(01) Find the range of Inequality
3x2 – 7x + 4 ≤ 0
(a) x > 0 (b) x < 0
(c) All x (d) None of these.
Read SolutionSol. 3x2 – 7x + 4 ≤ 0
3x2 – 4x – 3x + 4 ≤ 0
x (3x – 4) – 1 (3x – 4) ≤ 0
(x – 1) (3x – 4) ≤ 0
Case I – When (x – 1) ≤ 0 then (3x – 4) ≥ 0
x ≤ 1 and x ≥ 4/3
Hence, x belongs to [-∞, 1] and [4/3, ∞] which is not possible
Case II – When (x – 1) ≥ 0 then (3x – 4) ≤ 0
x ≥ 1 and x ≤ 4/3
Hence, x belongs to [1, 4/3]. It is possible but not in option.
Hence, option (d) is the correct answer.
(02) Solve the following inequality
3x2 – 7x – 6 < 0
(a) –0.66 < x < 3 (b) x < – 0.66 or x > 3
(c) 3 < x < 7 (d) –2 < x < 2
Read SolutionSol. 3x2 – 7x – 6 < 0
3x2 – 9x + 2x – 6 < 0
3x (x –3) + 2 (x –3) < 0
(3x + 2) (x – 3) < 0
Case I: When (3x + 2) < 0 then (x – 3) > 0
x < -2/3 and x > 3
Hence, x belongs to [-∞, -2/3] and [3, ∞] which is not possible.
Case II: When (3x + 2) > 0 then (x – 3) < 0
x > -2/3 and x < 3
Hence, -2/3 < x < 3 or –0.66 < x < 3
Hence, option (a) is the correct answer.
(03) Find the right answer for given inequality
3x2 – 7x + 6 < 0
(a) 0.66 < x < 3 (b) –0.66 < x < 3
(c) –1 < x < 3 (d) None of these
Read SolutionSol. 3x2 – 7x + 6 < 0
Putting values,
x = {- (-7) ± √(-7)2 – 4(3)(6)}/2(3)
Here, value of “d” = √49 – 72 = √-23 is negative.
Hence, the value of x will be an imaginary number (i.e., not real)
Hence, option (d) is the correct answer.
(04) Find the right value for given inequality
x2 – 14x – 15 > 0
(a) x < –1 (b) 15 < x
(c) Both (a) and (b) (d) –1 < x < 15
Read SolutionSol. x2 – 14x – 15 > 0
x2 – 15x + x – 15 > 0
X (x – 15) + 1 (x – 15) > 0
(x + 1) (x – 15) > 0
Case I: When (x + 1) > 0 and (x – 15) > 0
x > -1 and x > 15
Hence, x > 15
Case II: When (x + 1) < 0 and (x – 15) < 0
x < -1 and x < 15
Hence, x < -1
Hence, option (c) is the correct answer.
(05) Select the right answer for the given inequality
2 – x – x2 ≥ 0
(a) –2 ≤ x ≤ 1 (b) –2 < x < 1
(c) x < –2 (d) x > 1
Read SolutionSol. 2 – x – x2 ≥ 0
Multiplying by (-1)
x2 + x – 2 ≤ 0
x2 + 2x –x – 2 ≤ 0
x (x + 2) -1 (x + 2) ≤ 0
(x – 1) (x + 2) ≤ 0
Case I – When (x – 1) ≥ 0 and (x + 2) ≤ 0
x ≥ 1 and x ≤ -2
It is not possible.
Case II – When (x – 1) ≤ 0 then (x + 2) ≥ 0
x ≤ 1 and x ≥ -2
Hence, -2 ≤ x ≤ 1
Hence, option (a) is the correct answer.