Test: Algebra - 2 - CAT MCQ

As a₁, a₂, a₃ ... are in A.P.
then, 

{Where d = Common difference}.

Given, log5(x + y) + log5(x – y) = 3
⇒ log5[(x + y)(x – y)] = 3
⇒ (x + y)(x – y) = 5³ = 125
⇒ x² – y² = 125 ...(i)
And log2y – log2x = 1 – log2 3.

Then if x = 3k, then y = 2k
Now putting this value of x and y in (i), we get
(3k)² – (2k)² = 125
⇒ 5k² = 125, ∴ k = 5
∴ x × y = 3k × 2k = 6 × 25 = 150

We can transform each of the options for ‘n’ years.
(997)2¹⁴ + 3 ≡ (p – 3)2n ^{– 1} + 3
(997)¹⁵ – 3 ≡ (p – 3)n – 3
(1003)¹⁵ + 6 ≡ (p + 3)ⁿ + 6
(1003)2¹⁵ – 3 ≡ (p + 3)2n – 3
As per the condition, in one year, the population ‘p’ becomes ‘3 + 2p’
Putting the value of n = 1 in each option, and checking to get 3 + 2p,
we have (997)2¹⁴ + 3 ≡ p ≠ 3 + 2p
(1003)¹⁵ + 6 ≡ p + 9 ≠ 3 + 2p
(1003)2¹⁵ – 3 ≡ (p + 3)2n – 3 = 3 + 2p
(997)¹⁵ – 3 ≡ (p – 3) – 3 ≡ p – 6

The sequence (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280 is in A.P. with first term (a) = 2n + 1
common difference (d) = 2 and last term (l) = 2n + 47,
Let ‘m’ be the number of terms in this sequence, then
l = a + (m – 1)d
2n + 47 = (2n + 1) + (m – 1) (2) ⇒ m = 24
Now, (2n + 1) + (2n + 3) + (2n + 5) +....+ (2n + 47) = 5280


⇒ 24(2n + 1 + 23) = 48(n + 12) = 5280
⇒ 48(n + 12) = 5280
∴ n = 98 Now, 1 + 2 + 3 + ... +

We know that the quadratic equation ax² + bx + c = 0 has real and distinct roots, if b² – 4ac > 0 Hence for real and distinct roots of
x² – 4x – log₂ A = 0, D > 0
∴ (–4)² – 4 × 1 × (–log₂A) > 0
⇒ 16 + 4 log₂A > 0 ⇒ log₂A > – 4

The given expression will be real only if

⇒ 4x – x² ≥ 3 ⇒ x² – 4x + 3 ≤ 0
⇒ (x – 1) (x – 3) ≤ 0 ⇒ 1 ≤ x ≤ 3 

For n = 1 , a1 = 1
For n = 2 , a₁ – a₂ = 2 ⇒ a₂ = –1
For n = 3 , a₁ – a₂ + a₃ = 3 ⇒ a₃ = 1
For n = 4 , a₁ – a₂ + a₃ – a₄ = 4 ⇒ a₄ = –1
From the above shown pattern, we conclude that each odd term = 1 and each even term = –1
⇒ a₅₁ + a₅₂ + ... + a₁₀₂₂ + a₁₀₂₃
= (a₅₁ + a₅₂ + ....+ a₁₀₂₂) + a₁₀₂₃
= 0 + 1 = 1

Since, 5x, 16y, 12z are in AP.
∴ 32y = 5x + 12z …(1)
∵ x, y, z are in GP 
∴ y² = xz ...(2)
Squaring both sides of (1), we get
1024y² = 25x² + 144z² + 120xz
⇒ 1024xz = 25x² + 144z² + 120xz
⇒ 25x²+144z² - 904xz = 0
⇒ 25x2 - 900xz - 4xz + 144 z² = 0
⇒ 25x(x - 36z) - 4z(x - 36z) = 0
⇒ (25x - 4z) (x - 36z) = 0
∴
⇒[r is the common ratio]
⇒
But  because x, y, z > 0 and x < y < z
∴ common ratio = 5 / 2 

2x = 3log₅2
Taking logarithms to base 5 on both sides, we have
x (log₅2) = log₅2 . log₅3
x = log₅3 = log₅

x²⁰¹⁸ y²⁰¹⁷ = 1/2 and x²⁰¹⁶ y²⁰¹⁹ = 8

⇒ y⁴⁰³⁵ = 2⁴⁰³⁵

log₁₂81 = p ⇒ log₁₂3⁴ = p

5 + log₃ a = 2³ = 8 ⇒ log3a = 3  ⇒ a = 27 Similarly, 4a + 12 + log₂b = 5³ = 125
Since a = 27, 4(27) + 12 + log₂b = 125
⇒ log₂b = 5  ⇒ b = 32.
∴ a + b = 27 + 32 = 59

a₁ = 3, a₂ = 7, d = 4

n(6n + 1) = 610
6n² + n - 610 = 0
solved, n = 10
Now, m(a₁ + a₂ + ... a_n) > 1830

m > 8.7 → m > 9

(a + 6d)² = (a + 2d)(a + 16d)
a² + 12ad + 36d² = a² + 18ad + 32d² 4d² = 6ad

9^2x–1 – 81^x–1 = 1944

4x – 4 = 5 → x = 9/4

x² – x – 1 = 0
 (∵ x > 0)

If a positive number a is expressed as the sum of two positive numbers s₁ and s₂ then [a] could be at the most 1 more than [s₁] + [s₂], i.e., the fractional parts of s₁ and s₂ together, can provide at most 1.
Similarly, the fractional parts of s₁, s₂, s₃, s₄, s₅ can together, provide at most 4.
Conversely, if [a] is 4 more than [s₁] + [s₂] + [s₃] + [s₄] + [s_m], then m has to be at least 5.
Similarly, the least value of n is 4.
∴ (m + n)_min = 5 + 4 = 9

Let the number of balls with
P_i = a_i (i = 1 to 11)
a₁ + a₃ + a₅ ...... + a₁₁ = 6 (a₆) = 72.
As a₆ would be the arithmetic mean of these 11 numbers and
2(a₆) = (a₁ + a₁₁)
= (a₂ + a₁₀)
= (a₃ + a₉)
= (a₄ + a₈)
= (a₅ + a₇)
∴ a₁ + a₆ + a₁₁ = 3 (a₆)
= 36

According to question,

Let 2^x = a

⇒ a = 4 or 8
∴ x = 2 or 3
Hence, 2^x – 5 = –1,
when x = 2, which is not possible.
∴ x = 3.
Or
All numbers are in AP,
2 log₃ (2^x – 5) = log₃ 2 + log₃ (2^x – 7/2)
Suppose 2^x = t
2 log₃ (t – 5) = log₃ 2 + log₃ (t – 7/2)
→ log₃ (t – 5)² = log₃ (2t – 7)
→ (t – 5)² = 2t – 7
→ t² – 10t + 25 = 2t – 7
→ t² – 12t + 32 = 0
→ t = 4.8
2^x = 4.8
∴ x = 2, 3.

Take –y in place of y
Then equation is  

X = (log₁₀ 1 + log₁₀ 2 + ... + log₁₀ n) – (log₁₀ 1 + log₁₀ 2 + .... + log₁₀ p) – (log₁₀ 1 + log₁₀ 2 + .... + log₁₀ (n – p))
⇒ X = log₁₀ n! – log₁₀ p! –  log₁₀ (n – p)!
log (m × n) = log m + log n

X is maximum when   is maximum.

Given that x + y = 1
⇒ x + y – 1 = 0
⇒ x³ + y³ – 1 = – 3xy
(a³ + b³  + c³ = 3abc if a + b + c = 0)
⇒ x³ + y³ + 3xy = 1

 ...(i)
 ...(ii)
From equations (i) and (ii), we get

Let log₃6 be equal to k; therefore,

Let a = b² = c³ = d⁴