PE 1 - Surds & Indices | Algebra - Surds & Indices
Simplify:
- (a)
2
- (b)
64
- (c)
8
- (d)
None of these
Answer: Option A
Explanation :
Let x =
⇒ x =
⇒ x =
⇒ x =
⇒ x =
⇒ x =
⇒ x = = 2
Hence, option (a).
Workspace:
Find the value of , where a = 2 and b = 1.
- (a)
- (b)
- (c)
- (d)
None of these
Answer: Option B
Explanation :
Let x =
Rationalising with
⇒ x = ×
⇒ x =
⇒ x =
Now, a = 2 and b = 1
∴ x =
Now =
∴ =
Hence, option (b).
Workspace:
Find the value of & if a + b + c = 0.
- (a)
x3
- (b)
x
- (c)
x2
- (d)
xabc
Answer: Option A
Explanation :
Given &
=
=
=
Since, a + b + c = 0, it means a3 + b3 + c3 = 3abc
∴ = = x3
Hence, option (a).
Workspace:
If x = 3 + + , then find the value of x3 - 18x2 + 18x + 27?
- (a)
√108
- (b)
8
- (c)
√66
- (d)
66
- (e)
None of these
Answer: Option D
Explanation :
Given, x = 3 + +
⇒ x - 3 = +
Cubing both sides
⇒ x3 - 27 - 9x2 + 27x = 12 + 3 × × × ( + )
⇒ x3 - 9x2 + 27x - 27 = 12 + 3 × 3 × x = 12 + 9x
⇒ x3 - 18x2 + 18x = 39
∴ x3 - 18x2 + 18x + 27 = 66
Hence, option (d).
Workspace:
If 3x + 5y = 52 and 32x-2 + 5y+1 = 206 where x and y are integers, find x + y.
- (a)
3
- (b)
4
- (c)
5
- (d)
6
- (e)
Cannot be determined
Answer: Option C
Explanation :
Let 3x = a and 5y = b
⇒ 3x + 5y = 52
⇒ a + b = 52
Also, 32x-2 + 5y+1 = 206
⇒ 32x/9 + 5 × 5y = 206
⇒ 32x + 45 × 5y = 1854
⇒ a2 + 45b = 1854 ...(2)
(2) - 45 × (1)
⇒ a2 - 45a = 1854 - 2340
⇒ a2 - 45a + 486 = 0
⇒ (a - 18)(a - 27) = 0
⇒ a = 18 or 27
⇒ 3x = 18 or 27
Since x is an integer, 3x = 27 and x = 3.
Also, b = 5y = 52 - 27 = 25, hence, y = 2
∴ x + y = 3 + 2 = 5.
Hence, option (c).
Workspace:
If x is a positive integer, the greatest integer by which 11x + 11x+1 + 11x+2 is always divisible is
- (a)
11
- (b)
121
- (c)
133
- (d)
1463
Answer: Option D
Explanation :
Let a = 11x + 11x+1 + 11x+2
⇒ a = 11x × (1 + 111 + 112)
⇒ a = 11x × 133
Since x is a positive integer, the least value of 11x will always be a multiple of 11 and its least value will be 11.
∴ Least possible value of a = 11 × 133 = 1463
Hence, option (d).
Workspace:
Find square root of 5 + √6 + √10 + √15.
- (a)
√2 + √3 + √5
- (b)
1 + +
- (c)
(√2 + √3 + √5)/2
- (d)
None of these
Answer: Option B
Explanation :
Let x = 5 + √6 + √10 + √15
Now, 2x = 10 + 2√6 + 2√10 + 2√15
This can be written as:
2x = (√2)2 + (√3)2 + (√5)2 + 2 × √2 × √3 + 2 × √2 × √5 + 2 × √3 × √5
The RHS is of the form a2 + b2 + c2 + 2ab + 2bc + 2ca which is equal to (a + b + c)2
∴ 2x = (√2 + √3 + √5)2
∴ x =
⇒ √x =
∴ √x = 1 + +
Hence, option (b).
Workspace:
Given 2x = 8y+1 and 9y = 3x–9; the value of x + y is:
- (a)
18
- (b)
21
- (c)
24
- (d)
27
Answer: Option D
Explanation :
2x = 8y+1
∴ 2x = (23)y+1
∴ 2x = 23(y+1)
∴ x = 3y+3 ...(1)
Consider,
9y = 3x–9
∴ (32)y = 3x–9
∴ 32y = 3x–9
∴ 2y = x – 9 …(2)
Solving (1) and (2), we get,
x = 21 and y = 6
∴ x + y = 21 + 6 = 27
Hence, option (d).
Workspace:
Let r be the result of doubling both the base and the exponent of ab, b ≠ 0. If r equals the product of ab and xb, then x equals:
- (a)
2a
- (b)
4a
- (c)
2
- (d)
4
Answer: Option B
Explanation :
By the condition given in the question,
r = (2a)2b
⇒ r = 22b × a2b
⇒ r = (22)b × (ab)2
⇒ r = (4)b × (ab)2
Also, r = ab × xb
∴ ab × xb = 4b × (ab)2
∴ xb = 4b × ab
∴ xb = (4a)b
∴ x = 4a
Hence, option (b).
Workspace:
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