Inequalities MCQ Quiz in मराठी - Objective Question with Answer for Inequalities - मोफत PDF डाउनलोड करा

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

\(x\epsilon (-\infty ,-2)\cup (2,\infty )\)...(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)

Concept:

• The Modulus Function '| |' is defined as: \(\rm |x|=\left\{\begin{matrix}\rm \ \ \ x, &\rm x \geq 0\\ \rm -x, &\rm x<0\\\end{matrix}\right.\).
• 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.

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.

Concept:

Steps to find the solutions of inequality by the wavy curve method:

1. Find the roots of the numerator and denominator.
2. Plot those according to their values on the number line.
3. 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 \(\frac{(x-4)^{1991}(x-1)^{2020}}{x^{47}(x+3)^{101}} ≥0\)

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}

Concept:

Log of a power: log x^a = 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.

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, log_a x < log_a 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 x^a = 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, \(\log _{0.5}(x^{2}-25)-\log _{0.5}(x-5)<\log _{0.5}3\)

\(⇒ \log _{0.5}(x^{2}-25)<\log _{0.5}3+\log _{0.5}(x-5)\)

\(⇒ \log _{0.5}(x^{2}-25)<\log _{0.5}3(x-5)\)      (∵ log x + log y = log xy)

⇒ 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.

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

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 + ....a_nb_n)² ≤ (a₁² + a₂² + ...a_n²)(b12 + b22 + ...bn2)

= \(\displaystyle(\sum\limits_{i=1}^na_ib_i)^2≤\sum\limits_{i=1}^na_i^2\sum\limits_{i=1}^nb_i^2\)

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.

Concept:

1. Function y = √x is defined for x ≥ 0

2. Polynomial y = p/q is defined for q ≠ 0 

Calculation:

Given that,

\(y = \frac{x}{\sqrt{x^2 - 3x + 2}}\)

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, ∞)

Concept:

Cauchy Schwarz inequality:

It states that for all the sequences of real numbers a_i and b_i, we have \(\bigg(\displaystyle\sum_{i=1}^{n} a_i^2\bigg).\bigg(\displaystyle\sum_{i=1}^{n} b_i^2\bigg)\geq\bigg(\displaystyle\sum_{i=1}^{n} a_ib_i\bigg)^2 \) . Equality holds if and only if a_i = kb_i for a non-zero constant k ∈ R.

Calculations:

By Cauchy Schwarz inequality, we have for x, y, z ∈ R+

\(\sqrt{x(3x+y)}+\sqrt{y(3y+z)}+\sqrt{z(3z+x)} \)

\(\leq \sqrt{((\sqrt{x})^2+(\sqrt{y})^2+(\sqrt{z})^2)((\sqrt{3x+y})^2+(\sqrt{3y+z})^2+(\sqrt{3z+x})^2)} \)

\(\leq \sqrt{4(x+y+z)^2} \)

\(\leq2(x+y+z) \)

∴ \(\sqrt{x(3x+y)}+\sqrt{y(3y+z)}+\sqrt{z(3z+x)} \)  \(\leq2(x+y+z) \)

Hence, the correct answer is option 2)