# Free Algebra 01 Practice Test - CAT

### Question 1

What is the value of M(M(A(M(x, y), S(y, x)), x), A(y, x)) for x = 2, y = 3? (CAT 1996)

#### SOLUTION

Solution :D

M (M (A (M (x, y), S (y, x)), A (y, x)

M (M (A (6, 1), 2), A (3, 2))

M (M (7, 2), A (3, 2))

M (14, 5) = 70

### Question 2

What is the value of S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]?

^{2}+ b

^{2}

^{2}– b

^{2}

#### SOLUTION

Solution :B

Soln:(b)

S [M (D (A (a, b),2), D (A (a, b), 2)), M (D (S (a, b), 2), D (S (a, b), 2))]

S [M (D (a + b, 2), D (a + b, 2)), M (D (a – b, 2),D (a – b, 2))]

S [M(a + b)/2), (a + b) /2), M( )]

S [ (] = [(a +b)

^{2}– (a – b)^{2}]/ 2^{2}

### Question 3

Find the sum of + + ………… + (CAT 2008)

#### SOLUTION

Solution :A

= ; = ; = .

+ = + = ; = ->3 - (1/3)

Thus, on similar lines,

+.......... =;

### Question 4

The number of common terms in the two sequences 17, 21, 25,…417 and 16, 21, 26….466 is: (CAT 2008)

#### SOLUTION

Solution :C

Soln:

If we have 2 AP’s and are trying to find common terms, then the common terms also form a AP with a CD of LCM of the 2 common differences.

The first sequence is of the form 17 + 4a

And the second sequence is of the form 16 + 5b

By observation we see that 21 is the first term of the new sequence and the common difference is 20 ( LCM of 4 and 5)

Hence the sequence will be 21 + 20c

Now substitute values of c from the answer options and check that the last term should not cross 417.

Hence when we sub c=19 we see that the value is lesser than 417 and when we sub c=20 the value is 421 which is greater than 417.

Hence the number of terms will be 19 + first term = 20 terms or option ( c)

### Question 5

Consider the sequence of numbers a_{1}, a_{2}, a_{3},…. to infinity where a1 = 81.33 and a_{2} = -19 and a_{j} = a_{j} _{– 1} a_{j} _{– 2} , for j >= 3. What is the sum of the first 6002 terms of this sequence?(CAT 2004)

#### SOLUTION

Solution :C

Ans: (c)

The terms of the given sequence are as follows :

a

_{1}= 81.33 a_{7}= a_{1}a

_{2}= -19 a_{8}= a_{2}a

_{3}= a_{2}– a_{1 }a_{9}= a_{3}a

_{4}= - a_{1 }a_{10}= a_{4}= -a_{1}a

_{5}= -a_{2 }a_{11}= a_{4}= - a_{2}a

_{6}= - a_{2}+ a_{1 }a_{12}= a_{6}= -a_{3}and so on.The sum of the first six terms, the next six terms and so on is 0.

The sum of the first 6002 terms can be written as the sum of first 6000 terms + 6001

^{st}term + 6002^{nd}term. From the above explanation, the sum of the first 6000 terms is zero, 6001 term will be a1 and 6002^{nd}term will be a_{2}. The sum of the first 6002 terms will be a_{1}+ a_{2}= 81.33 + (-19) = 62.33.

### Question 6

If f (0, y) = y + 1, and f(x +1, y) = f (x, f(x, y)). Then, what is the value of f(1,2) ? (CAT 2000)

#### SOLUTION

Solution :D

SOLN:(d)

f (x + 1, y) = f(x, f(x, y))

put x = 0, f (1, y) = f(0, f(0, y)) = f(0, y + 1)

= y + 1 + 1 = y + 2

Put y = 2, f (1, 2) = 4.

### Question 7

The function f(x) = |x - 2| + |2.5 - x| + |3.6 - x|, where x is a real number, attains minimum at

#### SOLUTION

Solution :B

Option (b)

f(x) = | x – 2| + | 2.5 – x | + | 3.6 – x | can attain minimum value when either of the terms = 0.

Case I :

When | x – 2 | = 0 => x = 2, value of f(x) = 0.5 + 1.6 = 2.1.

Case II.

When | 2.5 – x | = 0 => x = 2.5

value of f(x)= 0.5 + 0 + 1.1 = 1.6.

Case III.

When | 3.6 – x | = 0 => x = 3.6

f(x) = 1.6 + 1.1 + 0 = 2.7. Hence the minimum value of f(x) is 1.6 at x = 2.5.

### Question 8

Let u _{n+1} = 2u_{n} +1, (n = 0, 1, 2,…) and u_{0} = 0. Then u_{10} would be nearest to: (CAT 1993)

#### SOLUTION

Solution :A

Ans: (a)

u

_{n+1}= 2u_{n}+ 1 (n = 0, 1, 2,…..)Put n = 0, u

_{1}= 1n = 1, u

_{2}= 3n = 2, u

_{3}= 7n = 3, u

_{4}= 15n = 4, u

_{5}= 31Seeing the pattern it is clear that u

_{n}= 2^{n}– 1Hence u

_{10}= (2)^{10}– 1 = 1,023.

### Question 9

If there are 10 positive real numbers n1 < n2 < n3 …… < n10…… How many triplets of these numbers (n1, n2, n3), (n2, n3, n4) …. can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number :(CAT 2002)

#### SOLUTION

Solution :C

Ans: (c)

Step 1: select the numbers

Step 2: arrange the selected numbers according to the given condition.

Three numbers can be selected and arranged out of 10 numbers in

^{10}C_{3}ways = 120.Now, only one arrangement can be possible as A>B>C, hence the solution will be 120.

### Question 10

Value of Ma [md (a), mn (md (b), a), mn (ab, md (ac))] where a = -2, b = -3, c = 4 is (CAT 1994)

#### SOLUTION

Solution :B

(b)

Ma [md (a), mn (md (b), a), mn (ab, md (ac))]

Ma [| - 2 |, mn ( | - 3 |, -2), mn (6, | -8 |)]

Ma [2, mn (3, -2), mn (6, 8)]

Ma[2, -2, 6] = 6

### Question 11

Give that a>b then the relation Ma [md (a) . mn (a, b)] = mn [a, md (Ma (a, b))] does not hold if (CAT 1994)

#### SOLUTION

Solution :A

Soln:(a)

Ma [md (a), mn (a, b)] = mn [a, md (Ma (a, b)]

Ma [2, -3] = mn [-2, md (-2)]

2 = mn ( -2 , 2)

2 = -2

Relation does not hold for a = -2 and b = -3

Or a < 0, b < 0

### Question 12

fog (x) =

#### SOLUTION

Solution :B

Option (b)

fog (x) = f{g(x)} = f {(x-3)/2} = 2 {(x-3)/2} + 3 = x

gof (x) = g{f(x)} = g (2x + 3) = (2x + 3 – 3)/2= x

therefore fog (x) = gof (x)

### Question 13

For what value of x; f(x) = g(x – 3)

#### SOLUTION

Solution :C

Option (c)

f(x) = g(x – 3)

2x + 3 = (x – 3 –3 )/2 = (x – 6)/2

4x + 6 = x – 6

3x = -12

x = -4

### Question 14

What is the value of (gofofogogof) (x) (fogofog) (x)

^{2}

#### SOLUTION

Solution :B

Option (b)

{go fo fo go go f(x)} { fo go fo g (x)} from Q. 3, We have fog (x) = gof (x) = x

Therefore above expression becomes (x) . (x) = x

^{2}

### Question 15

What is the value of fo(fog)o(gof) (x)

^{2}

#### SOLUTION

Solution :C

Option (c)

fo(fog)0(gof) (x)

We have, fog (x) = gof (x) = x

So given expression reduces to f(x), that is, 2x + 3