NCERT Solution Class XI Mathematics Linear Inequalities Question 3 (Ex 6.1)

Question 3:

Solve 5x– 3 < 7, when

(i) x is an integer (ii) x is a real number

Answer

The given inequality is 5x– 3 < 7.

−12x > 30

⇒ 5x – 3 + 3 < 7 + 3

⇒ 5x < 10

 

(i) The integers less than 2 are …, –4, –3, –2, –1, 0, 1.

Thus, when x is an integer, the solutions of the given inequality are …, –4, –3, –2, –1, 0, 1.

Hence, in this case, the solution set is {…, –4, –3, –2, –1, 0, 1}.

(ii) When x is a real number, the solutions of the given inequality are given by x < 2,

that is, all real numbers x which are less than 2.

Thus, the solution set of the given inequality is x ∈ (–∞, 2).

NCERT Solution Class XI Mathematics Linear Inequalities Question 2 (Ex 6.1)

Question 2:

Solve –12x > 30, when

(i) x is a natural number (ii) x is an integer

Answer

The given inequality is –12x > 30.

−12x > 30

 

(i) There is no natural number less than (−5/2).

Thus, when x is a natural number, there is no solution of the given inequality.

(ii) The integers less than (−5/2) are …, –5, –4, –3.

Thus, when x is an integer, the solutions of the given inequality are …, –5, –4, –3.

Hence, in this case, the solution set is {…, –5, –4, –3}.

NCERT Solution Class XI Mathematics Linear Inequalities Question 1 (Ex 6.1)

Question 1:

Solve 24x < 100, when (i) x is a natural number (ii) x is an integer

Answer

The given inequality is 24x < 100

24x < 100

 

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than 25/6.

Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.

Hence, in this case, the solution set is {1, 2, 3, 4}.

(ii) The integers less than 25/6 are … are …–3, –2, –1, 0, 1, 2, 3, 4.

Thus, when x is an integer, the solutions of the given inequality are …–3, –2, –1, 0, 1, 2, 3, 4.

Hence, in this case, the solution set is {…–3, –2, –1, 0, 1, 2, 3, 4}.

NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 19 (Mis Ex)

Question 19:

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

Answer

(a + ib) (c + id) (e + if) (g + ih) =A + IB’

∴ |(a + ib) (c + id) (e + if) (g + ih)| = |A + iB|

⇒ |(a + ib) ×|(c + id)| × |(e + if | × |(g + ih)| = |A + iB|   [|z1z2| = |z1||z2|]

On squaring both sides, we obtain

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2

Hence, proved.