Topic: Algebra

Solved? : Yes

Source: Previous CAT Questions ]]>

Topic: Algebra

Solved? : Yes

Source: Previous CAT Questions ]]>

(1) 7

(2) 13

(3) 14

(4) 18

(5) 20 (CAT 2006) ]]>

x can take all the values from 1.

Let's see the point where x becomes greater than or equal to y:

2x + x = 40

x=13, y=14

So, x can take values from 1-13 : 13 possible solutions. ]]>

Study the 1st graph (question, not the options carefully). These things can be observed.

When y - x = 0, y + x = 0

Take around the middle positive portion of the graph. When y-x is in the middle, y+x is not that high. It has just started off. We can assume that y-x is 5 and y+x will be around 1 on the same scale. ( need not be accurate values, just an approximate value)

Similarly, extending more in positive, we can see that y-x = 10, y+x = 2 (again, only approximate, the only thing to bear in mind is that, y-x varies more than y+x)

taking the negative portion, if we extend the graph (imaginary), y - x = -5, y + x = -1.

Similarly, extending more in the negative, y - x = -10, y + x = -2

Let us solve these simple equations, we have in hand :

y – x = 0, y + x = 0. This gives when y = 0, x = 0.

y – x = 5, y + x = 1. This gives y = 3, x = -2.

y - x = 12, y + x = 2. This gives when y = 6, x = -4

y - x = -5, y + x = -1. This gives y = -3, x = 2

y - x = -10, y + x = -2. This gives y = -6, x = 4

So, when y is 0, x is 0, and when y is negative, x is positive and vice versa. So the answer is definitely option d or option e. However, in option 5, we see that the absolute value of x is large, when y is small. But, we have just proved otherwise. Absolute value of y is more, and x is less, and hence option d is the right one.

P.S: One can stop when you feel you have derived enough equations. You will never need to derive more than 5 equations. I personally recommend deriving 3 equations (one for 0, another one for positive y-x and the last one for negative y-x). From that we can observe the relation between y&x (negative and positive), and the absolute value variation ( y varies more than x or vice versa). In theory, this variation of y-x with respect to y+x is nothing but the slope, and from this, the formulae are derived. :-)

]]>(1) 1/3

(2) 1/2

(3) 2/3

(4) 4/3

(5) 5/3 (CAT 2006) ]]>

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.

(1) –119

(2) –159

(3) –110

(4) -180

(5) -105 (CAT 2007) ]]>

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 * (x-1)^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 * (x-1)^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

(1) 80

(2) 240

(3) 200

(4) 100

(5) 120 (CAT 2007) ]]>

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.

(1) 5

(2) 3

(3) 2

(4) 6 (CAT 2005) ]]>

Let g (x) = p, g (x-1) = q.

Then, g (x+1) = g (x) – g (x-1) = p – q

g (x+2) = g (x+1) – g (x) = p – q – p = -q

g (x+3) = g (x+2) – g (x+1) = -q-p + 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 (x-1)

So, we see g (x+5) = g (x-1). Thus, entry repeats after 6 times. p = 6 ]]>

What is the other root of f(x) = 0?

(1) −7

(2) − 4

(3) 2

(4) 6

(5) cannot be determined

What is the value of a+b+c?

(1) 9

(2) 14

(3) 13

(4) 37

(5) cannot be determined (CAT 2008)

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 (x-3) (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.

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 ]]>

f(x) = ax^2 – b|x|

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 – b|x|

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.

a. –1 < p < 2

b. 0 < p < 3

c. –2 < p < 1

d. –3 < p < 0 ]]>

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. ]]>

]]>

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 = h

3rd 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

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 ]]>

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)

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 ]]>

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.

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 ]]>