NUMBERS.**________________________________________________________

Q) In order to quadruple the resistance of a wire of uniform cross section, a fraction of its length was stretched uniformly till the final length of the wire was 1.5 times the original length. The value of the fraction elongated of the wire in comparison to original length of the wire is ?

A) let length of wire = L

let length of fraction = x

let fraction x be stretched to 't' times its length

R = L/A

4R = (L-x)/A + tx/(A/t)

4(L/A) = L/A - x/A + t

^{2}x/A3L/A + x/A = t

^{2}x/A3L + x = t

^{2}xx/L = 3/ t

^{2}- 1 ----------------------(1)also given,

new length of wire/original length of wire = 1.5

L - x + tx / L = 1.5

0.5L = tx - x

x/L = 0.5 / t-1 ----------------------(2)

comparing (1) and (2)

0.5 / t-1 = 3 / t

^{2}-10.5(t+1) = 3

0.5t = 2.5

t = 5

now put this value of t in (1)

x/L = 3/ t

^{2}- 1x/L = 3 / 5^2 - 1

x/L = 3/24

x/L = 1/8

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Q) Let f(x)=(2x/2x^2+5x+2) AND g(x)=(1/1+x).Find the set of real values of x for which f(x)> g(x) .

A) f(x) > g(x)

f(x) - g(x) > 0

2x/ (2x^2 + 5x+ 2) - 1 / (1+x) > 0

2x / (2x+1)(x+2) - 1/1+x > 0

-3x - 2 / (2x-1)(x+2)(x+1) > 0

3x+2 / (2x-1)(x+2)(x+1) <>

A) f(x) > g(x)

f(x) - g(x) > 0

2x/ (2x^2 + 5x+ 2) - 1 / (1+x) > 0

2x / (2x+1)(x+2) - 1/1+x > 0

-3x - 2 / (2x-1)(x+2)(x+1) > 0

3x+2 / (2x-1)(x+2)(x+1) <>

x not equal to -1, -2, -1/2 as we get 0 in denominator... and x not equal to -2/3 as numerator becomes 0

Now apply wave curve method.**x E (-2, -1) U (-2/3, -1/2)**

Q) LET be defined on the interval [0 ,1] . The odd extension of F(x) on the interval [ -1 , 1 ] is ?

A) G(x) = { - F(-x) -1<= x < 0

{ F(x) 0<= x <= 1

where G(x) is the odd extension.

WHERE

- F(-x) = - [ -4sinx + 3cosx + log (|x| + sqrt ( 1 + x^2 ) ]

= 4 sin x - 3 cos x - log (|x| + sqrt (1 + x^2) ]

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A) G(x) = { - F(-x) -1<= x < 0

{ F(x) 0<= x <= 1

where G(x) is the odd extension.

WHERE

- F(-x) = - [ -4sinx + 3cosx + log (|x| + sqrt ( 1 + x^2 ) ]

= 4 sin x - 3 cos x - log (|x| + sqrt (1 + x^2) ]

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Q) One bag contains 4 white and 5 black balls. Another bag contains 6 white and 7 black balls. A ball is transferred from the first bag to the second bag. Find the probability that the ball drawn is white.

Case 1. white transferred :

Case 2. black transferred :

= 29/63

**A) Prob. of white ball being transferred = 4/9**

** Prob. of white ball being drawn from 2nd bag** ---->

Case 1. white transferred :

7 white 7 black so 1/2

Case 2. black transferred :

6 white 8 black so 6/14 = 3/7

Total Prob. = (4/9) (1/2) + (5/9)( 3/7)

= 29/63

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Q) An unbiased die is rolled until a number greater than 4 appears on the die . Find the probability that an even number of tosses are needed.

A) Prob. that a number greater than 4 appears = 2/6 = 1/3

Prob. that a number less than or equal to 4 appears = 2/3

Prob. that an even no. of tosses are needed =

(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......

= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)

= 2/9 [ 1 / ( 5/9) ]

= (2/9)*(9/5)

= 2/5

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A) Prob. that a number greater than 4 appears = 2/6 = 1/3

Prob. that a number less than or equal to 4 appears = 2/3

Prob. that an even no. of tosses are needed =

(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......

= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)

= 2/9 [ 1 / ( 5/9) ]

= (2/9)*(9/5)

= 2/5

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Q) An electric kettle takes 4A current at 220V. How much time will it take to boil 1 kg of water from temperature 20 degrees C ? (The temperature of boiling water is 100 degrees C).

A) P = IV = 880 Watts

P = work / time = 880 joules per sec

Q = m.c.del T

c is 4200 J per kg per K for water

Q = 1*4200* 80 = 336000

Time = Q / W = 336000 / 880 = 381.81 seconds

A) P = IV = 880 Watts

P = work / time = 880 joules per sec

Q = m.c.del T

c is 4200 J per kg per K for water

Q = 1*4200* 80 = 336000

Time = Q / W = 336000 / 880 = 381.81 seconds

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