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1.

The solution of the inequality `(|x+2|-|x|)/(sqrt(8-x^(3))) ge 0` isA. `[-1,2]`B. `[1,2]`C. `[-1,1]`D. `[0,3sqrt(4))`

Answer» Correct Answer - A
`(a)` `(|x+2|-|x|)/(sqrt(8-x^(3))) ge 0`
We must have `8-x^(3) gt 0`
`implies x^(3) lt 8 implies x lt 2`
Also `|x+2|-|x| ge 0`
`implies |x+2| ge |x|implies (x+2)^(2)-x^(2) ge 0 implies 2(x+1) ge 0`
`implies x ge -1`
So `x in [-1,2)`
Discussion
2.

The number of solutions of the equation `sqrt(x^(2))-sqrt((x-1)^(2))+sqrt((x-2)^(2))=sqrt(5)` isA. `0`B. `1`C. `2`D. More than `2`

Answer» Correct Answer - C
`(c )` `|x|-|x-1|+|x-2|=sqrt(5)`
For `x ge 2`,
`x-(x-1)+x-2=sqrt(5)impliesx=1+sqrt(5)`
For ` 1 le x le 2`,
`x-(x-1)+(2-x)=sqrt(5)`
`x=3-sqrt(5)` No solution
For `0 le x lt 1`,
`x-(1-x)+2-x=sqrt(5)`
`x=sqrt(5)-1` No solution
For `x lt 0`,
`-x-(1-x)+2-x=sqrt(5)`
`x=1-sqrt(5)`
Hence, `x=sqrt(5)+1` or `x=1-sqrt(5)`
Discussion
3.

The number of integers satisfying the equation `|x|+|(4-x^(2))/(x)|=|(4)/(x)|` isA. `5`B. `4`C. `6`D. `7`

Answer» Correct Answer - B
`(b)` `|x|+|(4-x^(2))/(x)|=|x+(4-x^(2))/(x)|`
As `|a|+|b|=|a+b|implies ab ge 0`
`:.x((4-x^(2))/(x)) ge 0`, `x ne 0`
`implies x^(2)-4 le 0`
`implies x in [-2,2]-{0}`
Discussion
4.

If `|(12x)/(4x^(2)+9)| le 1`, thenA. `x in R`B. `x in [-3,3]`C. `x in [-1,oo)`D. `x in (-oo,2]`

Answer» Correct Answer - A
`(a)` `|(12x)/(4x^(2)+9)| le 1`
`-1 le (12x)/(4x^(2)+9) le 1`
Since `4x^(2)+(9)` is postive for all real `x`, we can multiply by `4x^(2)+9`
`:. -(4x^(2)+(9)) le 12x le (4x^(2)+(9))`
`(2x+3)^(2) ge 0` and `(2x-3)^(2) ge 0`
Above is true for all real values of `x`.
Discussion
5.

The solution of `|x+(1)/(x)| gt 2` isA. `R-{0}`B. `R-{-1,0,1}`C. `R-{1}`D. `R-{-1,1}`

Answer» Correct Answer - B
`(b)` `|x+(1)/(x)|gt 2` (clearly `x ne 0`)
`implies|(x^(2)+1)/(x)| gt 2 implies (x^(2)+1)/(|x|) gt 2` (`:.x^(2)+1 gt 0`)
`impliesx^(2)+1 gt2|x|`
`implies|x|^(2)-2|x|+1 gt 0 implies (|x|-1)^(2) gt 0`
`implies |x| ne 1 implies x ne-1,1`
`:.xinR-{-1,0,1}`
Discussion
6.

The solution of `|2x-3| lt |x+2|` isA. `(-oo,1//3)`B. `(1//3,5)`C. `(5,oo)`D. `(-oo,1//3) uu (5,oo)`

Answer» Correct Answer - B
`(b)` `|2x-3| lt |x+2|`
Squaring both sides
`(2x-3)^(2)-(x+2)^(2) lt 0`
`(x-5)(3x-1) lt 0`
`implies x in ((1)/(3),5)`
Discussion
7.

Solve : `(|x-1|-3)(|x+2)-5) lt 0`.

Answer» Correct Answer - `x in (-7,-2) uu (3,4)`
We have `(|x-1|-3)(|x+2|-5) lt 0`
Case I : `x ge 1`
`:. (x-4)(x-3) lt 0`
`:.x in (3,4)` .......`(i)`
Case II : `-2 le x lt 1`
`:. (x-3)(x+2) gt 0`
`:.x in (-oo,-2)uu(3,oo)`
`:. Xin phi` (as `-2 le x lt 1`)
Case III : `x lt -2`
`:. (x+2)(x+7) lt 0`
`:.x in (-7,-2)` ..........`(ii)`
From `(i)` and `(ii)`, `x in (-7,-2)uu(3,4)`
Discussion
8.

Solve, `(1-sqrt(21-4x-x^(2)))/(x+1) ge 0`.

Answer» Correct Answer - `x in [-2-2sqrt(6),-1) uu [-2+2sqrt(6),3]`
We have `(1-sqrt(21-4x-x^(2)))/(x+1)ge0`
We must have `21-4x-x^(2) ge 0`
`implies x^(2)+4x-21 le 0`
`implies (x+7)(x-3) le 0`
`implies x in [-7,3]`
Case I : `x+1 gt 0 implies x gt -1` .......`(i)`
So given inequality reduces to
`1 ge sqrt(21-4x-x^(2))`
`implies 1 ge 21-4x-x^(2)`
`implies x^(2)+4x-20 ge 0`
`implies (x+2)^(2)-(2sqrt(6))^(2) ge 0`
`implies (x+2+2sqrt(6))(x+2-2sqrt(6)) ge 0`
`implies x in [-2+2sqrt(6),3]` ............`(ii)`
Case II : `x lt -1`
So given inequality reduces to
`1-sqrt(21-4x-x^(2)) le 0`
`implies 1 le sqrt(21-4x-x^(2))`
`implies 1 le 21-4x-x^(2)`
`implies x^(2)+4x-20 le 0`
`implies x in [-2-2sqrt(6),-1)` ...........`(iii)`
So from `(ii)` and `(iii)`,
`x in [-2-2sqrt(6),-1)uu[-2+2sqrt(6),3]`
Discussion
9.

Solve, `(sqrt(2x-1))/(x-2) lt 1`.

Answer» Correct Answer - `x in [1//2,2) uu (5,oo)`
We have `(sqrt(2x-1))/(x-2) lt 1`
We have `x ge (1)/(2)`
Obviously `x lt 2` satisfies the inequality.
For `x gt 2`, `sqrt(2x-1) lt x -2`
`implies 2x-1 lt x^(2)-4x+4`
`impliesx^(2)-6x+5 gt 0`
`implies x in (-oo,1)uu(5,oo)`
From `(i)`, `(ii)`, `(iii)` , `x in [1//2,2)uu(5,oo)`
Discussion
10.

`(sqrt(8+2x-x^2)>6-3x)`

Answer» Correct Answer - `x in (1,4]`
`sqrt(8+2x-x^(2)) gt 6-3x`
We must `8+2x-x^(2) ge 0`
`implies x in [-2,4 ]` …….`(i)`
For `6-2x lt 0` or `x gt 2` , inequality is satisfied.
Let `6-3x ge 0`
`implies x le 2`…….`(ii)`
`implies 8+2x-x^(2) gt 36+9x^(2)-36x`
`implies 5x^(2)-19x+14 lt 0`
`implies (5x-14)(x-1) lt 0`
`implies x in (1,(14)/(5))`.........`(iii)`
From `(ii)` and `(iii)`, `x in (1,2]`
`:. "finally" x in (1,4]`
Discussion
11.

Solve `x-sqrt(1-|x|) lt 0`.

Answer» Correct Answer - `x in [-1,-(1)/(2)+(sqrt(5))/(2))`
`x-sqrt(1-|x|) lt 0`
We must have `1-|x| ge 0`
`implies |x| le 1`
`implies x in [-1,1]`
Case I : `x in[0,1]`
So, we have `x lt sqrt(1-x)`
`implies x^(2)lt1-x`
`implies x^(2)+x-1 lt 0`
`implies (x+(1)/(2))^(2)-((sqrt(5))//(2))^(2) lt 0`
`implies x in[0,-(1)/(2)+(sqrt(5))/(2))` ..........`(i)`
Case II : `x in [-1,0)`
So, we have `x lt sqrt(1+x)`
`implies "negative " lt "positive"`
`implies x in [-1,0)`.........`(ii)`
Hence, from `(i)` and `(ii)`
`x in [-1,-(1)/(2)+(sqrt(5))/(2))`
Discussion
12.

Solve the simultaneous equations `|x+2|+y=5`, `x-|y|=1`Find x.

Answer» Correct Answer - `x=2`
`|x+2|+y=5` ……….`(i)`
`x-|y|=1` ……..`(ii)`
Eliminating `x`, we get
`|1+|y|+2|+y=5`
`:. (3+|y|)+y=5`
`:. y+|y|=2`
`:.y=2`
`:.x=2`
Discussion
13.

Solve `|x-1|-2|=|x-3|`.

Answer» Correct Answer - `[1,oo)`
`||x-1||-2|=|x-3|`
Case I : `x ge 3`
`:. |x-1-2|=x-3 implies x-3=x-3`
`:.x ge 3`
Case II : If `1 le x lt3`
`:. |x-1-2|=-(x-3)`
`:. -(x-3)=-(x-3)implies1le x lt3`
Case III : `-1 le x lt 1`
`:, |-x+1-2|=-(x-3)`
`:. |-x-1|=-(x-3) implies |x+1|=-x+3`
`x+1=-x+3 implies 2x=2impliesx=1`
Which is not possible
Case IV : IF `x lt -1`
`:. |-x+1-2|=-(x-3)`
`implies |x+1|=-(x-3)implies-(x+1)=-(x-3)`
`implies -x-1=-x+3`
`implies` No solution
Hence `x in [1,oo)`
Discussion
14.

Solve `|x|^(2)-|x|+4=2x^(2)-3|x|+1`.

Answer» Correct Answer - `x=+-3`
`|x|^(2)-|x|+4=2x^(2)-3|x|+1`
`implies |x|^(2)-|x|+4=2|x|^(2)-3|x|+1`
`implies |x|^(2)-2|x|-3=0`
`implies (|x|-3)(|x|+1)=0`
`implies |x|-3=0implies|x|=3impliesx=+-3`
Discussion
15.

Let `A={x:x^(2)-4x+3 lt 0,x in R }` `B={x: 2^(1-x)+p le 0 , x^(2)-2(p+7)x+5 le0}` If `B sube A`, then `p in `A. `[-4,-1]`B. `[-4,oo)`C. `(-oo,1)`D. `[0,1]`

Answer» Correct Answer - A
`(a)` `A=(1,3)`
For `BsubeA{:(2^(1-1)+p le0,p le -1),(2^(1-3)+p le 0, p le -1//4):}}p le -1`
`f(x)=x^(2)-2(p+7)x+5`
`{:(f(1) le0 implies p ge -4),(f(3) le 0 implies pge-1):}}P ge -4`
So `p in[-4,-1]`
Discussion
16.

The solution set of the inequation `|(1)/(x)-2| lt 4`, isA. `(-oo,-1//2)`B. `(1//6,oo)`C. `(-1//2,1//6)`D. `(-oo,-1//2) uu (1//6,oo)`

Answer» Correct Answer - D
`(d)` `|(1)/(x)-2| lt 4 hArr-4 lt (1)/(x)-2 lt 4 hArr -2 lt (1)/(x) lt 6`
Hence `x in (-oo,-(1)/(2))uu((1)/(6),oo)`
Discussion
17.

Let a,b,c,d be real numbers such that |a-b|=2, |b-c|=3, |c-d|=4 Then the sum of all possible values of |a-d|=A. `9`B. `18`C. `24`D. `30`

Answer» Correct Answer - B
`(b)` `|a-b|=2impliesa-b=+2`
`|b-c|=3impliesb-c=+-3`
`|c-d|=4impliesc-d=+-4`
Possible values of `a-d` are `+-9`, `+-5`, `+-3`, `+-1`
`:. |a-d|=9,5,3,1`
Sum `=18`
Discussion
18.

The number of integers satisfying `|2x-3|+|x+5| le |x-8|` isA. `5`B. `6`C. `7`D. `8`

Answer» Correct Answer - C
`(c )` `|2x-3|+|x+5| le |x-8|`
`implies |2x-3|+|x+5| le |(2x-3)-(x-5)|`
`implies (2x-3)(x+5) le 0 implies -5 le x le 3//2`
Discussion
19.

If `f(x)=ax^(2)+bx+c` and `f(-1) ge -4`, `f(1) le 0` and `f(3) ge 5`, then the least value of `a` isA. `1//4`B. `1//8`C. `1//3`D. `-1//3`

Answer» Correct Answer - B
`(b)` `f(-1) ge -4`
`implies a-b+c ge -4`………`(i)`
`f(1) le 0`
`implies a+b+c le 0`
`implies -a-b-c ge 0`………`(ii)`
`f(3) ge 5`
and `9a+3b+c ge 5`……….`(iii)`
From `(i)+(ii) implies -2b ge -4`…….`(iv)`
From `(ii)+(iii)+(iv) implies 8a ge 1 implies age 1//8`
Discussion
20.

The number of integral values of `x` satisfying the equation `|x-|x-4||=4` isA. `5`B. `7`C. `9`D. infinite

Answer» Correct Answer - D
`(d)` `f(x)=|x-|x-4||`
`={{:(2|x-2|,,,xlt4),(4,,,xge4):}`
For `|x-|x-4||=4`
`2|x-2|=4impliesx=0`,`4`
`:.` solution set is `{0}uu[4,oo)`.
Discussion
21.

Which of the following is not the solution of `|2x+5|-|x-3| ge |x+8| ` ?A. `(-oo,-8]`B. `[3,oo)`C. `[-8,3)`D. none of these

Answer» Correct Answer - C
`(c )` `|2x+5|-|x-3| ge |x+8|`
`implies |2x+5|-|x-3| ge |(2x+5)-(x-3)|`
`implies (2x+5)(x-3) ge 0 implies -x le -(5)/(2)` or `x ge 3` .........`(i)`
Also `|2x+5| ge |x-3|`
Squaring both sides,
`4x^(2)+20x+25 ge x^(2)-6x+9`
`implies3x^(2)+26x+16 ge 0`
`implies (3x+2)(x+8) ge 0`
`implies x le -8` or `x ge -(2)/(3)`..........`(ii)`
From `(i)` and `(ii)` , `x in (-oo,-8]uu[3,oo)`
Discussion
22.

Let `a`, `b gt 0` satisfies `a^(3)+b^(3)=a-b`. ThenA. `a^(2)+b^(2)=1`B. `a^(2)+ab+b^(2) lt 1`C. `a^(2)+b^(2) gt 1`D. none of these

Answer» Correct Answer - B
`(b)` `a^(2)+ab+b^(2)=(a^(3)-b^(3))/(a-b)=(a^(3)-b^(3))/(a^(3)+b^(3)) lt 1`.
Discussion
23.

The equation `|2ax-3|+|ax+1|+|5-ax|=(1)/(2)` possessesA. Infinite number of real solutions for some `a in R`B. finitely many real solutions for some `a in R'`C. no real solutions for some `a in R`D. no real solutions `a in R`

Answer» Correct Answer - D
`(d)` We have `|2ax-3|+|ax+1|+|5-ax|=(1)/(2)`
Now
`|2ax-3|+|ax+1|+|5-ax| ge |2ax-3-(ax+1)+5-ax|=1`
So the given equation has no solution.
Discussion