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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If `x = 2 + sqrt3, y = 2 - sqrt3`, then find the simpliest value of (a) `x - (1)/(x)`, (b) `y^(2) + (1)/(y^(2))`, (c) `x^(3) - (1)/(x^(3))`, (d) `xy + (1)/(xy)` |
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Answer» `x = 2 + sqrt3, " " :. (1)/(x) = (1)/(2 + sqrt3) = (1)/(2 + sqrt3) xx (2 - sqrt3)/(2 - sqrt3)` `= (2 - sqrt3)/((2)^(2) - (sqrt3)^(2)) = (2 - sqrt3)/(4 -3) = 2 - sqrt3` `y = 2 - sqrt3, " " :. (1)/(y) = (1)/(2 - sqrt3) = (1)/(2 - sqrt3) xx (2 + sqrt3)/(2 + sqrt3)` `= (2 + sqrt3)/((2)^(2) - (sqrt3)^(2)) = (2 + sqrt3)/(4 - 3) = (2 + sqrt3)/(1) = 2 + sqrt3` `:.` (a) `x - (1)/(x) = 2 + sqrt3 - 2 + sqrt3 = 2 sqrt3` (b) `y^(2) + (1)/(y^(2)) = (y + (1)/(y))^(2) -2.y.(1)/(y) = (2 - sqrt3 + 2 + sqrt3)^(2) -2` `= (4)^(2) -2 = 16 - 2 = 14` (c) `x^(3) - (1)/(x^(3)) = (x)^(3) - ((1)/(x))^(3) = (x - (1)/(x))^(3) + 3.x.(1)/(x) (x - (1)/(x))` `= (2 sqrt3)^(3) + 3 xx 2 sqrt2 = 24 sqrt3 + 6 sqrt3 = 30 sqrt3` (d) `xy + (1)/(xy) = (2 + sqrt3) (2 - sqrt3) + (1)/((2 + sqrt3) (2 - sqrt3))` `= (2)^(2) - (sqrt3)^(2) + (1)/((2)^(2) - (sqrt3)^(2)) = 4 - 3 + (1)/(4 -3)` `= 1 + (1)/(1) = 1 + 1 = 2` |
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| 2. |
If `m + (1)/(m) = sqrt3`, then find the simpliest value of (i) `m^(2) + (1)/(m^(2))` and (ii) `m^(3) + (1)/(m^(3))`: |
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Answer» (i) `m^(2) + (1)/(m^(2)) = (m)^(2) + ((1)/(m))^(2)` `= (m + (1)/(m))^(2) -2.m. (1)/(m) = (sqrt3)^(2) -2 = 3 -2 =1` `:. m^(2) + (1)/(m^(2)) = 1` (ii) `m^(3) + (1)/(m^(3)) = (m)^(3) + ((1)/(m))^(3) = (m + (1)/(m))^(3) -3.m. (1)/(m) (m + (1)/(m))` `= (sqrt3)^(3) -3 xx sqrt3 = 3 sqrt3 - 3 sqrt3 = 0` `:. m^(3) + (1)/(m^(3)) = 0` |
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| 3. |
The simpliest value of `(sqrt5 (sqrt(3 - sqrt5)))/(sqrt2 + sqrt(7 - 3 sqrt5))` isA. 1B. 5C. 10D. None of these |
| Answer» Correct Answer - A | |
| 4. |
Rationalise the denominator: (a) `(1)/(root(3)(3) + root(3)(2))`, (b) `(2)/(sqrt5 + sqrt3 + sqrt2)`, (c) `(x^(2))/(sqrt(x^(2) + y^(2)) - y)`, (d) `(1)/(sqrt6 + sqrt5 - sqrt11)` (e) `(sqrt(x + 2y) - sqrt(x -2y))/(sqrt(x + 2y) + sqrt(x - 2y))`, (f) `(sqrt10 + sqrt5 - sqrt3)/(sqrt10 - sqrt5 + sqrt3)` |
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Answer» (a) `(root(3)(9)- root(3)(6) + root(3)(4))/(5)`, (b) `(2 sqrt3 + 3 sqrt2 - sqrt30)/(6)`, (c) `sqrt(x^(2) + y^(2)) + y`, (d) `(6 sqrt5 + 5 sqrt6 + sqrt(330))/(60)`, (e) `(x - sqrt(x^(2) - 4y^(2)))/(2y)`, (f) `(1)/(7) (3 sqrt30 + 5 sqrt15 - 10sqrt2 - 12)` |
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| 5. |
Prove that `(12)/(3 + sqrt5 - 2 sqrt2) = 1 + sqrt5 + sqrt10 - sqrt2` |
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Answer» `(3 + sqrt5 -2 sqrt2) (1 + sqrt5 + sqrt10 - sqrt2)` `= 3 + sqrt5 - 2 sqrt2 + 3 sqrt5 + 5 -2 sqrt10 + 3 sqrt10 + 5sqrt2- 4sqrt5 - 3sqrt2 - sqrt10+4` = 12 `:. 1 + sqrt5 + sqrt10 - sqrt2 = (12)/(3 + sqrt5 - 2 sqrt2)` Hence, `(12)/(3 + sqrt5 -2 sqrt2) = 1 + sqrt5 + sqrt10 - sqrt2` |
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| 6. |
`sqrt7 - sqrt5` is a binomial surd |
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Answer» Correct Answer - 1 Since here the number of terms of surd is two, viz, `sqrt7 and sqrt5`. Hence the given statement is true |
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| 7. |
Which of `(sqrt15 + sqrt3) and (sqrt10 + sqrt8)` is greater ? |
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Answer» We have `(sqrt15 + sqrt3)^(2) = (sqrt15)^(2) + 2 xx sqrt15 xx sqrt3 + (sqrt3)^(2)` `= 15 + 2 sqrt45 + 3 = 18 + 2 xx sqrt45` Again, we have `(sqrt10 + sqrt8)^(2) = (sqrt10)^(2) + 2 xx sqrt10 xx sqrt8 + (sqrt8)^(2)` `= 15 + 2 sqrt45 + 3 = 18 + 2 xx sqrt45` Again, we have `(sqrt10 + sqrt8)^(2) = (sqrt10)^(2) + 2 xx sqrt10 xx sqrt8 + (sqrt8)^(2)` `= 10 + 2 xx sqrt(10 xx 8) + 8` `= 18 + 2 sqrt80` Now, `sqrt80 gt sqrt45 , :. (sqrt10 + sqrt8)^(2) gt (sqrt15 + sqrt3)^(2)` `rArr (sqrt10 + sqrt8) gt (sqrt15 + sqrt3)` Hence between `(sqrt10 + sqrt8) and (sqrt15 + sqrt3), (sqrt10 + sqrt8)` is greater |
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| 8. |
Between `(sqrt6 + sqrt7) and (sqrt5 + sqrt8)`, the greater one is-A. `sqrt5 + sqrt8`B. `sqrt6 + sqrt7`C. `sqrt5 + 2 sqrt2`D. None of these |
| Answer» Correct Answer - B | |
| 9. |
Which one is greater between `(sqrt7 - sqrt5) and (sqrt13 - sqrt11)` ? |
| Answer» Greater `= (sqrt7 - sqrt5)` | |
| 10. |
Express in the form of like surds: `root(3)(4), root(6)(5), 2sqrt3`. |
| Answer» `root(12)(256), root(12)(25), root(12)(2985984)` | |
| 11. |
Express the followings in the form of mixed surds: `root(5)(-2187 a^(6) b^(11)), root(3)(48 (sqrt3 -1)^(3))` |
| Answer» `-3ab^(2) root(5)(9ab), 2(sqrt3-1) root(3)(6)` | |
| 12. |
Find the product: `(5 - sqrt3) (sqrt3 -1) (5 + sqrt3) (sqrt3 +1)` |
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Answer» `(5 - sqrt3) (sqrt3 -1) (5 sqrt3) (sqrt3 +1)` `= {(5 - sqrt3) (5 + sqrt3)} {(sqrt3-1) (sqrt3+1)}` `= {(5)^(2) - (sqrt3)^(2)} {(sqrt3)^(2) - (1)^(2)} = (25 -3) (3-1)` `= 22 xx 2 = 44` Hence the required result = 44 |
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| 13. |
If `(sqrt(5 + 2sqrt6)- sqrt(5 - 2 sqrt6))/(sqrt(5 + 2 sqrt6)+ sqrt(5 - 2 sqrt6)) = sqrt((x)/(16))`, then x = _____ |
| Answer» Answer: `10 (2)/(3)` | |
| 14. |
Between the two numbers `4 root(3)(3) and 2 root(4)(3)`, the number _________ is greater |
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Answer» `4root(3)(3)` is greater, since `4 root(3)(3) = 2^(2) xx 3^((1)/(3)) and 2root(4)(3) = 2 xx 3^((1)/(4))` and we have `2^(2) gt 2 and 3^((1)/(3)) gt 3^((1)/(4))` So, `4root(3)(3)` is greater |
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| 15. |
Write which one is a rational number and which one is a surd : `root(6)((64)/(729)), sqrt((2)/(12))` |
| Answer» Not surd, surd | |
| 16. |
The conjugate surd of `(sqrt3 -5)` is _______ |
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Answer» `(sqrt5 -5) (-sqrt3 -5) = - (sqrt3-5) (sqrt3+5)` `= - {(sqrt3)^(2) - (5)^(2)} = - (3 - 25)` `= - (-22) =22`, which is a rational number Hence the conjugate surd of `(sqrt3 -5) " is " (sqrt3 + 5)` |
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| 17. |
The complete surd form of the surd `b root(q)(a)` is ____ |
| Answer» `root(q)(b^(a)a)`, since `root(q)(b^(q)a) = b^((q)/(q)).a^((1)/(q)) = broot(q)(a)` | |
| 18. |
`sqrtpi` is a quadratic surd |
| Answer» The given statement is false, since `pi` is a trancendental irrational number | |
| 19. |
What should be subtracted from `(7 - sqrt3)` to get `3 + sqrt3` ? |
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Answer» The required quantity `= (7 - sqrt3) - (3 + sqrt3)` `= 7 - sqrt3 - 3 - sqrt3` `=4 -2 sqrt3` `:. (4 -2 sqrt3)` should be subtracted from `(7 - sqrt3) " to get " (3 + sqrt3)` |
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| 20. |
If `a + b = sqrt5 and a -b = sqrt3`, then the value of `a^(2) + b^(2)` isA. 8B. 4C. 2D. 1 |
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Answer» Correct Answer - B We know that `a^(2) + b^(2) = (1)/(2) xx 2 (a^(2) + b^(2))` `= (1)/(2) {(a + b)^(2) + (a-b)^(2)}` (by formula) `= (1)/(2) {(sqrt5)^(2) + (sqrt3)^(2)} = (1)/(2) (5 +3)` `= (1)/(2) xx 8 4` `:.` (b) is correct |
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| 21. |
What should be added to `sqrt5 + sqrt3` so that the result is `2 sqrt5` ? |
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Answer» The required quantity `= 2 sqrt5 - (sqrt5 + sqrt3))` `= 2 sqrt5- sqrt5 - sqrt3` `= sqrt5 - sqrt3` Hence `(sqrt5 - sqrt3)` should be added to `sqrt5 + sqrt3` to get `2 sqrt5` |
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| 22. |
What should be added to `(1 + sqrt5 - sqrt3)` so that the result of addition is `2 sqrt3` ?A. `3 sqrt3 -1-sqrt5`B. `3 sqrt3+ 1 - sqrt5`C. `3 sqrt3 -1+ sqrt5`D. `3 sqrt3 + 1 + sqrt5` |
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Answer» Correct Answer - A We have, `2sqrt3 - (1 + sqrt5 -sqrt3)` `= 2 sqrt3 -1- sqrt5 + sqrt3` `= 3 sqrt3 -1- sqrt5` Hence (c) is correct |
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| 23. |
The result of subtraction of `sqrt5` from `sqrt(125)` isA. `sqrt80`B. `sqrt(120)`C. `sqrt(100)`D. None of these |
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Answer» Correct Answer - A The subtraction `= sqrt(125) - sqrt5` `= sqrt(25 xx 5) - sqrt5 = 5 sqrt5 - sqrt5` `=4 sqrt5 = sqrt(16 xx 5) = sqrt80` `:.` (a) is correct |
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| 24. |
Simplify : `sqrt12 + sqrt18 + sqrt27 - sqrt32` |
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Answer» `sqrt12 + sqrt18 + sqrt27- sqrt32 = sqrt(4 xx 3) + sqrt(9 xx 2) + sqrt(9 xx 3) - sqrt(16 xx 2)` `=2 sqrt3 + 3 sqrt2 + 3sqrt3 - 4 sqrt2` `= 5 sqrt3 - sqrt2` Hence the required simpliest result `= 5 sqrt3 - sqrt2` |
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| 25. |
`sqrt(75) and sqrt(147)` are like surds |
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Answer» Correct Answer - 1 `sqrt75 = sqrt(25 xx 3) = 5 sqrt3, and sqrt(147) = sqrt(49 xx 3) = 7 sqrt3` `:. Sqrt75 and sqrt(147)` are like surds Hence the given statement is true |
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| 26. |
`sqrt(147) - sqrt(75)`=A. `sqrt3`B. `2 sqrt3`C. `3 sqrt3`D. `4 sqrt3` |
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Answer» Correct Answer - B `sqrt(147) - sqrt(75)` `= sqrt(49 xx 3) - sqrt(25 xx 3)` `= 7 sqrt3 - 5 sqrt3 = 2 sqrt3` Hence (b) is correct |
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| 27. |
(a) `sqrt(175)` (b) `2sqrt(112)` (c) `sqrt(125)` (d) `5sqrt(119)` |
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Answer» (a) `sqrt(175) = sqrt(25 xx 7) = 5sqrt7` (b) `2sqrt(112) = 2sqrt(16 xx 7) = 2 xx 4 sqrt7 = 8 sqrt7` (c) `sqrt(125) = sqrt(25 xx 5) = 5 sqrt5` (d) `5sqrt(119) = 5 sqrt(119)` |
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| 28. |
Prove that `3sqrt(48) -4 sqrt(75) + sqrt(192) = 0` |
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Answer» `3 sqrt(48) -4 sqrt(75) + sqrt(192) = 3 sqrt(16 xx 3) -4 sqrt(25 xx 3) + sqrt(64 xx 3)` `= 3 xx 4 sqrt3 -4 xx 5 sqrt3 + 8sqrt3` `= 12 sqrt3 - 20 sqrt3 + 8 sqrt3` `= 20 sqrt3 - 20 sqrt3 = 0` Hence, `3 sqrt(48) - 4 sqrt(75) + sqrt(192) = 0` |
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| 29. |
Prove that `sqrt(108) - sqrt(75) = sqrt3` |
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Answer» `sqrt(108) - sqrt75 = sqrt(36 xx 3) - sqrt(25 xx 3)` `= 6 sqrt3 - 5 sqrt3 = sqrt3` |
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| 30. |
Prove that `sqrt(98) + sqrt8 - 2 sqrt(32) = sqrt2` |
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Answer» `sqrt(98) + sqrt8 -2 sqrt(32) = sqrt(49 xx 2) + sqrt(4 xx 2) -2 sqrt(16 xx 2)` `=7 sqrt2 + 2 sqrt2-2 xx 4 sqrt2 = 9 sqrt2 - 8sqrt2 = sqrt2` Hence `sqrt98 + sqrt8 -2 sqrt(32) = sqrt2` |
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| 31. |
Simplify : `(x + sqrt(x^(2) - 1))/(x - sqrt(x^(2) -1)) + (x - sqrt(x^(2) -1))/(x + sqrt(x^(2) -1))` If the result of the simplification is equal to 14, then find the value of x |
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Answer» `(x + sqrt(x^(2) -1))/(x - sqrt(x^(2) -1)) + (x - sqrt(x^(2) - 1))/(x + sqrt(x^(2) -1))` `= ((x + sqrt(x^(2) + 1))^(2) + (x - sqrt(x^(2) -1))^(2))/((x - sqrt(x^(2) -1)) (x + sqrt(x^(2) -1)))` `= (x^(2) + 2x sqrt(x^(2) -1) + x^(2) - 1 + x^(2) - 2x sqrt(x^(2) -1) + x^(2) -1)/((x)^(2) - (sqrt(x^(2) -1))^(2))` `= (2 (2x^(2) -1))/(x^(2) - x^(2) + 1) = 2 (2x^(2) -1) = 4x^(2) -2` As per question, `4x^(2) -2 = 14 or, 4x^(2) = 16 or, x^(2) = 4` `:. x = +- 2` Hence the given expression `= 4x^(2) -2 and x = +- 2` |
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| 32. |
Simplify: `= (4 sqrt3)/(2 - sqrt2) - (30)/(4 sqrt3 - sqrt18) - (sqrt18)/(3-sqrt12)` |
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Answer» Given expression `= (4 sqrt3)/(2 - sqrt2) - (30)/(4 sqrt3 - sqrt18) - (sqrt18)/(3-sqrt12)` `= (4 sqrt3 (2 sqrt2))/((2 - sqrt2) (2 + sqrt2)) - (30 (4 sqrt3 + sqrt18))/((4sqrt3 - sqrt18) (4 sqrt3 + sqrt18)) - (sqrt18 (3 + sqrt12))/((3-sqrt12) (3 + sqrt12))` `= (4 sqrt3 (2 + sqrt2))/((2)^(2) - (sqrt2)^(2)) - (30 (4 sqrt3 + sqrt18))/((4sqrt3)^(2) - (sqrt18)^(2)) - (sqrt18 (3 + sqrt12))/((3)^(2) - (sqrt12)^(2))` `= (4 sqrt3 (2 + sqrt2))/(4-2) - (30(4 sqrt3 + sqrt18))/(48 - 18) - (sqrt18 (3 + sqrt12))/(9 - 12)` `= (4 sqrt3 (2 + sqrt2))/(2)- (30 (4 sqrt3 + sqrt18))/(30) - (sqrt18 (3 + sqrt12))/(-3)` `= 2 sqrt3 (2 + sqrt2) - (4 sqrt3 + 3 sqrt2) - (3 sqrt2 (3 + 2 sqrt3))/(-3)` `=4 sqrt3 + 2 sqrt6 - 4sqrt3 - 3sqrt2 + 3sqrt2 + 2 sqrt6` `=4 sqrt6` Hence the given expression `= 4 sqrt6` |
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| 33. |
If `x = 2, y = 3 and z = 6`, then find the value of `(3sqrtx)/(sqrty + sqrtz) - (4 sqrty)/(sqrtz + sqrtx) + (sqrtz)/(sqrtx + sqrty)` |
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Answer» `(3 sqrtx)/(sqrty + sqrtz) - (4 sqrty)/(sqrtz + sqrtx) + (sqrtz)/(sqrtx + sqrty)` `= (3 sqrtx (sqrty - sqrtz))/((sqrty + sqrtz) (sqrty - sqrtz)) - (4 sqrty (sqrtz - sqrtx))/((sqrtz + sqrtx) (sqrtz - sqrtx)) + (sqrtz (sqrtx - sqrty))/((sqrtx + sqrty) (sqrtx - sqrty))` `= (3 sqrtx (sqrty - sqrtz))/(y - z) - (4 sqrty (sqrtz - sqrtx))/(z -x) + (sqrtz (sqrtx - sqrty))/(x -y)` `= (3 sqrt2 (sqrt3 - sqrt6))/(3 -6) - (4 sqrt3 (sqrt6 - sqrt2))/(6 -2) + (sqrt6 (sqrt2 - sqrt3))/(2 -3)` [Putting the value of x, y and z] `= (3 sqrt2 (sqrt3 - sqrt6))/(-3) - (4 sqrt3 (sqrt6 - sqrt2))/(4) + (sqrt6 (sqrt2 - sqrt3))/(-1)` `= -sqrt2 (sqrt3 - sqrt6) - sqrt3 (sqrt6 - sqrt2) - sqrt6(sqrt2 - sqrt3)` `= -sqrt6 + sqrt12 - sqrt18 + sqrt6 - sqrt12 + sqrt18` = 0 Hence the given expression = 0 |
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| 34. |
Simplify: `(3 sqrt2)/(sqrt3 + sqrt6) - (4 sqrt3)/(sqrt6 + sqrt2) + (sqrt6)/(sqrt2 + sqrt3)` |
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Answer» `(3 sqrt2)/(sqrt3 + sqrt6) - (4 sqrt3)/(sqrt6 + sqrt2) + (sqrt6)/(sqrt2 + sqrt3)` `= (3 sqrt2 (sqrt3 - sqrt6))/((sqrt3 + sqrt6) (sqrt3 - sqrt6)) - (4 sqrt3 (sqrt6 - sqrt2))/((sqrt6 + sqrt2) (sqrt6 - sqrt2)) + (sqrt6 ( sqrt2 - sqrt3))/((sqrt2 + sqrt3) (sqrt2 - sqrt3))` `= (3 sqrt2 (sqrt3 - sqrt6))/((sqrt3)^(2) - (sqrt6)^(2)) - (4 sqrt3 (sqrt6 - sqrt2))/((sqrt6)^(2) - (sqrt2)^(2)) + (sqrt6 (sqrt2 - sqrt3))/((sqrt2)^(2) - (sqrt3)^(3))` `= (3 sqrt2 (sqrt3 - sqrt6))/(3 -6) - (4 sqrt3 (sqrt6 - sqrt2))/(6-2) + (sqrt6 (sqrt2 - sqrt3))/(2-3)` `= (3 sqrt2 (sqrt3 - sqrt6))/(-3) - (4 sqrt3 (sqrt6 - sqrt2))/(4) + (sqrt6 (sqrt2 - sqrt3))/(-1)` `= -sqrt2 (sqrt3 - sqrt6) - sqrt3 (sqrt6 - sqrt2) - sqrt6 (sqrt2 - sqrt3)` `= - sqrt6 + sqrt12 - sqrt18 + sqrt6 - sqrt12 + sqrt18` = 0 Hence the expression = 0 |
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| 35. |
Simplify: `(sqrt2 (2 + sqrt3))/(sqrt3 (sqrt3 + 1)) - (sqrt2 (2 - sqrt3))/(sqrt3 (sqrt3 -1))` |
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Answer» `(sqrt2 (2 + sqrt3))/(sqrt3 (sqrt3 + 1)) - (sqrt2 (2 - sqrt3))/(sqrt3 (sqrt3 -1))` `= (sqrt2 (2 + sqrt3) (sqrt3 -1))/(sqrt3 (sqrt3 + 1) (sqrt3 -1)) - (sqrt2 (2 - sqrt3) (sqrt3 + 1))/(sqrt3 (sqrt3 -1) (sqrt3 + 1))` `= (sqrt2 (2 sqrt3 + 3 - 2- sqrt3))/(sqrt3 {(sqrt3)^(2) - (1)^(2)}) - (sqrt2 (2 sqrt3 -3 + 2 - sqrt3))/(sqrt3 {(sqrt3)^(2) - (1)^(2)})` `= (sqrt2 (sqrt3 + 1))/(sqrt3 (3 -1)) - (sqrt2 (sqrt3-1))/(sqrt3 (3-1))` `= (sqrt2 (sqrt3 + 1))/(2sqrt3) - (sqrt2 (sqrt3 -1))/(2sqrt3) = (sqrt2 (sqrt3 + 1- sqrt3+1))/(2sqrt3)` `= (sqrt2 xx 2)/(2sqrt3) = (sqrt2)/(sqrt3) xx (sqrt3)/(sqrt3) = (sqrt6)/(3)` `:.` Given expression `= (sqrt6)/(3)` |
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| 36. |
`5 root(5)(15625) =`A. `5root(5)(5)`B. `25root(5)(5)`C. `50root(5)(5)`D. `125root(5)(5)` |
| Answer» Correct Answer - B | |
| 37. |
If `2x = a + sqrt((4b^(3) - a^(3))/(3a)) and 2y = a - sqrt((4b^(3) - a^(3))/(3a))`, then prove that `x^(3) + y^(3) = b^(3)` |
| Answer» (a) `a = (4)/(sqrt19) , b = (sqrt3)/(sqrt19)` | |
| 38. |
If `x = [-(q)/(2) + sqrt((q^(2))/(4) + (p^(3))/(27))]^((1)/(3)) + [-(q)/(2) - sqrt((q^(2))/(4) + (p^(3))/(27))]^((1)/(3))` then prove that `x^(3) + pq + q = 0` |
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Answer» `x = [-(q)/(2) + sqrt((q^(2))/(4) + (p^(3))/(27))]^((1)/(3)) + [-(q)/(2) - sqrt((q^(2))/(4) + (p^(3))/(27))]^((1)/(3))` or `x^(3) = - (q)/(2) + sqrt((q^(2))/(4) + (p^(3))/(27)) - (q)/(2) - sqrt((q^(2))/(4) + (p^(3))/(27)) - (q)/(2) - sqrt((q^(2))/(4) + (p^(3))/(27)) + 3 [ - (q)/(2) + sqrt((q^(2))/(4) + (p^(3))/(27))]^((1)/(3)) xx [-(q)/(2) - sqrt((q^(2))/(4) + (p^(3))/(27))]^((1)/(3)) [(-(q)/(2) + sqrt((q^(2))/(4) + (p^(3))/(27)))^((1)/(3)) + (-(q)/(2) - sqrt((q^(2))/(4) + (p^(3))/(27)))^((1)/(3))]` or, `x^(3) = - q + 3 [(-(q)/(2))^(2) - (sqrt((q^(2))/(4) + (p^(3))/(27)))^(2) ]^((1)/(3)) xx (x)` or, `x^(3) = - q + 3 [(q^(2))/(4) - (q^(2))/(4) - (p^(3))/(27)]^((1)/(3)) xx(x)` or `x^(3) = -q + 3 [-(p^(3))/(27)]^((1)/(3)) xx x` or `x^(3) = - q + 3 xx - (p)/(3) xx x or, x^(3) = -q - px` or, `x^(3) = -q + 3 xx -(p)/(3) xx x or, x^(3) = -q - px` or, `x^(3) + px + q = 0` Hence, `x^(3) + px + q = 0` |
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| 39. |
If n is a positive integer and `U_(n) = (3 + sqrt5)^(n) + (3 - sqrt5)^(n)`, then prove that `U_(n + 1) = 6U_(n) - 4U_(n -1), n ge 2` |
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Answer» L.H.S = `U_(n+1)` `= (3 + sqrt5)^(n + 1) + (3 - sqrt5)^(n+1)` `= (3 +sqrt5)^(n) (3 + sqrt5) + (3 - sqrt5)^(n) (3 - sqrt5)` `= 3 (3 + sqrt5)^(n) + sqrt5 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) - sqrt5 (3 - sqrt5)^(n)` `= 3 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + sqrt5 (3 + sqrt5)^(n) - sqrt5 (3 - sqrt5)^(n)` `= 3 [(3 + sqrt5)^(n) + (3 -sqrt5)^(n)] + sqrt5 [(3 + sqrt5)^(n) - (3 - sqrt5)^(n)]` L.H.S `= 6U_(n) - 4U_(n -1)` `= 6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] - 4 [(3 + sqrt5)^(n-1) + (3-sqrt5)^(n-1)]` `=6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] -4 [((3 + sqrt5)^(n))/((3 + sqrt5)) + ((3 - sqrt5)^(n))/((3 - sqrt5))]` `=6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)]-4 [((3 - sqrt5) (3 + sqrt5)^(n) + (3 + sqrt5) (3 - sqrt5)^(n))/((3 + sqrt5) (3 - sqrt5))]` `=6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] -4 [(3 (3 + sqrt5)^(n) - sqrt5 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + sqrt5 (3 - sqrt5)^(n))/((3)^(2) - (sqrt5)^(2))]` `= 6 [(3 + sqrt5)^(n) + 6(3 - sqrt5)^(n)] -4 [(3(3 + sqrt5)^(n) - sqrt5 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + 3 (3 -sqrt5)^(n) + sqrt5(3 - sqrt5)^(n))/(4)]` `= 6 (3 + sqrt5)^(n) + 6 (3 - sqrt5)^(n) -3 (3 + sqrt5)^(n) + sqrt5(3 + sqrt5)^(n) -3 (3 - sqrt5)^(n) - sqrt5 (3 - sqrt5)^(n)` `= 3 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + sqrt5 [(3 + sqrt5)^(n) - (3 - sqrt5)^(n)]` `=3 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] + sqrt5 [(3 + sqrt5)^(n) - (3 - sqrt5)^(n)]` `:.` LHS = RHS Hence `U_(n+1) = 6U_(n) - 4U_(n-1)` |
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| 40. |
If `x = a^((1)/(3)) b^((1)/(3)) + a^(-(1)/(3)) + a^(-(1)/(3)) b^((1)/(3))` then prove that `a(bx^(3) - 3bx - a) = b^(2)` |
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Answer» `x = a^((1)/(3)) b^((1)/(3)) + a^(-(1)/(3)) + a^(-(1)/(3)) b^((1)/(3))` `:. x^(3) = (a^((1)/(3)) b^(-(1)/(3)) + a^(-(1)/(3)) b^((1)/(3)))^(3)` `= (a^((1)/(3)) b^(-(1)/(3)))^(3) + (a^(-(1)/(3)) b^((1)/(3)))^(3) + 3a^((1)/(3)) b^(-(1)/(3)).a^(-(1)/(3)) b^((1)/(3)) (a^((1)/(3)) b^(-(1)/(3)) + a^(-(1)/(3)) b^((1)/(3)))` `= ab^(-1) + a^(-1) b + 3.a^(0).b^(0).x` `= (a)/(b) + (b)/(a) + 3x = (a^(2) + b^(2) + 3abx)/(ab)` `:. abx^(3) = a^(2) + b^(2) + 3abx` or, `abx^(3) - 3abx = a^(2) + b^(2)` or, `abx^(3) - 3abx - a^(2) = b^(2)` or, `a (bx^(3) - 3bx - a) = b^(2)` Hence, `a (bx^(3) - 3bx -a) = b^(2)` |
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