Question Bank  100 Algebra Questions From Previous CAT Papers (Solved)

Solution: For an equation with power 3, the coefficient of x is the sum of the individual product of the roots.
Let (n1), n, (n+1) be the roots of the equation. So, n(n1) + n(n+1) + (n1)(n+1) = b
n^2 – n + n^2 + n + n^2 – 1 = b
3n^2 – 1 = b.
We know that n^2 is positive, the minimum value of n^2 will be 0 when n = 0. So, the minimum value of b is 1 when n = 0.

Q14. (CAT 2005)
For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?
x^2 – y^2 = 0
(x – k)^2 + y^2 = 1
(1) 2
(2) 0
(3) √2
(4) √2

Solution: From 1st equation x^2 = y^2.
Substituting for y^2 in the 2nd equation, we get, (x – k)^2 + x^2 = 1
x^2 + k^2 – 2kx + x^2 = 1
2x^2  2kx + (k^2 – 1) = 0
For unique solution, Discriminant D should be 0. (D = b^2 – 4ac)
D = 4k^2 – 8 (k^2 – 1) = 8 – 4k^2
So, for unique solution, 8 – 4k^2 is 0
4k^2 = 8 => k^2 = 2 => k = √2 or √2
We require the positive solution, substituting, we get positive solution for k = √2

Q15. (CAT 2005)
If x ≥ y and y > 1, then the value of the expression logx(x/y) + logy(y/x) can never be
(1) 1
(2) 0.5
(3) 0
(4) 1

Solution: Let us take everything to the same base, so that it is easier.
log_x(x/y) = (log x – log y)/ log x
log_y(y/x) = (log y – log x)/ log y
Adding these 2, we have (log x – log y)/ log x + (log y – log x)/ log y,
1 – (log_y/log x) + 1 – (log x/ log y)
2  (logy/log x)  (log x/ log y)
2 – (log_x y + log_y x) (As, log y/log x = log_x y and log x/ log y = logy x)
Now, we know y > 1 and x ≥ y. So, log_x y < 1, and logy x > 1 (You can try out by taking examples as x = 100 and y = 10 or similar).
So, we have the expression 2  (log x y + log y x) as 2 – (a number greater than 1).
So, our result is lesser than 1 always.

Q16. (CAT 2004)
f1 (x) = x when 0 ≤ x ≤ 1
= 1 when x ≥ 1
= 0 otherwise
f2 (x) = f1(–x) for all x
f3 (x) = –f2(x) for all x
f4 (x) = f3(–x) for all xHow many of the following products are necessarily zero for every x:
f1(x) f2 (x)
f2 (x) f3 (x)
f2(x) f4 (x)
a. 0
b. 1
c. 2
d. 3Which of the following is necessarily true?
a. f4(x) = f1(x) for all x
b. f1(x) = –f3(–x) for all x
c. f2(–x) = f4(x) for all x
d. f1(x) + f3(x) = 0 for all x

Solution: We have to consider positive and negative numbers for all the cases.
f1(x) is positive for positive numbers, and 0 for negative numbers. (0 for x = 0)
f2(x) is 0 for positive numbers, and positive for negative numbers. (0 for x = 0)
f3(x) is 0 for positive numbers, and negative for negative numbers. (0 for x = 0)
f4(x) is negative for positive numbers, and 0 for negative numbers. (0 for x = 0)
So, we see that, out of the 3 products in the question, f1(x) * f2(x) and f2(x) * f4(x) are always zero, for any x.Second sub question, f4(x) = f3(x) = f2(x) = f1(x).
Hence, 1st option is false.
–f3(x) = f2(x) = f1(x). Hence this is true.

Q17. (CAT 2003)
If 1/3 log_3M + 3 log_3N = 1 + log_0.0008 (5), then:
(1) M^9 = N/9
(2) N^9 = 9/M
(3) M^3 = 3/N
(4) N^9 = 3/M

Solution: 1st let us find the value of log_0.0008 (5).
Let log_0.0008 (5) = y.
So 5 = (0.008)^y.
5 = (8/1000)^y
5 = (2/10)^3y
5 = (1/5)^3y
5 = (5)^(3y)
3y = 1. y = 1/3.
So, 1 + log_0.0008 (5) = 1 – 1/3 = 2/3.Now, the options need us to find a relation between M & N, and hence it should be true for all values of M & N.
We have the equation :1/3 log_3 (M) + 3 log_3 (N) = 2/3.
For simplicity, let us split and give, 1/3 log_3 (M) = 1/3 and 3 log_3 (N) = 1/3.
That means log_3 (M) = 1, and M = 3.
Also, it means, log_3 (N) = 1/9. N = 3^(1/9).
N^9 = 3 and M = 3.
So M * N^9 = 3 * 3 = 9.
This is correct as per the 2nd option.

Q18. (CAT 2003)
If both a and b belong to the set {1, 2, 3, 4}, then the number of equations of the form ax^2 + bx + 1 = 0 having real roots is
a. 10
b. 7
c. 6
d. 12

Solution: For real roots, discriminant should be greater than or equal to 0.
D = b^2 – 4ac. Here, c = 1.
So D = b^2 – 4a ≥ 0.
If b = 1, a cannot be any number among 1, 2, 3, 4 as D < 0.
If b = 2, a can be 1 from the set. For other numbers D < 0.
If b = 3, a can be 1 or 2 from the set. For other numbers D < 0.
If b = 4, a can be 1, 2, 3 or 4.
totally there are 0 + 1 + 2 + 4 = 7 numbers possible for which D ≥ 0, and hence real roots.

Q19. (CAT 2003)
If log_10 (x) – log_10 (√x) = 2 log_x (10), then a possible value of x is given by:
a. 10
b. 1/100
c. 1/1 000
d. None of these

Solution: If you are not sure of the basics of logarithms, follow this method :
Let us suppose log10x = y.
Then x = 10y.
x^(1/y) = 10
logx10 = 1/yIf you are pretty thorough with logarithms, then you know that if log_10 (x) = y, then log_x (10) = 1/y. Also, we know that log_10 (√x) is same as log_10 (x)^1/2, which is same as 1/2 log_10 (x).
So, we have log_10 (x) – log_10 (√x) = 2 log_x (10)
y  y/2 = 2/y
y/2 = 2/y
y^2 = 4
y = 2 or y = 2.
If y = 2, then log_10 (x) = 2, and x = 10^2 = 100.
If y = 2, then log_10 (x) = 2, and x = 10^(2) = 1/100

Q20. (CAT 2003)
What is the sum of ‘n’ terms in the series: log m + log(m^2/n) + log(m^3/n^2) + log (m^4/n^3) + …
(1) log (n^(n1)/ m^(n + 1))^n/2
(2) log (m^m/ n^n)^n/2
(3) log (m^(1  n)/ n^(1  m))^n/2
(4) log (m^(n+1)/ n^(n  1))^n/2

Solution: log m + log(m^2/n) + log(m^3/n^2) + log (m^4/n^3) + ….
= log m + 2 log m – log n + 3 log m – 2 log n + 4 log m – 3 log n + …
= log m ( 1 + 2 + …n ) – log n ( 1 + 2 + ….(n1)) (This is an AP)
= (n)(n+1)/2 * log m  (n1)(n)/2 * log n
= log [m^(n)(n+1)/2  log n^((n1)n/2)]
=log (m^[(n+1)/2]/n^[(n1)/2])^n
= log (m^(n+1)/n^(n1))^(n/2)

Q21. (CAT 2003)
If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y?
a. 2^1/3
b. 2^2/3
c. 2^1/4
d. 2^3/4

Solution: Since all the options of the answer are in the power of 2, let us take x = 2^a and z = 2^b.
So, we get y = 2^[(a + b)/2] (As y is AM of x and z)
We know xyz = 4.
So, 2^a * 2^b * 2^[(a + b)/2] = 2^2.
i.e. 2^[(3a +3b)/2] = 2^2.
So, (3a + 3b)/2 = 2.
3a + 3b = 4.
a + b = 4/3.
(a + b)/2 = 2/3.
We know that y = 2(a + b)/2 , and (a + b)/2 is 2/3.

Q22. (CAT 2002)
For three numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?
(1) 3 √13
(2) √19
(3) 13/3
(4) √15

Solution: (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2xz
25 = x^2 + y^2 + z^2 + 2(xy +yz + xz)
25 = x^2 + y^2 + z^2 + 6
x^2 + y^2 + z^2 = 19.
For x to be maximum, y^2, z^2 should be minimum. Minimum value of a “square” integer is 0.
Hence y^2, z^2 = 0. x^2 will be 19, and x will be √19.

Q23. (CAT 2002)
If x^2 + 5y^2 + z^2 = 2y (2x + z), then which of the following statements are necessarily true?
I. x = 2y II. x = 2z III. 2x = z
(1) Only I
(2) Only II
(3) Only III
(4) Only I and II