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

f(x) is the maximum value of 2x + 1 and 3 – 4x. Here, we see that in 1 place, x is added to something else, while in the other equation (3 – 4x), x is subtracted from something else. So, in the 1st equation, as x goes higher, f(x) will become bigger, while in the second case, when x goes bigger, the value of f(x) goes smaller, and it will be viceversa when x goes lower. So, we need to find an optimum solution, which we can obtain by equating both the equations.
2x + 1 = 3 – 4x. On solving, we get x = 1/3 and f(x) = 5/3.
If we put any value more than 1/3, 2x + 1 becomes high, and f(x) increases.
If we put any value less than 1/3, 3 – 4x becomes high, and f(x) increases.
So for x = 1/3, f(x) = 5/3 is the minimum possible value.

Q4) A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?
(1) –119
(2) –159
(3) –110
(4) 180
(5) 105 (CAT 2007)

Method 1: As f(x) is a quadratic function and f(x) attains max value of 3 at x=1, so we can express f as in this way, f(x)= k*(x 1)^2 +3.
Remember this useful property, if f is a quadratic function and f attains max/min of b at x=b, then we can write f(x)= k * (x  a)^2 + b where k is a constant.
Now, f(x)= k * (x1)^2 + 3.
Put x=0 in this eq, f(0)= k * (1)^2 + 3
or, 1= k +3
or, k = 2.
So, f(x)= 2 * (x1)^2 +3.
Now, put x=10 in this eq, f(10) = 2 * 9^2 + 3 = 162 + 3 = 159 is the ans. :)Method 2: Let the quadratic equation be ax^2 + bx + c. So, f(x) = ax^2 + bx + c
At x = 0, the value of function is 1.
x = 0, f(x) = 1
ax^2 + bx + c = a * 0 + b * 0 + c = c
c = 1.
At x = 1, f(x) = 3
x = 1, f(x) = 3
a * 1 + b * 1 + c = 3
Since c = 1, a + b = 2.
Also, we know that f(x) is maximum when x = 1. If f(x) is maximum, (dx/dt)(f(x)) = 0
Differentiating f(x), we have d/dt (ax2 + bx + c) = 2ax + b
At x = 1, 2ax + b = 0
2a + b = 0.
b = 2a
Substituting we have a + b = 2, or a + 2a = 2. a = 2. So, b = 4. So the equation is 2 x 2 + 4x + 1.
At x = 10, the value is 2 * 100 + 4 * 10 + 1 = 159

Q5) A function f (x) satisfies f (1) = 3600, and f (1) + f (2) + ... + f (n) = n² f(n), for all positive integers n >1. What is the value of f (9) ?
(1) 80
(2) 240
(3) 200
(4) 100
(5) 120 (CAT 2007)

Solution: There is more than 1 approach to solve the problem. You can either group f(n), find f(n1), gather some formula and find the solution, which might be tricky, time consuming, and “not so easy to click”. Another way is by finding each one of f(2), f(3), f(4)….till f(9), which might take upto 23 minutes depending on your speed, but is a much better choice, as you will definitely get the solution here. I approach these kinds of questions by examples initially, and then try to generalize to a solution. This method will be less time consuming than the above 2 methods, and might prove be less tedious as well.
f(1) = 3600. f(1) + f(2) = 4 f(2)
f(2) = f(1)/3. We can write this as f(2) = f(1) * 1/3.
Now f(1) + f(2) + f(3) = 9 * f(3)
f(3) = f(1) + f(2) / 8. We know f(2) = f(1)/3.
So, f(3) = f(1) + (f(1)/3) / 8.
We can write this as f(3) = f(1) * 1/3 * 2/4.
Since we already have f(1) * 1/3, we write it in such a way that the equation has f(1) * 1/3 and then what comes rest, so that we can generalize.
Calculating similarly for f(4), we get f(4) = f(1) * 1/3 * 2/4 * 3/5.
So f(9) will be f(1) * 1/3 * 2/4 * 3/5 * 4/6 * 5/7 *6/8 * 7/9 * 8/10.
f(9) = (f(1) * 1 * 2) / (9 * 10)
f(9) = 3600 * 2/ 9 * 10 = 80.

Q6) Let g (x) be a function such that g (x + 1) + g (x – l) = g (x) for every real x. Then for what value of p is the relation g (x + p) = g (x) necessarily true for every real x ?
(1) 5
(2) 3
(3) 2
(4) 6 (CAT 2005)

Solution: g (x+1) = g (x) – g (x1)
Let g (x) = p, g (x1) = q.
Then, g (x+1) = g (x) – g (x1) = p – q
g (x+2) = g (x+1) – g (x) = p – q – p = q
g (x+3) = g (x+2) – g (x+1) = qp + q = p
g (x+4) = g (x+3) – g (x+2) = q  p
g (x+5) = g (x+4) – g (x+3) = q –p +p = q = g (x1)
So, we see g (x+5) = g (x1). Thus, entry repeats after 6 times. p = 6

Q7) Let f(x) = ax^2 + bx + c, where a, b and c are certain constants and a ≠ 0. It is known that f (5) = −3f (2) and that 3 is a root of f(x) = 0.
What is the other root of f(x) = 0?
(1) −7
(2) − 4
(3) 2
(4) 6
(5) cannot be determinedWhat is the value of a+b+c?
(1) 9
(2) 14
(3) 13
(4) 37
(5) cannot be determined (CAT 2008)

Solution: First part of the question :
f(3) = 0. So 9a + 3b + c = 0
f(5) = 3f(2). So 25a + 5b + c = 3 (4a + 2b + c). Solving, we get 37a + 121b + 4c = 0.
So we have 2 equations:
9a + 3b + c = 0 > (1)
37a + 11b + 4c = 0 > (2)
Multiply (1) with 4 , we get, 36a + 12b + 4c = 0 > (3)
Subtract (3) from (2), we get, a – b = 0 or a = b. So, we got a = b.
Sum of the roots of a quadratic equation is –b/a. Here it is –a/a = 1
one of the root is 3. Sum of roots is 1. Hence the other root is 4.Second part :
We know the roots are 3 and 4.
Hence the equation is (x3) (x+4) = 0, or x^2 + x – 12 = 0.
However, even 2(x^2 + x – 12) = 0 or 100 (x^2 + x – 12) = 0 and so on will have the same roots.
Hence, we cannot find unique values of a,b,c.

Q8) Let f(x) = ax^2 – b x , where a and b are constants. Then at x = 0, f(x) is:
a. maximized whenever a > 0, b > 0
b. maximized whenever a > 0, b < 0
c. minimized whenever a > 0, b > 0
d. minimized whenever a > 0, b < 0

Method 1 :
f(x) = ax^2 – bx
x^2 and x are always non negative.
If a > 0 and b < 0, f(x) > = 0
So at x = 0, f(x) is minimum when a > 0 and b < 0
Option d.Method 2 :
f(x) = ax^2 – bx
To find max and minimum value, we have to double differentiate.
First differential f ' (x) = 2ax – b * ( x/x ). At x = 0, f ' (x) = b.
Second differential f '' (x) = 2a.
If a > 0, then f '' (x) is positive and hence the equation is minimal.
Also for the minimal value at x = 0, b is less than 0.

Q9) If f (x) = x^3 – 4x + p, and f (0) and f (1) are of opposite signs, then which of the following is necessarily true?
a. –1 < p < 2
b. 0 < p < 3
c. –2 < p < 1
d. –3 < p < 0

f(0) = p
f (1) = 1 – 4 + p = p + 3.
p and p+3 are of different signs.
Substitute for the options:
1st option, if p is a negative number, then p – 3 is also negative, and both are same sign.
For 3rd and 4th option, also it is the same.
For the 2nd option, value of p is between 0 and 3, and p – 3 is negative. Hence it agrees.
Even without substituting the answer can be found out. As p and p – 3 are different signs; p should be between 0 and 3.

Q10)

Solution: 1st sub question:
We need to multiply g with g till we reach e as the solution
g^2 = g * g = h
g^3 = g * g * g = h * g =f
g^4 = h * g * g = h * h = e
So g^4 = e.2nd sub question:
Solve from the innermost bracket.
f * f = h
f ⊕ h = e
f * e = f
f + f = h3rd sub question:
g^9 = g^2 * g^2 * g^2 * g^2 * g = h * h * h * h * g = e * e * g = e * g = g
f ^10 = f^2 * f^2 * f^2 * f^2 * f^2 = h * h * h * h * h = e * e * h = e * h = h
a^10 = a^2 * a^2 * a^2 * a^2 * a^2 = a * a * a * a * a = a * a * a = a * a = a
e^8 = e^2 * e^2 * e^2 * e^2 = e * e * e * e = e * e = e
So, the equation reduces to [a * (h ⊕g)] + e = [a * f ] + e = a + e = e

Q11. (CAT 2002)
If u, v, w and m are natural numbers such that u^m + v^m = w^m, then which of the following is true?
(1) m < Min (u, v, w)
(2) m > Max (u, v, w)
(3) m < Max (u, v, w)
(4) None of these

The question does not specify that u, v, m and w are distinct. The best way to go about this is by trial and error method.
Let us have m = 1. If u = 1, v = 1, then w = 2.
We get the 1st 2 options wrong hence. So, we can either have 3rd option or 4th option right here.
Now, let us have m = 2. If m = 2, we have the equation as u^2 + v^2 = w^2
This is basically a Pythagorean triplet. Examples are u = 3, v =4, w = 5 or u = 5, v = 12, w = 13 etc etc. Hence here also, we see that m < Max (u, v, w). If we put m = 3, we see that the lowest of the two numbers cannot be any combination of 2, 3, 4. i.e.(u, v) cannot be (2,2) as w will not be natural number.
(u, v) cannot be (2,3) as w will not be natural number.
(u, v) cannot be (2,4) as w will not be natural number.
(u, v) cannot be (3,3) as w will not be natural number.
(u, v) cannot be (3,4) as w will not be natural number.
Hence m will be lesser than u, v and w for sure.Hence the answer would be m < Max (u, v, w).
However, if it is mentioned that u, v, w and m are distinct, then the most suited will be Pythagorean triplets or the next case discussed. i.e. We will not have the 1st case where m = 1 and u = 1. u has to be minimum 2 then (distinct). Hence, m is always less than min (u, v, w). (If distinct is mentioned in the question). So, basically,
If the numbers are distinct, then m < Min (u, v, w)
If the numbers are not distinct, then m < Max (u, v, w)

Q12. (CAT 2006)
If logy(x) = a * logz(y) = b * logx(z) = a * b, then which of the following pairs of values for (a, b) is not possible?
(1) 2, 1/2
(2) 1, 1
(3) 0.4, 2.5
(4) π, 1/π
(5) 2, 2

Solution: A little understanding on the theoretical side of logarithms is necessary for solving such questions. If you do not have much theoretical understanding of log, you should know the following at least :
log_10(100) = 2 (You have to know this)
log_10(1000) = 3 (It is easy to remember such things, compared to the formulas)
This means, 10^2 = 100, 10^3 = 1000. That means log_x(y) = a means x^a = y.Also you need to know, log_x(y) = log y/log x (Both to the same base, 10 or e or anything)
Now, moving on to the question,
We know, a * log_z (y) = a * b. So, log_z (y) = b.
We know, b * log_x (z) = a * b. So, log_x (z) = a.
We know, log_y (x) = a * b
Substituting equations 1 and 2 in equation 3, we get,
log_y (x) = log_z(y) * log_x (z)
log x/log y = log y/log z * log z/log x
log x/ log y = log y / log x
log x^2 = log y^2
log x = log y or log x =  log y
x = y or x = 1/y
We know that log y x = a * b
So a * b = log y y or log y 1/y (Substituting for x = 1/y, in the above equation)
So a * b = 1 or a * b = 1.
See from the options, which value of (a, b) does not obey the above.

Q13. (CAT 2008)
If the roots of the equation x^3 – ax2 + bx – c =0 are three consecutive integers, then what is the smallest possible value of b ?
(1) 1/√3
(2) 1
(3) 0
(4) 1
(5) 1/√3