Test: Algebra - 1

If n = 1, then a₁ = 3(2^{1 + 1} – 2) = 6 = 3 × 2¹
If n =  2, then a₁ + a₂ = 3(2^{2 + 1} – 2) = 18 ⇒ a₂
= 18 – a1 = 18 – 6 = 12 = 3 × 2²
If n = 3, then a₁ + a₂ + a₃ = 3(2^{3 + 1} – 2) = 42 ⇒ a₃
= 42 – (a₁ + a₂) = 24 = 3 × 2³
Similarly, a_n = 3 × 2ⁿ
Therefore, a₁₁ = 3 × 2¹¹ = 6144

Case I: x = 0.
Clearly, x = 0 satisfy the equation
Case II: x > 0 |x| (6x² + 1) = 5x² ⇒ x(6x² + 1) = 5x²
⇒ x(6x² + 1 – 5x) = 0
∴ x ≠ 0
∴ 6x² + 1 – 5x = 0
On solving the quadratic equation, we get

Case III: x < 0
|x| (6x² + 1) = 5x² ⇒ –x(6x² + 1) = 5x2
⇒ –x(6x² + 1 + 5x) = 0
⇒ x(6x² + 5x + 1) = 0
∵ x ≠ 0, ∴ 6x² + 5x + 1 = 0
On solving the quadratic equation, we get

Hence there are 5 solutions.

x² – x – 6 = (x + 2)(x – 3)
Case 1: x² – x – 6 < 0 ⇒ (x + 2)(x – 3) < 0
⇒ –2 < x < 3 and |x² – x – 6| = –(x² – x – 6)
∴ |x² – x – 6| = x + 2 ⇒ –(x + 2)(x – 3) = x + 2
⇒ (x – 3) = –1 ⇒ x = 2
Case 2: x² – x – 6 ≥ 0 ⇒ (x + 2) (x – 3) ≥ 0
⇒ x ≤ –2 or x ≥ 3
Checking for boundary conditions:
For x = –2, |x² – x – 6| = x + 2, therefore,x = -2 is also the root.
But for x = 3, |x² – x – 6| ≠ x + 2
Hence x = 3 is NOT the root.
And for the interval x < –2 or x > 3 the expression |x² – x – 6|
= x² – x – 6
∴ |x² – x – 6| = x + 2
⇒ (x + 2)(x – 3) = x + 2
⇒ (x – 3) = 1 ⇒ x = 4
Hence, the root are –2, 2, and 4.
∴ Required product = (2)(–2)(4) = –16

Given: 2 cos (x(x + 1)) = 2^x + 2^–x
Now, by AM – GM inequality, we get


∴ 2 cos (x(x + 1)) ≥ 2
We know, –1 ≤ cosθ ≤ 1
∴ 2 cos(x(x + 1)) = 2
Hence, the expression is valid only if 2^x + 2–x = 2, which is true for only one value of x i.e. 0.
Therefore, the expression has only one real solution.

Case 1: m is even
Given, 8f(m + 1) – f(m) = 2
⇒ 8(m + 1 + 3) – m(m + 1) = 2
⇒ 8m + 32 – m2 – m = 2
⇒ m² – 7m – 30 = 0
⇒ (m – 10)(m + 3) = 0
⇒ m = 10 or –3
As m is a positive integer, therefore m = 10.
Case 2: If m is odd, then
8f(m + 1) – f(m) = 2
⇒ 8(m + 1)(m + 2) – (m + 3) = 2
⇒ 8(m² + 3m +  2) – m – 3 = 2
⇒ 8m² + 24m + 16 – m – 3 = 2
⇒ 8m² + 23m + 11 = 0,
Which is not possible and hence no solution.

Given,
f(a + 1) + f(a + 2) + .. + f(a + n) = 16(2n – 1)
⇒ f(a) f(1) + f(a) f(2) + ... + f(a) f(n) = 16(2n – 1)
⇒ f(a) (f(1) + f(2) + ... + f(n)) = 16 (2n – 1)
When n = 1, then f(a)f(1) = 16(21 – 1) = 16
⇒ f(a) × 2 = 16 ⇒ f(a) = 8
∴ f(a) (f(1) + f(2) + ... + f(n)) = 16(2n – 1)
⇒ f(1) + f(2) + ... + f(n) = 2(2n – 1)
When n = 2, then f(1) + f(2) = 2(22 – 1) = 6
⇒ f(2) = 6 – f(1) = 6 – 2 = 4
When n = 3, then f(1) + f(2) + f(3) = 2(23 – 1) = 14
⇒ f(3) = 14 – f(1) – f(2) = 14 – 2 – 4 = 8 = f(a)
∴ a = 3

2^6x + 2^3x+2 – 21 = 0
⇒ 2^6x + 2^3x × 2² – 21 = 0
Put 2^3x = y, then y² + 4y – 21 = 0
⇒ (y – 3) (y + 7) = 0 ⇒ y = 3 or y = –7
⇒ 2^3x = 3 or 2^3x = –7 (No solution)
⇒ 3x = log₂3 ⇒ x = log₂ 3 / 3

The expression is positive integer if  is non-negative integer.
For this (n – 4) must be factor of 8.
For n to be largest, n – 4 = 8
∴ n = 12

Consider the second given equation
5^x–1 + 3^y+1 = 9686 ...(i)
Last digit of 5^x–1 will always be 5 for all positive integral values of (x – 1)
The power cycle of 3 is
3^4k+1 ≡ 3
3^4k+2 ≡ 9
3^4k+3 ≡ 7
3^4k ≡ 1
Clearly 3^y+1 must be in the form of 34k as the unit digit of
R.H.S. = 6 of eq. (i)
We have 3⁴ = 81, and 3⁸ = 6561
Also, 9686 – 81 = 9605 and 9686 – 6561 = 3125
We observe that 3125 = 5⁵
∴ 5^x–1 = 5⁵ ⇒ x = 6
and 3^y+1 = 3⁸ ⇒ y = 7
x = 6 and y = 7 also satisfies the first given equation 5^x – 3^y = 13438
∴ x + y = 6 + 7 = 13

Sum of roots = 4a + 3a = 7a = –b
∴ b = –7a
Product of roots = 4a × 3a = c
∴ c = 12a²
Now, b2 +  c = (–7a)² + 12a² = 61a²
Comparing the options
Option (a) : 61a² = 3721 ⇒ a² = 61, clearly a is not an integer.
Option (b) : 61a² = 361 ⇒ a² = 361 / 61, clearly a is not an integer.
Option (c) : 61a² = 549 ⇒ a² = 9, we can have a = –3 or 3 (an integer)
Option (d) : 61a² = 427 ⇒ a² = 7, clearly a is not an integer

We can take a = 5, b = 0, x = 13 and y = 0 as these values that satisfies all three equations
∴ k = ay – bx = 5 × 0 – 0 × 13 = 0

Cost of table of Amal = 1.2p
Cost of table for Asim = 0.8p
Amal sells to Bimal at 1.3 × 1.2p = 1.56p
= cost of table for Bimal = x
Asim sells table to Barun at 0.7 × 0.8p = 0.56p
= cost of table for Barun = y

g (f(f(3))) = (3)² – 3 × 2 – 1 = 9 – 6 – 1 = 9 – 7 = 2

x = 25 + y + z
Possible value table below.
27 ≤ x ≤ 40

The no. of solution is 1 + 2 + ... + 12 + 11 + 10 = 99

log₃x = a, x = 3^a
log₁₂y = a, y = 12^a
Therefore, the geometric mean of x and y equals 

Hence, 6^a or log₆ G = a

Subtracting,

∵ S = 0.2 + 0.06 + 0.012 + 0.0020 + 0.00030 + 0.000042 + ...
⇒ S = 0.274342 + ...............
Going from answer choices, choice (a) = 0.2666 ....
Choice (b) = 0.268888 .....; choice (c) = 0.27222 ...
Choice (d) = 0.27434 ...
So, choice (d) is answer.

Here, (x + 1) (x + 9) + 8 = 0
x² + 10x + 17 = 0
The roots of the equation are a and b
∴ a + b = – 10
ab = 17
(x + a) (x + b) – 8 = 0
x² + (a + b) x + ab – 8 = 0
x² – 10x  +  9 = 0
Therefore, roots of (x + a) (x + b) – 8 = 0 are 1 and 9.

x² + ( x + 1) (x + 2) (x + 3) (x + 6) = 0
(x + 1) (x + 2) (x + 3) (x + 6) = –x²
(x² + 7x + 6) (x²  + 5x + 6) = –x²


then, (y + 7) (y + 5) = –1
⇒ y² + 12y + 35 = –1
y² + 12y + 36 = 0
⇒ (y + 6)²  = 0
⇒ y = – 6

By observing we can find that x > 1 and x < 2.
Else the RHS ≠ 8.
So the combinations are [x] = 1, [2x] = 2 or 3.
[3×] = 4 or 5
The combinations that give RHS = 8 are 1 + 2 + 5 or 1 + 3 + 4.
For any value of x, the case of "1 + 2 + 5" is not possible. Hence it has to be the case of "1 + 3 + 4".
Which will occur

Hence the solution is 

3x + y + 4 = 2x y
⇒ 3x + 4 = y (2x – 1)

When x = 6 ⇒ y = 2
When x = 1 ⇒ y = 7
These two are the only possible pairs of values of x and y. Where x and y are natural numbers.
∴  Required ratio 

By assuming the values of x, y, z and t, (a) and (b) can be very easily ruled out.
Checking option (c), if x > y + z. then x > y and x > z (since all numbers are positive).
So, using statements I and II, x > z > t > y.
So, option (c) is correct.

Take roots as 2, 2
⇒ r = – 4 & s = 4

According to question,
x (x – 3) = –1
cubing on both side.
⇒ x³ (x – 3)³ = (–1)³
⇒ x³ (x³– 27 – 9x² + 27x) = – 1
⇒ x³ (x³ – 18) + x³ (–9 – 9x² + 27x) = –1
⇒ x³ (x³ – 18) – 9x³ (x² – 3x + 1) = –1
⇒ x³ (x³ – 18) – 9x³ (–1 + 1) = –1
⇒ x³ (x³ – 18) = –1

nth term of the series can be written as

The equations formed by the roots of the equation
(x – a) (x – b)(x – c) can be as follows:
(i)  (x – a)(x – b) ⇒ Roots are a, b
(ii) (x – b)(x – c) ⇒ Roots are b, c
(iii) (x – c)(x – a) ⇒ Roots are c, a
(iv) (x – a)² ⇒ Roots are a, a
(v) (x – b)² ⇒ Roots are b, b
(vi) (x – c)² ⇒ Roots are c, c
Adding all these roots, we get 4(a + b + c).

We obtain the sum of all the coefficients of a polynomial by equating all the variables to 1. Here by putting x = 1 in the polynomial, the required sum comes out to be zero.

Here, (x – 1) (x  – 2) (x – 3) = 0
⇒ (x² – 3x + 2) (x – 3) = 0
⇒ x³ – 6x² + 11x – 6 = 0
According to question,
x = 3, a_n = –6, a₁ = –6
Substitute the above values in the option,
Option (a) : (–6)³ ≥ 3³ × – 6
⇒ – 216 ≥ – 168
This is incorrect, thus option (a) is incorrect.
Opiton (b) : 3³ ≥ (–6)³ × –6                       27 ≥ 6⁴
This is incorrect, thus option (b) is incorrect.
Option (c) : (–6)³ ≥ 3³ × – 6
⇒  –216 ≥ – 168
This is incorrect, thus option (c) is incorrect.
Thus, none of the options is necessarily true.

General term expansion of
(a² + b)¹³ = ⁿC_r x^n-r y^r
To get coefficient of a¹²  b⁸
(a²)⁶ b⁸ where 6 + 8 ≠ 13.
So the term is not possible

(x – 2) is a factor of R(x)
∴ R (2) = 0
⇒ m(2)³ – 100 (2)³ + 3n = 0
⇒ 8m – 400 + 3n = 0

∵ m and n are positive integers, so n must be a multiple of 8. i.e. n = 8, 16, 24, ....., 128, then we get m = 47, 44, 41, ....,2 respectively.
So, the number of ordered pairs (m, n) is 16.