Inequalities MCQ Quiz in मल्याळम - Objective Question with Answer for Inequalities - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 23, 2025
Latest Inequalities MCQ Objective Questions
Top Inequalities MCQ Objective Questions
Inequalities Question 1:
The smallest negative integer satisfying both the quadratic inequalities x2 < 4x + 77 and x2 > 4 is
Answer (Detailed Solution Below)
Inequalities Question 1 Detailed Solution
Explanation:
Given that, x2 < 4x + 77 and x2 > 4
x2 - 4x - 77 < 0
⇒ x2 - 11x + 7x - 77 < 0
⇒ x(x - 11) + 7(x - 11) < 0
⇒ (x + 7)(x - 11) < 0
∴ -7 < x < 11
x ∈ (-7, 11)...(1)
Also, x2 > 4
⇒ x2 - 4 > 0
⇒ (x + 2)(x - 2) > 0
∴ x < -2 or x > 2
Using (1) and (2), Mark these regions on a real line and take their intersection.
The intersection of the two regions is given by -7 < x < -2 or 2 < x < 11
We are to find the smallest negative integer.
Hence we choose -7 < x < -2.
Therefore all negative integers -6 to -3 will satisfy the above and the smallest amongst these is -6.
∴ The correct option is (2)
Inequalities Question 2:
How many numbers in the set {-4, -3, 0, 2} satisfy the conditions |y - 4| < 6 and |y + 4| < 5?
Answer (Detailed Solution Below)
Inequalities Question 2 Detailed Solution
Concept:
- The Modulus Function '| |' is defined as:
. - If a > b, then a + x > b + x for any x.
- If a > b, then ax > bx for positive x and ax < bx for negative x.
Calculation:
First let us solve both the equations for the values of y.
For |y - 4| < 6:
If y - 4 ≥ 0 ⇒ y ≥ 4, then:
y - 4 < 6
⇒ y < 10.
If y - 4 < 0 ⇒ y < 4, then:
-(y - 4) < 6
⇒ y - 4 > -6
⇒ y > -2.
∴ The solution in this case is [4, 10) ∪ (-2, 4) = (-2, 10).
For |y + 4| < 5:
If y + 4 ≥ 0 ⇒ y ≥ -4, then:
y + 4 < 5
⇒ y < 1.
If y + 4 < 0 ⇒ y < -4, then:
-(y + 4) < 5
⇒ y + 4 > -5
⇒ y > -9.
∴ The solution in this case is [-4, 1) ∪ (-9, -4) = (-9, 1).
Finally, the values of y which satisfy both the in-equations will be (-2, 10) ∩ (-9, 1) = (-2, 1).
Now, the numbers in the set {-4, -3, 0, 2} which are also in the solution set (-2, 1) are: {0} only.
∴ Only 1 number in the set {-4, -3, 0, 2} satisfies the conditions |y - 4| < 6 and |y + 4| < 5.
Inequalities Question 3:
If x satisfies the inequality
|x - 1| + |x - 2| + |x - 3| ≥ 6
then
Answer (Detailed Solution Below)
Inequalities Question 3 Detailed Solution
Concept:
Consider a function f(x) = |x| then
f(x) = x if x > 0 and
f(x) = -x if x < 0
Calculation:
Given that,
|x - 1| + |x - 2| + |x - 3| ≥ 6
Case:1 x < 1
|x - 1|, |x - 2| and |x - 3| will open as negative.
⇒ -(x - 1) - (x - 2) - (x - 3) ≥ 6
⇒ -3x + 6 ≥ 6
⇒ -3x ≥ 0 or x ≤ 0
Case:2 1 < x < 2
⇒ x - 1 - (x - 2) - (x - 3) ≥ 6
⇒ - x + 4 ≤ 6
⇒ x ≤ -2
As, we have taken x ∈ (1, 2) and x is come out to be in (-∞, -2].
Hence, no value will be possible.
Case:3 2 < x < 3
⇒ x - 1 + x - 2 - x + 3 ≥ 6
⇒ x ≥ 6
No value will will be possible.
Case:4 x > 3
⇒ x - 1 + x - 2 + x - 3 ≥ 6
⇒ 3x - 6 ≥ 6
⇒ x ≥ 4
By taking union of all solution,
x ∈ (-∞, 0] ∪ [4, ∞)
The solution of inequality is x ≤ 0 or x ≥ 4.
Inequalities Question 4:
The value of x satisfying the inequality
Answer (Detailed Solution Below)
Inequalities Question 4 Detailed Solution
Concept:
Steps to find the solutions of inequality by the wavy curve method:
- Find the roots of the numerator and denominator.
- Plot those according to their values on the number line.
- Find the sign (positive or negative) of the segments separated by the roots and answer according to the condition of the question.
Calculations:
Given inequality is
Roots of numerator:
(x - 4)1991 ≥ 0
⇒ (x - 4) ≥ 0
Root = 4
(x - 1)2020 ≥ 0
⇒ (x - 1) ≥ 0
Root = 1
Roots of Denominator:
x47 ≥ 0
Root = 0
(x + 3)101 ≥ 0
⇒ (x + 3) ≥ 0
Root = - 3
Now,
Using wavy curve method:
∴ x ∈ (-3, 0) U [ 4, ∞) U {1}
Inequalities Question 5:
The greatest possible value of n could be if 27n < 1010, given that log 3 = 0.4771 and n ϵ N:
Answer (Detailed Solution Below)
Inequalities Question 5 Detailed Solution
Concept:
Log of a power: log xa = a.log x ---(1)
- log x + log y = log xy
- log (x/y) = log x - log y
- log (e) = 1
- log (1) = 0
- log 10 = 1
-
log (1/x) = - log x
Calculation:
We have,
27n < 1010
⇒ 33n < 1010
Taking log both sides, we get,
⇒ 3n.log 3 < 10.log 10
⇒ 3n × 0.4771 < 10 (∵ log 10 = 1 and log 3 = 0.4771)
⇒ n × 1.4313 < 10
⇒ n < 10/1.4313
⇒ n < 6.9867
⇒ n = 6
Hence, the greatest possible value of n = 6.
Inequalities Question 6:
The real values of x satisfying the inequalities:
Answer (Detailed Solution Below)
Inequalities Question 6 Detailed Solution
Concept:
Solving logarithmic inequalities, it is important to understand that the direction of inequality changes if the base of the logarithmic is less than 1.
If, loga x < loga y
Then,
- x < y, when 'a' is more than 1.
-
x > y, when 'a' is less than 1.
Formula used:
log x + log y = log xy
- log xa = a.log x
- log (x/y) = log x - log y
- log (e) = 1
- log (1) = 0
- log 10 = 1
-
log (1/x) = - log x
Calculation:
Domain: x2 - 25 > 0 or x - 5 > 0
⇒ x > 5 ---(1)
Now,
⇒ x2 - 25 > 3 (x - 5) (Direction changed as base is less than 1)
⇒ x2 - 25 > 3x - 15
⇒ x2 - 3x - 10 > 0
⇒ (x - 5) (x + 2) > 0
⇒ x > 5 or x > -2 ---(2)
From equation (1) & (2), we get,
x > 5
Hence, x > 5.
Inequalities Question 7:
Solve the system of inequalities (x + 5) - 7(x - 2) > 4x + 9, 2(x - 3) - 7(x + 5) < 3x - 9
Answer (Detailed Solution Below)
Inequalities Question 7 Detailed Solution
Calculation:
Given:
(x + 5) - 7(x - 2) > 4x + 9, 2(x - 3) - 7(x + 5) < 3x - 9
⇒ (x + 5) - 7(x - 2) > 4x + 9
⇒ x + 5 - 7x + 14 > 4x + 9
⇒ - 6x + 19 > 4x + 9
⇒ - 6x - 4x > 9 - 19
⇒ - 10x > - 10
⇒ x < 1
⇒ x € (- ∞, 1)
2(x - 3) - 7(x + 5) < 3x - 9
⇒ 2x - 6 - 7x - 35 < 3x - 9
⇒ - 5x - 41 < 3x - 9
⇒ -5x - 3x < 41 - 9
⇒ - 8x < 32
⇒ - x < (32/8) = 4
⇒ x > - 4
⇒ x € (- 4, ∞)
Hence the solution set is
[- 4, ∞) ∩ (- ∞, 1] = [- 4, 1]
Which means - 4 < x < 1
Inequalities Question 8:
If x2 + y2 + z2 = 5, then maximum value of (x + 2y + 3z)2 is
Answer (Detailed Solution Below)
Inequalities Question 8 Detailed Solution
Concept:
Cauchy - Schwarz Inequality:
a1, a2, b1, b2 are non-zero real numbers
If a1, . . . , an and b1, . . . , bn are real numbers, then
(a1a2 + b1b1 + ....anbn)2 ≤ (a12 + a22 + ...an2)(b12 + b22 + ...bn2)
=
Calculation:
Given that,
x2 + y2 + z2 = 5 ---(1)
Let,
P = (x + 2y + 3z)2
⇒ P = (1.x + 2.y + 3.z)2
Applying Cauchy - Schwarz Inequality
(a1a2 + b1b1 + ....anbn)2 ≤ (a12 + a22 + ...an2)(b12 + b22 + ...bn2)
⇒ (x + 2y + 3z)2 ≤ (12 + 22 + 32)(x2 + y2 + z2)
⇒ (x + 2y + 3z)2 ≤ 14 × 5 [From equation (1)]
⇒ (x + 2y + 3z)2 ≤ 70
Hence, required maximum value is 70.
Inequalities Question 9:
The domain of the function
Answer (Detailed Solution Below)
Inequalities Question 9 Detailed Solution
Concept:
1. Function y = √x is defined for x ≥ 0
2. Polynomial y = p/q is defined for q ≠ 0
Calculation:
Given that,
As discussed above, y is defined for
x2 - 3x + 2 > 0
⇒ (x - 2)(x - 1) > 0
This represents, a product of the two-term is positive, which can be possible in two cases.
Case:1 (x - 2) and (x - 1), both positive
⇒ x > 2 or x ϵ (2, ∞) ----(1)
Case:2 (x - 2) and (x - 1), both negative
⇒ x < 1 or x ϵ (-∞, 1) ----(2)
Hence, from (1) & (2)
⇒ x ϵ (-∞, 1) ∪ (2, ∞)
Inequalities Question 10:
For x, y, z ∈ R+, then
Answer (Detailed Solution Below)
Inequalities Question 10 Detailed Solution
Concept:
Cauchy Schwarz inequality:
It states that for all the sequences of real numbers ai and bi, we have
Calculations:
By Cauchy Schwarz inequality, we have for x, y, z ∈ R+
∴
Hence, the correct answer is option 2)