Question Bank  Equations  Set 2

First of all we open a bracket and try to it a identity so ..
p(p^2 + 3p + 3) + 1 = p^3 + 3p^2 + 3P + 1 = (p + 1)^3
Its cube root will be = (p + 1)
And p= 124
Ans = 125#Alter
For better approach you have to take p=1 and solve u will get 8 and cube root of 8 is 2
It means if we take a value of p =1 , we get 2 , similarly if p = 125 we will get p = 125
No need so solve further for identity

Q9) The unit digit of 27^16 – 18^48 ?

This question seems easier but it is little tricky, now solve
Unit digit of any number can be find out by cyclicity.
cycle of unit digit is four so expression reduces into 7^4k – 8^4k = …..1  ….6 = (5) but unit digit can not in negatives so we take carry as 10  5 = 5

Q10) If a/b = c/d = e/f = 3/1 , then (2a^2 + 4e^2 + 3c^2)/(2b^2 + 3d^2 + 4f^2) ?

For smarter approach you have to take values in this manner
Let a/b = 3/1 => a=3b
c = 3d , e = 3f
Put these values u will get 9 .

Q11) Find all ordered pairs of real numbers which satisfy 17a^2 + 2b^2  10ab  6a + 2 = 0

Write equation in this way as 2b^2 – 10ab + (17a^2  6a +2) = 0
Eqn is quadratic
Find discriminant D = b^2 – 4ac ( ax^2 + b x +c = 0 )
D = (10a)^2 – 4*2 (17^2 6a+2)= 4(3a2)^2 , this is negative unless a= 2/3 , now solving for b , it is 5/3
(a,b)= (2/3,5/3)

Q12) Find the sum of digits in the product 199999997 * 200000003 ?

Use identity ( ab) (a+b)= a^2 –b^2 , take a= 2 * 10^8 , b = 3 , so the product is
39999…(15 times)1
Sum = 3 + 15 * 9 + 1 = 139

Q13) A three digit natural number xyz is written side by side to form a six digit natural number n = xyzxyz . then it must be divisible by ?

N = abc * (1001)
It must be divisible by 7, 11,13 because 7 * 11 * 13= 1001
Also it must be divisible by factor of abc

Q14) If (x7)^{ (x^2  29x + 154)/(x^2  12x + 32)} = 1 , how many real values of x satisfy the given equation ?

(x70^{(x7)(x22)/(x8)(x4)} , now let we take f(x) = (x7) and G(x) = (x7)(x22)/ (x8)(x4) , now we take three cases .
i) If f(x) = 1 and g(x) can be anything
so x = 8 , for this value g(x) can not be defined
ii) if f(x)=1 , g(x) is an even , then x = 6 , g(x) is defined
iii) if f(x) does not zero and g(x) is zero ,
(x7)(x22) = 0 => x=7,22 , but for x=7 the eqn is not defined , so eqn is defined for two reals x i.e 6, 22

Q15) Find the number of integral solutions for (a, b, c) for if a^2 = bc + 4 and b^2 = ca + 4

OA  12 triplets are possible.

Q16) Find a natural number n for which 3^9 + 3^12 + 3^15 + 3^n is a perfect cube

First of all we will try to find out cube
Now, 3^9+ 3^12 +3^15 +3^n = 3^9( 1 + 3^3 + 3^6 +3^n9 )
= (3^3)^3 [ 1+ 3.3^2 + 3.(3^2)^2 +(3^2)^3 + 3^(n9) – 3(3^2)^2]
= (3^3)^3[(1+3^2)^3]
Since, 3^(n9) – 3^5 = 0
3^(n9) = 3^5
n  9 = 5
so, n = 14
hence given number is a pefect cube for n = 14

Q17) If a, b, c, d, e are positive real numbers such that a+b+c+d+e = 8 and a^2 + b^2 + c^2 + d^2 + e^2 = 16 find range of e.

Concept  We know that these of type of questions can be solved by inequality.
If a, b, c, d are positive real numbers then [( a+ b+ c + d)/4]^2 ≤ (a^2 +b^2 + c^2 + d^2)/4 ... eqn(1) { by Techebycheff’s Inequality}
Now, a + b + c + d + e = 8
a + b + c + d = 8  e
a^2 + b^2 + c^2 + d^2 = 16 – e^2
putting above value in equation (1)
(8e)^2/ 16 ≤ ( 16 – e^2)/ 4
On solving this ,
e(5e – 16) ≤ 0
Using number line rule .
0 ≤ e ≤ 16/5

Q18) Find the period of f(x) = Sin^4 x + cos^4 x