Basic Mathematics for JEE Main – Inequality | Concepts, Formulas & Questions

Q 1 :

If $\frac{| x + 3 | + x}{x + 2} > 1$ , then $x \in$

$(- 5, - 2)$
$(- 1, \infty)$
$(- 5, - 2) \cup (- 1, \infty)$
none of these

1

2

3

4

(3)

$We have,$ $\frac{| x + 3 | + x}{x + 2} > 1$

$\Rightarrow$ $\frac{| x + 3 | + x}{x + 2} - 1 > 0 \Rightarrow \frac{| x + 3 | - 2}{x + 2} > 0$

$Now, two cases arise:$

$Case I: When x + 3 \geq 0, i.e., x \geq - 3 .$

$Then,$ $\frac{| x + 3 | - 2}{x + 2} > 0 \Rightarrow \frac{x + 3 - 2}{x + 2} > 0 \Rightarrow \frac{x + 1}{x + 2} > 0$

$\Rightarrow$ ${x + 1 > 0 and x + 2 > 0}$ $or$ ${x + 1 < 0 and x + 2 < 0}$

$\Rightarrow$ ${x > - 1 and x > - 2}$ $or$ ${x < - 1 and x < - 2}$

$\Rightarrow$ $x > - 1 or x < - 2$

$\Rightarrow$ $x \in (- 1, \infty) or x \in (- \infty, - 2)$

$\Rightarrow$ $x \in [- 3, - 2) \cup (- 1, \infty)$ $[Since x \geq - 3]$ ...(i)

$Case II: When x + 3 < 0, i.e., x < - 3 .$

$Then, \frac{| x + 3 | - 2}{x + 2} > 0 \Rightarrow \frac{- x - 3 - 2}{x + 2} > 0$

$\Rightarrow \frac{- (x + 5)}{x + 2} > 0 \Rightarrow \frac{x + 5}{x + 2} < 0$

$\Rightarrow$ ${x + 5 < 0 and x + 2 > 0}$ $or$ ${x + 5 > 0 and x + 2 < 0}$

$\Rightarrow$ ${x < - 5 and x > - 2}$ $which is not possible,$

$or$ ${x > - 5 and x < - 2}$

$\Rightarrow$ $x \in (- 5, - 2)$

$\Rightarrow$ $x \in (- 5, - 3)$ $[Since x < - 3]$ ...(ii)

$Combining (i) and (ii), the required solution is$

$x \in (- 5, - 2) \cup (- 1, \infty)$

Q 2 :

Solution of $| 3 x + 2 | < 1$ is

$[- 1, - \frac{1}{3}]$
${- \frac{1}{3}, - 1}$
$(- 1, - \frac{1}{3})$
None of these

1

2

3

4

(3)

$We have,$

$| 3 x + 2 |$ $< 1 \Leftrightarrow - 1 < 3 x + 2 < 1$ $\Leftrightarrow - 3 < 3 x < - 1 \Leftrightarrow - 1 < x < - \frac{1}{3}$

Q 3 :

Solution of $2 x - 1 =$ $| x + 7 |$ is

- 2
8
- 2, 8
None of these

1

2

3

4

(2)

$2 x - 1 =$ $| x + 7 |$ $=$ ${\begin{matrix} x + 7, & if x \geq - 7 \\ - (x + 7) & if x < - 7 \end{matrix}$

$∴$ $If x \geq - 7, 2 x - 1 = x + 7 \Rightarrow x = 8$

$If x < - 7, 2 x - 1 = - (x + 7) \Rightarrow 3 x = - 6$

$\Rightarrow$ $x = - 2,$ $Not possible.$

Q 4 :

Solution of $0 <$ $| 3 x + 1 |$ $< \frac{1}{3}$ is

$(- \frac{4}{9}, - \frac{2}{9})$
$[- \frac{4}{9}, - \frac{2}{9}]$
$(- \frac{4}{9}, - \frac{2}{9}) - {- \frac{1}{3}}$
$[- \frac{4}{9}, - \frac{2}{9}] - {- \frac{1}{3}}$

1

2

3

4

(3)

$Let us first solve$ $| 3 x + 1 |$ $< \frac{1}{3}$

$\Leftrightarrow - \frac{1}{3} < 3 x + 1 < \frac{1}{3} \Leftrightarrow - \frac{4}{3} < 3 x < - \frac{2}{3}$

$\Leftrightarrow - \frac{4}{9} < x < - \frac{2}{9}$

$Also, 0 <$ $| 3 x + 1 |$ $is satisfied by each x except when$

$3 x + 1 = 0, i.e., x = - \frac{1}{3}$

$∴$ $Solution is (- \frac{4}{9}, - \frac{2}{9}) - {- \frac{1}{3}}$

Q 5 :

The set of real values of $x$ satisfying the inequality

$| x^{2} + x - 6 |$ $> 6,$ is

$| x - 2 |$ $+$ $| x + 2 |$ $< 4$ is

$(- 4, 3)$
$(- 3, 2)$
$(- 4, - 3) \cup (2, 3)$
$(- 4, - 1) \cup (0, 3)$

1

2

3

4

(4)

$We have,$ $| x^{2} + x - 6 |$ $< 6$

$\Rightarrow$ $- 6 < x^{2} + x - 6 < 6$

$\Rightarrow$ $- 6 < x^{2} + x - 6 and x^{2} + x - 6 < 6$

$\Rightarrow$ $x^{2} + x > 0 and x^{2} + x - 12 < 0$

$\Rightarrow$ $x (x + 1) > 0 and (x + 4) (x - 3) < 0$

$\Rightarrow$ $x \in (- \infty, - 1) \cup (0, \infty) and - 4 < x < 3$

$\Rightarrow$ $x \in (- 4, - 1) \cup (0, 3)$

Q 6 :

For $\frac{| x - 1 |}{x + 2} < 1$ , $x$ lies in the interval

$(- \infty, - 2) \cup (- \frac{1}{2}, \infty)$
$(- \infty, 1) \cup [2, 3]$
$(- \infty, - 4)$
$[- \frac{1}{2}, 1]$

1

2

3

4

(1)

$Case I:$ $| x - 1 |$ $= x - 1,$ $x \geq 1$

$∴$ $\frac{| x - 1 |}{x + 2} < 1 \Rightarrow \frac{x - 1}{x + 2} < 1 \Rightarrow \frac{x - 1}{x + 2} - 1 < 0 \Rightarrow \frac{- 3}{x + 2} < 0$

$\Rightarrow$ $\frac{3}{x + 2} > 0$ $which is possible when x > - 2$

$But it is true for x \geq 1, so common solution is x \geq 1 or x \in [1, \infty)$ ...(i)

$Case II:$ $| x - 1 |$ $= - x + 1,$ $x < 1$

$∴$ $\frac{| x - 1 |}{x + 2} < 1 \Rightarrow \frac{- x + 1}{x + 2} < 1 \Rightarrow \frac{x - 1}{x + 2} + 1 > 0 \Rightarrow \frac{2 x + 1}{x + 2} > 0$

$\Rightarrow$ $x < - 2 or x > - \frac{1}{2}$

$But x < 1, so x \in (- \infty, - 2) \cup (- \frac{1}{2}, 1)$ ...(ii)

$∴$ $From (i) and (ii), x lies in the interval$

$(- \infty, - 2) \cup (- \frac{1}{2}, 1) \cup [1, \infty)$ $or x \in (- \infty, - 2) \cup (- \frac{1}{2}, \infty)$

Q 7 :

The set of all real $x$ satisfying the inequality $\frac{3 - | x |}{4 - | x |} \geq 0$ is

$[- 3, 3] \cup (- \infty, - 4) \cup (4, \infty)$
$(- \infty, - 4) \cup (4, \infty)$
$(- \infty, - 3) \cup (4, \infty)$
$(- \infty, - 3) \cup (3, \infty)$

1

2

3

4

(1)

$Case I: When x \geq 0$

$∴$ $| x |$ $= x$

$\Rightarrow$ $\frac{3 - | x |}{4 - | x |} = \frac{3 - x}{4 - x} \geq 0 \Rightarrow \frac{(3 - x) (4 - x)}{(4 - x)^{2}} \geq 0$

$\Rightarrow$ $(x - 3) (x - 4) \geq 0 and x \neq 4 \Rightarrow x \leq 3 or x > 4$

$But x \geq 0, so x \in [0, 3] \cup (4, \infty)$ ...(i)

$Case II: When x < 0$

$∴$ $| x |$ $= - x$

$\Rightarrow$ $\frac{3 - | x |}{4 - | x |} = \frac{3 + x}{4 + x} \geq 0 \Rightarrow \frac{(x + 3) (x + 4)}{{(x + 4)}^{2}} \geq 0$

$\Rightarrow$ $(x + 3) (x + 4) \geq 0 and x \neq - 4$

$\Rightarrow$ $x < - 4 or x \geq - 3$

$But x < 0, so x \in (- \infty, - 4) \cup [- 3, 0)$ ...(ii)

$From (i) and (ii), x lies in the interval$

$(- \infty, - 4) \cup [- 3, 3] \cup (4, \infty)$

Q 8 :

Solution of $| x - 1 |$ $\geq$ $| x - 3 |$ is

$x \leq 2$
$x \geq 2$
$[1, 3]$
None of these

1

2

3

4

(2)

$We have,$ $| x - 1 |$ $-$ $| x - 3 |$ $\geq 0$

$Also,$ $| x - 1 |$ $-$ $| x - 3 |$ $=$ ${\begin{matrix} - 2, & if x < 1 \\ 2 x - 4, & if 1 \leq x \leq 3 \\ 2, & if x > 3 \end{matrix}$

$If x < 1 \Rightarrow - 2 \geq 0, not possible$

$If 1 \leq x \leq 3$

$\Rightarrow 2 x - 4 \geq 0 \Rightarrow x \geq 2 \Rightarrow x \in [2, \infty) \Rightarrow x \in [2, 3]$

$If x > 3$

$\Rightarrow 2 \geq 0 \Rightarrow x \in (3, \infty)$

$Combining above, we get$ $x \in [2, \infty)$

Q 9 :

If $| 2 x - 3 |$ $<$ $| x + 5 |$ , then $x$ lies in the interval

$(- 3, 5)$
$(5, 9)$
$(\frac{- 2}{3}, 8)$
$(- 8, \frac{2}{3})$

1

2

3

4

(3)

$We have,$ $| 2 x - 3 |$ $<$ $| x + 5 |$ ...(i)

$Let x \geq \frac{3}{2}, then 2 x - 3 \geq 0 and x + 5 \geq 0$

$Thus, 2 x - 3 < x + 5 \Rightarrow x < 8$

$Thus x \in [\frac{3}{2}, 8) satisfies the above inequality for x \geq \frac{3}{2}$

$Now, let - 5 \leq x < \frac{3}{2}$

$Then, 2 x - 3 < 0 and x + 5 \geq 0$

$So, 3 - 2 x < x + 5 \Rightarrow 3 x > - 2 \Rightarrow x > - \frac{2}{3}$

$But, - 5 \leq x < \frac{3}{2}$

$Thus, x \in (- \frac{2}{3}, \frac{3}{2}) satisfies the given inequality (i).$

$Also, let x < - 5 . Then, 2 x - 3 < 0 and x + 5 < 0$

$\Rightarrow 3 - 2 x < - 5 - x \Rightarrow x > 8$ ...(ii)

$However, x < - 5$

$∴$ $The above inequality (ii) does not hold.$

$We have checked all the required intervals.$

$∴$ $x \in [\frac{3}{2}, 8) \cup (- \frac{2}{3}, \frac{3}{2}) \Rightarrow x \in (- \frac{2}{3}, 8)$

Q 10 :

The solution of $\frac{6 x}{4 x - 1} < \frac{1}{2}$ is

$x < - \frac{1}{8}$
$- \frac{1}{8} < x < \frac{1}{4}$
$x < \frac{- 1}{8} and x > \frac{1}{4}$
$x > \frac{1}{8}$

1

2

3

4

(2)

Given inequality is, $\frac{6 x}{4 x - 1} < \frac{1}{2}$ $\Rightarrow \frac{12 x}{4 x - 1} < 1$

$\Rightarrow$ $\frac{12 x}{4 x - 1} - 1 < 0 \Rightarrow \frac{8 x + 1}{4 x - 1} < 0$ $\dots (i)$

For above inequality (i), we have two cases either

$(I)$ $8 x + 1 > 0, 4 x - 1 < 0$ or $(II)$ $8 x + 1 < 0, 4 x - 1 > 0$

From (I), we get $x > - \frac{1}{8}, x < \frac{1}{4}$

$∴$ $x \in (- \frac{1}{8}, \frac{1}{4})$ $\dots (i i)$

From (II), $x < - \frac{1}{8}, x > \frac{1}{4},$ both of which can not be true

$∴$ (ii) represents the solution set of given inequality.

Q 11 :

If $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$ , then the values of $x$ lie in the interval

$(- \infty, - 2]$
$[- 2, 6]$
$(- 3, 7)$
$(- 2, \infty)$

1

2

3

4

(2)

Here two cases arises,

Case 1: When $- \infty < x \leq 1$

Then, $(x - 1) \leq 0, x - 3 < 0$

$\Rightarrow$ $| x - 1 |$ $= - (x - 1)$ and $| x - 3 |$ $= - (x - 3)$

Now, $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$ $\Rightarrow$ $- (x - 1) - (x - 3) \leq 8$

$\Rightarrow$ $- 2 x + 4 \leq 8 \Rightarrow x \geq - 2$ $\dots (i)$

Case 2: When $3 \leq x < \infty$

$∴$ $| x - 1 |$ $= (x - 1)$ and $| x - 3 |$ $= (x - 3)$

Now, $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$

$\Rightarrow$ $x - 1 + x - 3 \leq 8 \Rightarrow 2 x - 4 \leq 8 \Rightarrow x \leq 6$ $\dots (i i)$

From (i) and (ii), we get $x \in [- 2, 6]$

Q 12 :

If $| x - 2 |$ $\leq 1$ , then

$x \in (1, 3)$
$x \in (- 1, 3)$
$x \in [1, 3]$
$x \in [- 1, 3)$

1

2

3

4

(3)

Here, $| x - 2 |$ $\leq 1$

$\Rightarrow$ $- 1 \leq x - 2 \leq 1 \Rightarrow 1 \leq x \leq 3 \Rightarrow x \in [1, 3]$

Q 13 :

If $\frac{| x - 3 |}{x - 3}$ $> 0$ , then

$x \in (- 3, \infty)$
$x \in (3, \infty)$
$x \in (2, \infty)$
$x \in (1, \infty)$

1

2

3

4

(2)

Given, $\frac{| x - 3 |}{x - 3}$ $> 0$

$\Rightarrow$ $| x - 3 |$ $> 0 \Rightarrow x - 3 > 0 \Rightarrow x > 3$

$∴$ $x \in (3, \infty)$

Q 14 :

If $| x + 5 |$ $\geq 10$ , then

$x \in (- 15, 5]$
$x \in (- 5, 5]$
$x \in (- \infty, - 15] \cup [5, \infty)$
$x \in [- \infty, - 15] \cup [5, \infty)$

1

2

3

4

(3)

We know that $| x - a |$ $\geq r$

$\Leftrightarrow$ $x \leq a - r or x \geq a + r$

$∴$ $| x + 5 |$ $\geq 10 \Leftrightarrow x \leq - 5 - 10, or x \geq - 5 + 10$

$\Leftrightarrow$ $x \leq - 15 or x \geq 5 \Leftrightarrow x \in (- \infty, - 15] or x \in [5, \infty)$

$\Leftrightarrow$ $x \in (- \infty, - 15] \cup [5, \infty)$

Q 15 :

The solution set of the rational inequality $\frac{x + 9}{x - 6} \leq 0$ is

$(- \infty, 9) \cup (6, \infty)$
$(- \infty, 9] \cup (6, \infty)$
$(- \infty, 9] \cup [6, \infty)$
$[- 9, 6)$

1

2

3

4

(4)

We have, $\frac{x + 9}{x - 6} \leq 0$

Case I: $x + 9 \geq 0$ and $x - 6 < 0$

$\Rightarrow$ $x \geq - 9$ and $x < 6$ $\Rightarrow - 9 \leq x < 6$ $\dots (i)$

Case II: $x + 9 \leq 0$ and $x - 6 > 0$ $\Rightarrow x \leq - 9$ and $x > 6$

which is not possible simultaneously.

$∴$ Required solution set is $[- 9, 6)$ .

Q 16 :

The solution set of the inequality $- 2 \leq \frac{3 x + 2}{2} < 7$ is

${x : 3 \leq x < 4}$
${x : - 2 \leq x < 3}$
${x : - 2 \leq x < 4}$
${x : 0 \leq x < 6}$

1

2

3

4

(3)

We have, $- 2 \leq \frac{3 x + 2}{2} < 7$

$\Rightarrow$ $- 4 \leq 3 x + 2 < 14 \Rightarrow - 6 \leq 3 x < 12 \Rightarrow - 2 \leq x < 4$

Q 17 :

Which of the following is incorrect for the solution set of the inequation $| | x | - 1 |$ $<$ $| 1 - x |$ $, x \in R$ , is

$(- 1, 1)$
$(0, \infty)$
$(- 1, \infty)$
$(- \infty, - 1)$

1

2

3

4

Select one or more options

(1, 2, 3)

Let us consider the following cases:

Case I: When $x \geq 0$

In this case, we have $| x |$ $= x$

$∴$ $| | x | - 1 |$ $<$ $| 1 - x |$

$\Rightarrow$ $| x - 1 |$ $<$ $| x - 1 |$ , which is not true for $x \geq 0$

Case II: when $x < 0$

$| x |$ $= - x$

$∴$ $| | x | - 1 |$ $<$ $| 1 - x |$

$=$ $| - x - 1 |$ $<$ $| 1 - x |$ $=$ $| x + 1 |$ $<$ $| x - 1 |$

If $x < - 1$ , then $| x + 1 |$ $= - (x + 1)$ and $| x - 1 |$ $= - (x - 1)$

$∴$ $| x + 1 |$ $<$ $| x - 1 |$ $= - (x + 1) < - (x - 1)$

$\Rightarrow$ $- 1 < 1$ , which is always true.

Thus, $(- \infty, - 1)$ is the solution set.

If $- 1 < x < 0,$ then $| x + 1 |$ $= x + 1,$ $| x - 1 |$ $= - (x - 1)$

$∴$ $| x + 1 |$ $<$ $| x - 1 |$ $= x + 1 < - (x - 1) \Rightarrow 2 x < 0 \Rightarrow x < 0$

But, $- 1 < x < 0 .$

Therefore, $x \in (- 1, 0)$ .

Hence, $(- \infty, 0)$ is the solution set.

Q 18 :

If $m x^{2} - 9 m x + 5 m + 1 > 0$ and $m \in [p, \frac{q}{16})$ , then $p + q =$ _________ .

(4)

$m x^{2} - 9 m x + (5 m + 1) > 0$

$\Rightarrow$ $m > 0 and D < 0 \Rightarrow m (61 m - 4) < 0$

$\Rightarrow$ $m \in [0, \frac{4}{61}) \Rightarrow p = 0, q = 4$

$∴$ $p + q = 4$

Q 19 :

If $n = 1983!$ , then the value of expression $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \frac{1}{\log_{4} n} + \dots + \frac{1}{\log_{1983} n}$ is _________.

(1)

$\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \frac{1}{\log_{4} n} + \dots + \frac{1}{\log_{1983} n}$

$= \log_{n} 2 + \log_{n} 3 + \dots + \log_{n} 1983$

$= \log_{n} (2 \cdot 3 \cdot 4 \dots 1983) = \log_{n} 1983! = \log_{n} n = 1$

Q 20 :

The value of ${(\frac{1}{\sqrt{27}})}^{2 - (\frac{\log_{5} 16}{2 \log_{5} 9})} = 0 . p q$ , then $p + q =$ ________ .

(1)

$2 - \frac{\log_{5} 16}{2 \log_{5} 9} = 2 - \frac{\log_{5} 16}{\log_{5} 81} = 2 - \log_{81} 16$

$= 2 - \log_{3} 2 = \log_{3} 9 - \log_{3} 2 = \log_{3} \frac{9}{2}$

${(\frac{1}{\sqrt{27}})}^{\log_{3} \frac{9}{2}} = {(3^{- \frac{3}{2}})}^{\log_{3} \frac{9}{2}} = 3^{\log_{3}}$ ${(\frac{9}{2})}^{- \frac{3}{2}}$

${(\frac{9}{2})}^{- \frac{3}{2}} = {(\frac{2}{9})}^{\frac{3}{2}} = \frac{2 \sqrt{2}}{27} = 0.10 = 0 . p q$

$∴$ $p + q = 1 + 0 = 1$

Q 21 :

If ${(3 x)}^{\log 3} = {(4 y)}^{\log 4}$ , $4^{\log x} = 3^{\log y}$ then $\frac{x + y}{x - y} =$ __________.

(7)

$Taking log on both sides, we get$

$\log 3 (\log 3 + \log x) = \log 4 (\log 4 + \log y)$

$\log x \cdot \log 4 = \log y \cdot \log 3$

$\Rightarrow$ $\log x = - \log 3,$ $\log y = - \log 4$

$\Rightarrow$ $x = \frac{1}{3},$ $y = \frac{1}{4} \Rightarrow \frac{x + y}{x - y} = 7$

Q 22 :

Number of integers in the solution set of $| x - 1 |^{\log_{2} (4 - x)}$ $<$ $| x - 1 |^{\log_{2} (1 + x)}$ are ________.

(2)

$4 - x > 0, 1 + x > 0 \Rightarrow x \in (- 1, 4)$ $\dots (i)$

Let $| x - 1 |$ $< 1 \Rightarrow x \in (0, 2)$ $\dots (i i)$

The inequality implies $\log_{2} (4 - x) > \log_{2} (1 + x)$

$\Rightarrow$ $4 - x > 1 + x$

$\Rightarrow$ $x < \frac{3}{2}$ $\dots (i i i)$

From (i), (ii) and (iii) $\Rightarrow$ $x \in (0, \frac{3}{2})$

Let $| x - 1 |$ $> 1 \Rightarrow x \in (- \infty, 0) \cup (2, \infty)$ $\dots (i v)$

The inequality implies $\log_{2} (4 - x) < \log_{2} (1 + x)$

$\Rightarrow$ $4 - x < 1 + x \Rightarrow x > \frac{3}{2}$ $\dots (v)$

(i), (iv), (v) $\Rightarrow x \in (2, 4)$

Finally, we have $x \in (0, \frac{3}{2}) \cup (2, 4)$

$∴$ Number of integers = 2

Q 23 :

A manufacturer has 800 L of a 10% solution of acid. The range of 35% acid to be added to it is $(a, b)$ such that the acid content in the resultant mixture will be more than 15% but less than 25%, then $a + b =$ ________.

(1400)

$Let x L of 35 % acid solution is added to the solution.$

$Total mixture = (x + 800) L$

$According to question,$

$35 % of x + 10 % of 800 > 15 % of (x + 800)$

and $35 % of x + 10 % of 800 < 25 % of (x + 800)$

$∴$ $\frac{15}{100} (800 + x) < \frac{35}{100} x + \frac{10}{100} \times 800$

and $\frac{25}{100} (800 + x) > \frac{35}{100} x + \frac{10}{100} \times 800$

$\Rightarrow$ $12000 + 15 x < 35 x + 8000$ and $20000 + 25 x > 35 x + 8000$

$\Rightarrow$ $4000 < 20 x$ and $12000 > 10 x$

$\Rightarrow$ $200 < x$ and $1200 > x$ i.e., $200 < x < 1200$

Thus, $a = 200,$ $b = 1200$

Hence, $a + b = 200 + 1200 = 1400$

Topic Question Set

Q 1 :

If $\frac{| x + 3 | + x}{x + 2} > 1$ , then $x \in$

Q 2 :

Solution of $| 3 x + 2 | < 1$ is

Q 3 :

Solution of $2 x - 1 =$ $| x + 7 |$ is

Q 4 :

Solution of $0 <$ $| 3 x + 1 |$ $< \frac{1}{3}$ is

Q 5 :

The set of real values of $x$ satisfying the inequality

$| x^{2} + x - 6 |$ $> 6,$ is

$| x - 2 |$ $+$ $| x + 2 |$ $< 4$ is

Q 6 :

For $\frac{| x - 1 |}{x + 2} < 1$ , $x$ lies in the interval

Q 7 :

The set of all real $x$ satisfying the inequality $\frac{3 - | x |}{4 - | x |} \geq 0$ is

Q 8 :

Solution of $| x - 1 |$ $\geq$ $| x - 3 |$ is

Q 9 :

If $| 2 x - 3 |$ $<$ $| x + 5 |$ , then $x$ lies in the interval

Q 10 :

The solution of $\frac{6 x}{4 x - 1} < \frac{1}{2}$ is

Q 11 :

If $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$ , then the values of $x$ lie in the interval

Q 12 :

If $| x - 2 |$ $\leq 1$ , then

Q 13 :

If $\frac{| x - 3 |}{x - 3}$ $> 0$ , then

Q 14 :

If $| x + 5 |$ $\geq 10$ , then

Q 15 :

The solution set of the rational inequality $\frac{x + 9}{x - 6} \leq 0$ is

Q 16 :

The solution set of the inequality $- 2 \leq \frac{3 x + 2}{2} < 7$ is

Q 17 :

Which of the following is incorrect for the solution set of the inequation $| | x | - 1 |$ $<$ $| 1 - x |$ $, x \in R$ , is

Q 18 :

If $m x^{2} - 9 m x + 5 m + 1 > 0$ and $m \in [p, \frac{q}{16})$ , then $p + q =$ _________ .

Q 19 :

If $n = 1983!$ , then the value of expression $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \frac{1}{\log_{4} n} + \dots + \frac{1}{\log_{1983} n}$ is _________.

Q 20 :

The value of ${(\frac{1}{\sqrt{27}})}^{2 - (\frac{\log_{5} 16}{2 \log_{5} 9})} = 0 . p q$ , then $p + q =$ ________ .

Q 21 :

If ${(3 x)}^{\log 3} = {(4 y)}^{\log 4}$ , $4^{\log x} = 3^{\log y}$ then $\frac{x + y}{x - y} =$ __________.

Q 22 :

Number of integers in the solution set of $| x - 1 |^{\log_{2} (4 - x)}$ $<$ $| x - 1 |^{\log_{2} (1 + x)}$ are ________.

Q 23 :

A manufacturer has 800 L of a 10% solution of acid. The range of 35% acid to be added to it is $(a, b)$ such that the acid content in the resultant mixture will be more than 15% but less than 25%, then $a + b =$ ________.

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Topic Question Set

Q 1 : If |x+3|+xx+2>1, then x∈

Q 2 : Solution of |3x+2|<1 is

Q 3 : Solution of 2x-1=|x+7| is

Q 4 : Solution of 0<|3x+1|<13 is

Q 5 : The set of real values of x satisfying the inequality |x2+x-6|>6, is |x-2|+|x+2|<4 is

Q 6 : For |x-1|x+2<1, x lies in the interval

Q 7 : The set of all real x satisfying the inequality 3-|x|4-|x|≥0 is

Q 8 : Solution of |x-1|≥|x-3| is

Q 9 : If |2x-3|<|x+5|, then x lies in the interval

Q 10 : The solution of 6x4x-1<12 is

Q 11 : If |x-1|+|x-3|≤8, then the values of x lie in the interval

Q 12 : If |x-2|≤1, then

Q 13 : If |x-3|x-3>0, then

Q 14 : If |x+5|≥10, then

Q 15 : The solution set of the rational inequality x+9x-6≤0 is

Q 16 : The solution set of the inequality -2≤3x+22<7 is

Q 17 : Which of the following is incorrect for the solution set of the inequation ||x|-1|<|1-x|, x∈R, is

Q 18 : If mx2-9mx+5m+1>0 and m∈[p,q16), then p+q= _________ .

Q 19 : If n=1983!, then the value of expression 1log2n+1log3n+1log4n+⋯+1log1983n is _________.

Q 20 : The value of (127)2-(log5162log59)=0.pq, then p+q= ________ .

Q 21 : If (3x)log3=(4y)log4, 4logx=3logy then x+yx-y= __________.

Q 22 : Number of integers in the solution set of |x-1|log2(4-x)<|x-1|log2(1+x) are ________.

Q 23 : A manufacturer has 800 L of a 10% solution of acid. The range of 35% acid to be added to it is (a, b) such that the acid content in the resultant mixture will be more than 15% but less than 25%, then a+b= ________.

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Q 1 :

If $\frac{| x + 3 | + x}{x + 2} > 1$ , then $x \in$

Q 2 :

Solution of $| 3 x + 2 | < 1$ is

Q 3 :

Solution of $2 x - 1 =$ $| x + 7 |$ is

Q 4 :

Solution of $0 <$ $| 3 x + 1 |$ $< \frac{1}{3}$ is

Q 5 :

The set of real values of $x$ satisfying the inequality

$| x^{2} + x - 6 |$ $> 6,$ is

$| x - 2 |$ $+$ $| x + 2 |$ $< 4$ is

Q 6 :

For $\frac{| x - 1 |}{x + 2} < 1$ , $x$ lies in the interval

Q 7 :

The set of all real $x$ satisfying the inequality $\frac{3 - | x |}{4 - | x |} \geq 0$ is

Q 8 :

Solution of $| x - 1 |$ $\geq$ $| x - 3 |$ is

Q 9 :

If $| 2 x - 3 |$ $<$ $| x + 5 |$ , then $x$ lies in the interval

Q 10 :

The solution of $\frac{6 x}{4 x - 1} < \frac{1}{2}$ is

Q 11 :

If $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$ , then the values of $x$ lie in the interval

Q 12 :

If $| x - 2 |$ $\leq 1$ , then

Q 13 :

If $\frac{| x - 3 |}{x - 3}$ $> 0$ , then

Q 14 :

If $| x + 5 |$ $\geq 10$ , then

Q 15 :

The solution set of the rational inequality $\frac{x + 9}{x - 6} \leq 0$ is

Q 16 :

The solution set of the inequality $- 2 \leq \frac{3 x + 2}{2} < 7$ is

Q 17 :

Which of the following is incorrect for the solution set of the inequation $| | x | - 1 |$ $<$ $| 1 - x |$ $, x \in R$ , is

Q 18 :

If $m x^{2} - 9 m x + 5 m + 1 > 0$ and $m \in [p, \frac{q}{16})$ , then $p + q =$ _________ .

Q 19 :

If $n = 1983!$ , then the value of expression $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \frac{1}{\log_{4} n} + \dots + \frac{1}{\log_{1983} n}$ is _________.

Q 20 :

The value of ${(\frac{1}{\sqrt{27}})}^{2 - (\frac{\log_{5} 16}{2 \log_{5} 9})} = 0 . p q$ , then $p + q =$ ________ .

Q 21 :

If ${(3 x)}^{\log 3} = {(4 y)}^{\log 4}$ , $4^{\log x} = 3^{\log y}$ then $\frac{x + y}{x - y} =$ __________.

Q 22 :

Number of integers in the solution set of $| x - 1 |^{\log_{2} (4 - x)}$ $<$ $| x - 1 |^{\log_{2} (1 + x)}$ are ________.

Q 23 :

A manufacturer has 800 L of a 10% solution of acid. The range of 35% acid to be added to it is $(a, b)$ such that the acid content in the resultant mixture will be more than 15% but less than 25%, then $a + b =$ ________.