Ans.

(i)

Sol. The equation can be written as,

(x –2y) (2x + y) = 7

Since x and y are integers and 7 is a prime number, the

above equation is possible only in four cases.

(i) x – 2y = 7, 2x + y = 1

(ii) x – 2y = –7, 2x + y = –1

(iii) x – 2y = 1, 2x + y = 7

(iv) x – 2y = –1, 2x + y = –7

On solving each case for x and y we note that only the

cases (iii) are (iv) yield integer solutions, which are

x = 3, y = 1; x = –3, y = –1

⇒ Choice is correct.

(ii)

Sol. Since y + > |x 2 –2x|

⇒ y + > 0

⇒ y > –

Again |x –1| < 2 – y

⇒ 2 – y > 0

⇒ y < 2

Thus from the inequalities given it follows that – < y

< 2

Since y is an integer y = 0, 1

If y = 0 then the inequalities become – |x 2 –2x|

+ > 0, |x–1| < 2

The second inequality is satisfied by only three integers

0, 1 and 2 (else we can write –2 < x –1 < 2 of –1 < x <

3 ⇒ x = 0, 1, 2)

But out of these only 0 and 2 satisfy the first inequality.

⇒ The ordered pairs (0, 0) and (2, 0) are solutions of the system.

Again for y = 1, we will get

– |x 2 –2x| > 0, |x –1| < 1

Proceeding as earlier, we will get the ordered pair (1, 1)

as another solution.

Thus, the system has 3 solutions.

(iii)

Sol. The inequation has a meaning if

x – 5 ≥ 0, 9 – x ≥ 0

⇒ x ∈ [5, 9]

Since x is a positive integer, we must have x = 5, 6, 7, 8, 9

We easily note that for x = 5, 6, the LHS of the

inequality is negative.

For x= 7 it is zero and for x = 8 it is equal to –1

which is not greater than 1

x = 9 satisfies the given inequality, whence choice

follows.