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At x = 0, inequality is not satisfied. Thus, option (c) is rejected. Also x = 0 is not a solution of the equation. Since, this is a continuous function, the solution cannot start from 0. Thus options (a) and (b) are not right. Further, we see that the given function is quadratic with real roots. Hence, option (d) is also rejected.
At x = 0, inequality is not satisfied.
Hence, options (b), (c) and (d) are rejected. At x = 2, inequality is not satisfied. Hence, option (a) is rejected.
Thus, option (d) is correct.
At x = 0 inequality is not satisfied. Thus option (d) is rejected. x = 1 and x = 15 are the roots of the quadratic equation. Thus, option (c) is correct.
At x = 0 inequality is satisfied, option (b) is rejected.
At x = 2, inequality is satisfied, option (c) is rejected.
At x = 5, LHS = RHS.
Thus, option (d) is correct.
At x = 1 and x = 3 LHS = RHS.
At x = 2 inequality is satisfied.
At x = 0.1 inequality is not satisfied.
At x = 2.9 inequality is satisfied.
At x = 3.1 inequality is not satisfied.
Thus, option (a) is correct.
At x = 0, inequality is not satisfied, option (a) is rejected.
At x = 5, inequality is not satisfied, option (b) is rejected.
At x = 2 inequality is not satisfied.
Options (d) are rejected.
Option (c) is correct.
At x = 2, inequality is satisfied.
At x = 0, inequality is not satisfied.
At x = 1, inequality is not satisfied but LHS = RHS. At x = 3, inequality is not satisfied but LHS = RHS. Thus, option (b) is correct.
Solve other questions of LOD I and LOD II in the same fashion.
At x = 0, inequality is satisfied. Hence, options (b) and (c) are rejected. x = 3 gives LHS = RHS.
and x = – 0.66 also does the same. Hence. roots of the equation are 3 and – 0.66.
Thus, option (a) is correct.
At x = 0 inequality is not satisfied. Thus option (d) is rejected.
x = –1 and x = 15 are the roots of the quadratic equation. Thus, option (c) is correct.
x^{2}  2x  3 ≥ 0
(x3) (x+1) ≥ 0
x belongs to (∞,3]∪[3,∞)
Therefore, x belongs to (1,3)
=> x^{2}  2x  3 > 0
x^{2}  2x  3< 3x  3
x^{2}  5x < 0
x(x5) < 0
x belongs to (0,5)........(1)
x^{2}  2x  3 < 0
x^{2}  2x  3 < 3x  3
x^{2} + x  6 > 0
(x+3)(x2) > 0
x belongs to (∞,3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
At x = 0, inequality is not satisfied, option (a) is rejected.
At x = 5, inequality is not satisfied, option (b) is rejected.
At x = 2 inequality is not satisfied.
Options (d) are rejected.
Option (c) is correct
At x = 2, inequality is satisfied.
At x = 0, inequality is not satisfied.
At x = 1, inequality is not satisfied but LHS = RHS.
At x = 3, inequality is not satisfied but LHS = RHS.
Thus, option (b) is correct.
Solve other questions of LOD I and LOD II in the same fashion.
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