Equations  Vikas Saini

Q.If log_{10}x – log_{10}(x)^{1/3} = 6log_{x}10 then find the value of x (IIFT 2013)
(a) 10 (b) 30 (c) 100 (d) 1000.
Solution :
log_{10}x – (1/3)log_{10}(x) = 6log_{x}10.
= > (2/3)log_{10}x = 6 / log_{10}x
Put log_{10}x = m.
= > (2/3) m = 6 / m
= > m^{2} = 9.
= > m = 3, 3.
We need to remove m = 3.
m = 3
log_{10}x = 3.
x = 1000.
Hence option d.
Q. The sum of the possible values of x in the equation x + 7 + x8 = 16. (XAT 2014)
(a) 0 (b) 1 (c) 2
(d) 3 (e) none of the above.
Solutioin :
Firstly put extreme value of x = 8.
At x = 8, equation = 15.
At x = 8.5, equation = 16.
Now put extreme negative value x = 7.
At x = 7, equation = 15.
At x = 7.5, equation = 16.
Sum = 8.5 – 7.5 = 1.
Q. An equation with all positive root is written as x^{n}+a_{n}x^{n1}+a_{n1}x^{n2}……..a_{1 }= 0.
Which of the following is necessarily true ?
(a) a_{n}x^{n } > n^{n} x a_{1 }(b) n^{n } > a_{n}n x a_{1}
(c) a_{1}^{n } > n^{n} x a_{n }(d) none of these
Solution :
Put n = 2.
x^{2}+a_{2}x+a_{1 }= 0.
Suppose a_{2 }= 2, a_{1 }= 1
Then x = 1.
Option (a) says a_{2}x^{2 } > 2^{2} x a_{1}
2 > 4
Not possible
Option (b) says n^{n} > a_{n}^{n} x a_{1.}
4 > 4
Not possible
Option (c) says a_{1}^{n} > n^{n} x a_{n.}
1 > 8.
Not possible.
Hence option D is correct choice.
Q. ax^{2 }+ bx + c = 0 is a quadratic equation with rational coefficients such that a+b+c = 0, then which of the following is necessarily true ?
(a) Both the roots of this equation are less than 1.
(b) One of the roots of the equation is c/a.
(c) Exactly one of the root is 1.
(d) b & c both.
Solution :
let f(x) = ax^{2 }+ bx + c
f(1) = a+b+c = 0.
Product of roots = c/a.
Hence option d.
Q. (x – 7)^(x^{2 }– 29x +154 / x^{2} – 12x+32) = 1.
How many real values of ‘x’ satisfy the given equation ?
Solution :
We know a^{0 } = 1.
We need to put x^{2} – 29x+154 = 0.
X = 22, 7.
We can’t put x = 7.
Hence x = 22.
We know 1^n = 1.
So x – 7 = 1
= > x = 8.
But it also won’t work.
And last property (1)^{even} = 1.
(1)^{odd } = 1.
So x – 7 = 1
= > x = 8,
But for x = 8, we know it won’t work.
X = 6.
For x =6, we will get required value.
Hence real values of x = 6 ,22.
Q. When ‘2’ is added to each of the three roots of x^{3}Ax^{2}+BxC = 0, we get the roots of x^{3}+Px^{2}+Qx18 = 0.
A,B,C,P and Q are all non zero real numbers. What is the value of (4A+2B+C).
Solution :
Let three roots of first equation are l, m, n respectively.
18 = (l+2)(m+2)(n+2)
18 = (lm+2l+2m+4)(n+2)
18 = lmn + 2lm + 2ln + 4l +2mn + 4m + 4n + 8.
10 = lmn + 2(lm+ln+mn) + 4(l+n+m)
10 = C+2B+4A.
Q. [x] =Greatest integer less than or equal to x
{x} = x – [x]
How many real values of x satisfy the equation ?
5[x] + 3{x} = 6 + x.
(a) 0 (b) 1
(c) 2 (d) 3.
Solution :
5[x] + 3(x – [x] ) = 6 + x.
5[x] + 3x – 3[x] = 6 + x.
2[x] + 3x = 6 + x.
2 [x] = 6 – 2x.
[x] = 3 – x
Hence 0.
Q. If log_{5}log_{4}log_{3}(x^{2}11x+1) = 0 (where x > 0) then what is value of x ?
Solution :
log_{5}log_{4}log_{3}(x^{2}11x+1) = 0.
log_{4}log_{3}(x^{2}11x+1) = 1.
log_{3}(x^{2}11x+1) = 4.
x^{2}11x+1 = 81.
x^{2}11x80 = 0
x = 16, 5.
Given x > 0.
Hence x = 16.
Q. If ‘x’ is a real number then what is the number of solutions for the equation x^{4}+16 = (x^{2}4)^{2} ?
(a) 0 (b) 1
(c) 2 (d) 4
Solution :
X^{4}+16 = x^{4}8x+16.
8x = 0
x = 0.
Option (a)
Q. All three roots of the cubic equation x^{3}10x^{2}+31xk = 0 are prime numbers. What is the value of K ?
Solution :
Let’s three roots l,m and n.
l + m + n = 10.
lm + mn + nl = 31.
lmn = k.
all roots are prime.
Hence l = 2, m = 3, n = 5.
lmn = 30.
Q. [ log (1) + log (1+3) + log (1+3+5)………………log(1+3+5…….19)
 2 (log 1 + log 2…………log 7) = m + nx + ay.
If log 2 = x and log 3 = y, then find the value of m, n, a ?
Solution :
(log 1 + log 4 + log 9……..log 100) – 2(log 1 + log 2…….log 7)
= 2(log 1 + log 2 + log 3……..log 10) – 2( log 1 + log 2…………log 7)
= 2 (log 8 +log 9 + log 10)
= 2(3log2 + 2log3 + 1)
= 2 + 6log 2 + 4log 3
= 2 + 6x + 4y
m = 2, x =6, y =4.
Q. x^{2}y^{3} = 8, where x,y > 0.
What is the minimum value of 4x+3y = ?
Solution :
4x+3y = P.
4x = 2k
x = k/2.
3y = 3k.
y = k.
(k/2)^{2} k^{3} = 8.
K^{5} / 4 = 8.
K = 2.
4x+3y = 2k+3k = 5k = 10.
Q. If a,b are integers then how many ordered pairs (a,b) satisfy the equation a^{2 }+ ab + b^{2} = 1 ?
Solution :
a^{2}+ab+ab+b^{2} – ab = 1
(a+b)^{2 }= 1 + ab.
a = 0, b = 1.
a = 1, b = 0.
a =0, b = 1.
a = 1, b = 0.
a = 1, b = 1.
a = 1, b=1.
6 ordered pairs.
Q. If a,b,c are root of the equation 3x^{3}+42x+93 = 0, then what is value of a^{3}+b^{3}+c^{3 }?
Solution :
Here coefficient of x^{2} = 0.
It means a+b+c = 0.
Then a^{3}+b^{3}+c^{3} = 3abc = 93
Q. Find the number of integer solutions of the equation x^{2} / y = 4x – 3, where x and y are non zero real numbers.
Solutions :
Suppose x^{2 }/ y = t.
t = 4x – 3.
x = 1, t = 1, y = 1.
x = 3, t = 9, y = 1.
Only two integer solutions.
Q. If all the roots of the equation (x – m)^{2 }(x – 10)+4 = 0 are integers, find the number of distinct values that ‘m’ can have ?
Solution :
(x – m)^{2}(x10) = 4.
4 = 4 x (1) , 1 x (4),
(x – m)^{2 }(x 10) = 4 x (1)
x – 10 = 1.
x = 9.
(9 – m)^{2} = 4.
( 9 – m) = 2, 2.
m = 7, 11.
(x – m)^{2}(x – 10) = 1 x (4)
x – 10 = 4.
x = 6.
( 6 – m)^{2} = 1.
6 – m =1, 1.
m = 5, 7.
Hence m can take three values.