1. The smallest value of k, for which both the roots of the equation x² − 8kx + 16(k² − k + 1) = 0 are real, distinct and both the roots are greater than or equal to 4.

∴ k ≥ 1                         ...(1)

Now, both roots are greater than 4.

Hence, sum of these roots must be greater than 8.

As both the roots are more than or equal to 4, so the value of the equation at x = 4 must be greater or equal to zero.
∴ 16 – 32k + 16(k² – k + 1) ≥ 0


∴ Minimum value of k is 2.


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2. Let abc be the three digit number. If bca + bac + cab + cba + acb = 2982, then find a × b × c.

 222(a + b + c) = 2982 + abc

Hence, 2982 + abc is divisible by 222.

Now, 2982 = 13 × 222 + 96

And, 96 + abc is divisible by 222.

Hence, abc = 126 + 222k, here k is an integer greater than or equal to 0.

∴ 222(a + b + c) = 13 × 222 + 96 + 126 + 222k

∴ a + b + c = 14 + k

Now, for k = 0, abc = 126, and a + b + c = 14

But both statements are contradictory.

Now for k = 1, 

abc = 126 + 222 = 348, and a + b + c = 15

Now, this is possible.

Hence, abc can be 348.

Similarly for k = 2 and k = 3 we can show that there is no possible value of abc.

For k > 3, abc will be greater 1000.

Thus the only possible value of abc = 348.

∴ a × b × c = 96

3. If f is a function such that f(x + y) = f(xy) for all values of x, y ≥ 4. Domain of f is [8, ∞), andf(8) = 9, then find f(100).


f(20) = f(16 + 4) = f(64)

4.  If;

 

Find the value of

 

4.x² = qy + rz … (I)

y² = px + rz … (II)

z² = px + qy … (III)

Adding all three equation, we get,

x² + y² + z² = 2(px + qy + rz)

Now, by (II) + (III) – (I), we get,

2px = (y² + z² – x²)

Similarly,

2qy = (x² + z² – y²), and

2rz = (x² + y² – z²)

Hence, we have,

 

 

 

Now, 

 

 

Similarly,

 

 

 

Similarly,

 

 

Hence, 

 

 

5.Let N be the smallest natural number for which (33N + 4)/(22N + 3) is reducible. Find the remainder when N is divided by 11.

Let p be the common divisor of (33N + 4) and (22N + 3).

Hence, (33N + 4 ) is divisible by p.

6.  

In the figure shown above, AB = 11, BC = 7 and CA = 9. AB, BC and CA are extended to D, E, and F respectively such that AD = 33, BE = 28 and CF = 27.

Find ratio of areas of ΔABC and ΔDEF.

Using sine rule of areas, we have,

 

 

Let A(ΔABC) = pNow,

= 8 × A(ΔABC) = 8p            …(from equation 1)

Similarly,



= 6 × A(ΔABC) = 6p           …(from equation 2)



= 9 × A(ΔABC) = 9p            …(from equation 3)

∴ (66N + 8) is also divisible by p.

Similarly, (22N + 3) is divisible by p.

∴ (66N + 9) is also divisible by p.

Hence, (66N + 9) – (66N + 8) is divisible by p.

i.e. 1 is divisible by p.

∴ p must be 1.

∴ 33N + 4 and 22N + 3 cannot have common divisor greater than 1.

Hence, the given fraction is irreducible for all N.

So, no such N exists

Let a, b be positive integers such that



7. Find the number of ordered pairs (a, b) which satisfy the above equation?

 
∴ 35(a + b) = ab

∴ ab – 35a – 35b = 0

Adding 1225 on both sides we get;

a(b – 35) – 35(b – 35) = 1225

∴ (a – 35)(b – 35) = 1225

But, 1225 = 5² × 7²

So, 1225 has 9 factors.

This implies that (a – 35) and (b – 35) can take 9 different values.

So, there are 9 ordered pairs (a, b).

7. If x = 1/9, then what is the value of 1 + 2x + 3x² + 12x³ + 33x⁴ + 102x⁵ … up to infinite terms? (17-05)

S = 1 + 2x + 3x² + 12x³ + 33x⁴ + 102x⁵…

Multiplying the above given equation by 3x, we get,

3x × S = 3x + 6x² + 9x³ + 36x⁴ + 99x⁵ + 306x⁶ … 

Subtracting the second equation from the first one, we get,

(1 – 3x) × S = 1 – x – 3x² + 3x³ – 3x⁴ +3x⁵ … 

(1 – 3x) × S = 1 – x – 3x²(1 – x + x² – x³ …)

Now, (1 – x + x² – x³ …) is an infinite GP with common ratio –x.

As x = 1/9 i.e. the value of x is less than 1, the formula for sum of infinite terms of GP can be used on this expression.

 

Substituting the value of x in the above equation, we get,