Q13/100) 2 ^ sin x + 2 ^ cos x is greater than

A> 1/2

B> 2^(1/(2^(1/2)))

C> 2^(1/2)

D> 2^(1-(1/(2^(1/2))))

Solution

by AM GM inequality we have

(a+b) /2 >= sqrt(ab)

so 2^ sin x + 2^ cos x >= 2 sqrt(2^( sinx + cos x))

sin x + cos x = sqrt(2) ( sin x cos pi/4 + cos x sin pi/4) = sqrt(2) sin (x+pi/4)

lowest value is - sqrt(2)

so 2^ sin x + 2^ cos x >= 2 sqrt(2^-(sqrt(2)) = 2^(1-(1/2^(1/2))

hence D

 

Q13/099) If /(z - 5i)/ (z +5i)/ = 1 then prove that z is a real number

let z = x + iy

|z + 5i| = | z - 5i|

or | x + iy + 5i | = | x + iy - 5i|

or | x + iy + 5i |^2 = | x + iy - 5i|^2

or x^2 + (y + 5)^2 = x^2 + (y- 5)^2

or (y + 5)^2 = (y- 5)^2 or y^2 + 10y + 25 = y^2 - 10y + 25 => y =0 or z is real

Q13/098) If α and β are two different values of θ ( between 0 and 2π ) which satisfy the equation .................? . 6 cos θ + 8 sin θ = 9, Find the value of ..... sin ( α + β )

α and ß are the roots of the equation : 6 cosΘ + 8 sinΘ = 9
=> 6 cosα + 8 sinα = 9 ------------- (1) and
.....6 cos ß + 8 sin ß = 9 ------------- (2)
Subtracting (2) from (1), we have :
6(cosα - cos ß) + 8(sin α - sin ß) = 0
=> 4(sinα - sin ß) = 3(cos ß - cos α)
=> (sin α - sin ß)/(cos ß - cos α) = 3/4
=> [2 cos{(α + ß)/2} sin {(α - ß)/2}] / [2 sin {(α + ß)/2} sin {(α - ß)/2}] = 3/4
=> cot {(α + ß)/2} = 3/4
=> cot²{(α + ß)/2} = 9/16
=> 1 + cot²{(α + ß)/2} = 1 + 9/16 = 25/16
=> cosec²{(α + ß)/2} = 25/16
=> sin² {(α + ß)/2} = 16/25
=> sin {(α + ß)/2} = ± 4/5 --------- (3)
=> 1 - sin²{(α + ß)/2} = 1 - 16/25 = 9/25
=> cos²{(α + ß)/2} = 9/25
=> cos {(α + ß)/2} = ± 3/5 ---------- (4)
=> sin (α + ß) = 2 sin {(α + ß)/2} cos {(α + ß)/2} = ± 2(4/5)*(3/5) = ± 24/25

However as 6 cosα + 8 sinα = 9

so cosα and sinα both > 0 ( as if one is -ve then sum < 8) and so α is in 1st quadrant and similarly β

so ( α + β ) is either in 1st or 2nd quadrant so sin ( α + β )is positive and hence -24/25 need to be ruled out and hence it is 24/25.

Another solution

6 cos θ + 8 sin θ = 9
=> (3 / 5) cos θ + (4 / 5) sin θ = 9 / 10
=> sin (θ + φ) = 9/10

Clearly if
sin x = sin a
=> x = a or π - a
for given range

=> θ + φ = α + φ or π - (α + φ)
Clearly one of them is equal to ß + φ
=> π - (α + φ) = ß + φ
=> α + ß = π - 2φ
=> sin (α + ß) = sin (π - 2φ)
=> sin (α + ß) = sin 2φ
=> sin (α + ß) = 2 sin φ cos φ = 24 / 25

Q13/097) The LCM of 2numbers is 45 times their HCF.Sum of HCF and LCM is 1150, find the numbers?

LCM = 45 *HCF
LCM + HCF = 46 * HCF = 1150

so HCF = 25 and LCM = 25 * 45

now both x and y ( the 2 numbers are) multiple of 25 and coprime

so we x = 25m, y = 25n and LCM(m,n) = 45, HCF(m,n) = 1

without loss of generality let m >= n

mn = 45 = 1 * 45 = 5 * 9
so m = 9, n =5 or m = 45 , n= 1

numbers are (225,125) or (1125,25)

Q13/096) Find sum of n terms ---- 1 + ( 1 + x ) + ( 1 + x + x² ) + ( 1 + x + x² + x³ ) + ..... + up to nth term.?

y = 1 + (1+ x) + (1+ x + x^2) + ...
y ( 1-x) = (1-x) + 1(1-x^2) + .. ( 1- x^(n))
= n - ( x + x^2 + .. x^(n)) as
= n - x ( 1 + x ... x^(n-1))
= n - x ( 1- x^(n))/(1-x)

or y = n/(1-x) - x (1-x^(n)/(1-x)^2

the above ans is correct if x is not 1

if x = 1 then we have 1 + 2 + 3 + .. n = n(n+1)/2

Q13/095) Find coefficient of xⁿ in the expansion of ( 1 + 2x + 3x² + 4x³ + ........ inf. )^(1/2)....?

let y = ( 1 + 2x + 3x² + 4x³ + ........ inf. ...(1)
y converges for |x| < 1

now xy = x + 2x^2 + 3x^3 ....(2)

subtract (2) from (1)

y - xy = 1 + x + x^2 ... = 1/(1-x)

so y(1-x) = 1/(1-x)

or y = 1/(1-x)^2

so ( 1 + 2x + 3x² + 4x³ + ........ inf. )^(1/2)..= 1/(1-x) = 1 + x + x^2 ...

so coefficient of x^n = 1 for all n

Q13/094) Solve |x+1|+|x-2| < 5

As -1 < 2 and in a line the distance between -1 and 2 is 3 so all the points from -1 to 2 satisfy the relation as the sum is the distance from -1 to 2

Now if it to the left of -1 the distance from -1 shall be added twice once for |x+1| and another time for |x-2| the sum < 2 and hence x > - 2

For the point to the right of 2 similarly x < 3

So -2 < x < 3

Q13/093) Find twin primes p and q(=p + 2) such that pq-2 is also a prime

P q are twin primes so they are 3 and 5 or p = 6n+ 1 and q = 6n – 1

If p = 6n- 1 and q = 6n + 1 then pq - 2 = 6n^2 – 1 – 2 = 36n^2 – 3 = 3(12n^2-1)

it is > 3 (minimum value = 33)  and 3 is a factor so cannot be prime

That leaves us with p = 3 , q = 5 and pq-2 = 13 meeting the condition

Q13/092) Prove 4 cos 36 sin 18 = 1

We have

sin 72 
= 2 cos 36 sin 36 using sin 2A formula
= 2 cos 36 ( 2 sin 18 cos 18) using sin 2A formula again
= 4 cos 36 sin 18 sin 72 as cos18 = sin(90-18) = sin 72

or

4 cos 36 sin 18 = 1

Q13/091) Prove that for every prime p > 7, p^6 - 1 is divisible by 504.

we have 504 = 9 * 56 = 7 * 8 * 9

we need to show that it is divisible by 7 , 8 and 9 as they are pairwise coprime

now as p is prime and 7 is also prime

so p^6 = 1 mod 7 by Fermat's Little theorem so p^6-1 is divisible by 7

p^6-1= (p^3-1)(p^3+1)

as these 2 are 2 successive even numbers Reason p is odd so p^3 is odd

so one is divisible by 4 and another by 2 so product is divisible by 8

one of p^3-1 and p^3 + 1 is divisible by 9

as p > 7 so p mod 3 = 1 or -1

to prove is if p = 1 mod 3 then p = (3n+1)

so p^3 = 27n^3 + 27n^3 + 9n + 1 or p^3-1 = 9(3n^3+3n^2 + 1) or dibsible by 9

if p = 3n-1 then p^3-1= 9(3n^3-3n^2 + 1) or divisible by 9

so p^6 -1 is divisible by 9

hence is divisible by 7 * 8 * 9 or 504

 

Q13/090) if M={x,xy,log(xy)}, N={0,∣x∣ ,y} given: M=N find : ( x + 1/y) + (x^2 + 1/y^2) + (x^3 + 1/y^3) …. + (x^2001 + 1/y^2001)

x and y cannot be zero as we have log(xy) and log zero is undefined .

so we have xy = 1 as log(xy) = 0

as x and y both are positive or –ve.

We have |x| in N . now x = |x| and xy = y => x = y = 1

but x and xy have to be different and |x| and y has to be different

so y = x = - 1 giving M={- 1,1,0 } and N={0, 1 ,0-1}

So x = y =-1 and sum = x^n + 1/y^n = -2 for n = odd and 2 for n = even

So sum = -2 as 1st 1000 pairs cancel leaving with x^2001 + 1/y^2001 = - 2

Q13/089) Find the real solution(s) to the equation 36/x^(1/2) +9/y^(1/4) = 42−9x^(1/2) −y^(1/2)

We have

36/x^(1/2) + 9x^(1/2) +     9/y^(1/4)+ y^(1/2) = 42

Complete the square for x and y terms giving

36/x^(1/2) + 9x^(1/2) - 36  +    9/y^(1/4)+ y^(1/2) = 42

Hence 9( 2/x^1/4 – x^(1/4))^2 + (3/y^(1/4) – y^(1/4))^2 = 0
so  2/x^(1/4) – x^(1/4) = 0 or x = 2^2 = 4
And 3/y^(1/4) – y^(1/4) = 0 or y = 9

Q13/ 088) Determine the smallest integer that is square and starts with the first four digits 3005.

we can have even number of digits or odd number of digits and they have to be treated separately

if even then we have

sqrt (3005 * 10^2n) = 54.8177 * 10^n
sqrt (3006* 10^2n) = 54.8270 * 10^n

if odd digits then

sqrt (30050 * 10^2n) = 173.34 * 10^n

sqrt (30060 * 10^2n)= 173.37 * 10 ^n

from the 1st set we get square root  5482 as smallest where a digit is different and from the second at least

17335

so it is 5482^2 = 30052324