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QUESTION: 1

If a_{1}, a_{2}, ... are in A.P., then, is equal to

(2019)

Solution:

As a_{1}, a_{2}, a_{3} ... are in A.P.

then,

{Where d = Common difference}.

QUESTION: 2

Let x and y be positive real numbers such that

log_{5} (x + y) + log_{5} (x – y) = 3, and log_{2} y – log_{2} x = 1 – log_{2} 3. Then xy equals

(2019)

Solution:

Given, log5(x + y) + log5(x – y) = 3

⇒ log5[(x + y)(x – y)] = 3

⇒ (x + y)(x – y) = 5^{3} = 125

⇒ x^{2} – y^{2} = 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

⇒ 5k^{2} = 125, ∴ k = 5

∴ x × y = 3k × 2k = 6 × 25 = 150

QUESTION: 3

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

(2019)

Solution:

We can transform each of the options for ‘n’ years.

(997)2^{14} + 3 ≡ (p – 3)2n ^{– 1} + 3

(997)^{15} – 3 ≡ (p – 3)n – 3

(1003)^{15} + 6 ≡ (p + 3)^{n} + 6

(1003)2^{15} – 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)2^{14} + 3 ≡ p ≠ 3 + 2p

(1003)^{15} + 6 ≡ p + 9 ≠ 3 + 2p

(1003)2^{15} – 3 ≡ (p + 3)2n – 3 = 3 + 2p

(997)^{15} – 3 ≡ (p – 3) – 3 ≡ p – 6

QUESTION: 4

If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + ... + n?

(2019)

Solution:

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 + ... +

QUESTION: 5

Let A be a real number. Then the roots of the equation x^{2} – 4x – log_{2}A = 0 are real and distinct if and only if

(2019)

Solution:

We know that the quadratic equation ax^{2} + bx + c = 0 has real and distinct roots, if b^{2} – 4ac > 0 Hence for real and distinct roots of

x^{2} – 4x – log_{2} A = 0, D > 0

∴ (–4)^{2} – 4 × 1 × (–log_{2}A) > 0

⇒ 16 + 4 log_{2}A > 0 ⇒ log_{2}A > – 4

QUESTION: 6

If x is a real number, thenis a real number if and only if

Solution:

The given expression will be real only if

⇒ 4x – x^{2} ≥ 3 ⇒ x^{2} – 4x + 3 ≤ 0

⇒ (x – 1) (x – 3) ≤ 0 ⇒ 1 ≤ x ≤ 3

QUESTION: 7

Let a_{1}, a_{2}, ... be integers such that a_{1} – a_{2} + a_{3} – a_{4} + ... + (–1)^{n–1}. an = n, for all n ≥ 1.

Then a_{51} + a_{52} + . . . + a_{1023} equals

(2019)

Solution:

For n = 1 , a1 = 1

For n = 2 , a_{1} – a_{2} = 2 ⇒ a_{2} = –1

For n = 3 , a_{1} – a_{2} + a_{3} = 3 ⇒ a_{3} = 1

For n = 4 , a_{1} – a_{2} + a_{3} – a_{4} = 4 ⇒ a_{4} = –1

From the above shown pattern, we conclude that each odd term = 1 and each even term = –1

⇒ a_{51} + a_{52} + ... + a_{1022} + a_{1023 }

= (a_{51} + a_{52} + ....+ a_{1022}) + a_{1023}

= 0 + 1 = 1

QUESTION: 8

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

(2018)

Solution:

Since, 5x, 16y, 12z are in AP.

∴ 32y = 5x + 12z …(1)

∵ x, y, z are in GP

∴ y^{2} = xz ...(2)

Squaring both sides of (1), we get

1024y^{2} = 25x^{2} + 144z^{2} + 120xz

⇒ 1024xz = 25x^{2} + 144z^{2} + 120xz

⇒ 25x^{2}+144z^{2} - 904xz = 0

⇒ 25x2 - 900xz - 4xz + 144 z^{2} = 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

QUESTION: 9

If x is a positive quantity such that 2^{x} = 3log _{5}^{2}, then × is equal to

(2018)

Solution:

2x = 3log_{5}2

Taking logarithms to base 5 on both sides, we have

x (log_{5}2) = log_{5}2 . log_{5}3

x = log_{5}3 = log_{5}

QUESTION: 10

Given that X^{2018}Y^{2017} = 1/2 and X^{2016}Y^{2019} = 8, the value of x^{2} + y^{3} is

(2018)

Solution:

x^{2018} y^{2017} = 1/2 and x^{2016} y^{2019} = 8

⇒ y^{4035} = 2^{4035}

QUESTION: 11

If log_{12}81 = p, thenis equal to

(2018)

Solution:

log_{12}81 = p ⇒ log_{12}3^{4} = p

QUESTION: 12

If log2(5 + log_{3} a) = 3 and log5(4a + 12 + log_{2} b) = 3, then a + b is equal to

(2018)

Solution:

5 + log_{3} a = 2^{3} = 8 ⇒ log3a = 3 ⇒ a = 27 Similarly, 4a + 12 + log_{2}b = 5^{3} = 125

Since a = 27, 4(27) + 12 + log_{2}b = 125

⇒ log_{2}b = 5 ⇒ b = 32.

∴ a + b = 27 + 32 = 59

QUESTION: 13

Let a_{1}, a_{2}...., a_{2n} be an arithmetic progression with a_{1} = 3 and a_{2} = 7. If a_{1} + a_{2} + ... + a_{3n }= 1830, then what is the smallest positive integer m such that m(a_{1} + a_{2 }+ ... + a_{n}) > 1830?

(2017)

Solution:

a_{1} = 3, a_{2} = 7, d = 4

n(6n + 1) = 610

6n^{2} + n - 610 = 0

solved, n = 10

Now, m(a_{1} + a_{2} + ... a_{n}) > 1830

m > 8.7 → m > 9

QUESTION: 14

If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

(2017)

Solution:

(a + 6d)^{2} = (a + 2d)(a + 16d)

a^{2} + 12ad + 36d^{2} = a^{2} + 18ad + 32d^{2} 4d^{2} = 6ad

QUESTION: 15

If 9^{2x–1} – 81^{x–1} = 1944, then x is

(2017)

Solution:

9^{2x–1 }– 81^{x–1} = 1944

4x – 4 = 5 → x = 9/4

QUESTION: 16

The value of log _{0.008} √5 + log _{√3} 81 - 7 is equal to

(2017)

Solution:

QUESTION: 17

If x + 1 = x^{2} and x > 0, then 2x^{4 }is

(2017)

Solution:

x^{2} – x – 1 = 0

(∵ x > 0)

QUESTION: 18

Two positive real numbers, a and b, are expressed as the sum of m positive real numbers and n positive real numbers respectively as follows:

a = s_{1} + s_{2} +…+ s_{m} and b = t_{1} + t_{2} +…+ t_{n}

If [a] = [s_{1}] + [s_{2}] +…+ [s_{m}] + 4 and [b] = [t_{1} ] + [t_{2} ] +…+ [t_{n}] + 3,

Where [x] denotes the greatest integer less than or equal to x, what is the minimum possible value of m + n?

(2016)

Solution:

If a positive number a is expressed as the sum of two positive numbers s_{1} and s_{2 }then [a] could be at the most 1 more than [s_{1}] + [s_{2}], i.e., the fractional parts of s_{1} and s_{2} together, can provide at most 1.

Similarly, the fractional parts of s_{1}, s_{2}, s_{3}, s_{4}, s_{5} can together, provide at most 4.

Conversely, if [a] is 4 more than [s_{1}] + [s_{2}] + [s_{3}] + [s_{4}] + [s_{m}], then m has to be at least 5.

Similarly, the least value of n is 4.

∴ (m + n)_{min} = 5 + 4 = 9

QUESTION: 19

P_{1}, P_{2}, P_{3}, ..., P_{11} are 11 friends. The number of balls with P_{1 }through P_{11} in that order is in an Arithmetic Progression. If the sum of the number of balls with P_{1}, P_{3}, P_{5}, P_{7}, P_{9} and P_{11} is 72, what is the number of balls with P_{1}, P_{6} and P_{11} put together?

(2014)

Solution:

Let the number of balls with

P_{i} = a_{i} (i = 1 to 11)

a_{1} + a_{3} + a_{5} ...... + a_{11} = 6 (a_{6}) = 72.

As a_{6} would be the arithmetic mean of these 11 numbers and

2(a_{6}) = (a_{1 }+ a_{11})

= (a_{2} + a_{10})

= (a_{3} + a_{9})

= (a_{4} + a_{8})

= (a_{5} + a_{7})

∴ a_{1} + a_{6} + a_{11} = 3 (a_{6})

= 36

QUESTION: 20

If log_{3}2, log_{3}(2^{x} – 5) and log_{3} are in Arithmetic Progression, then x is equal to

(2014)

Solution:

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 log_{3} (2^{x} – 5) = log_{3} 2 + log_{3} (2^{x} – 7/2)

Suppose 2^{x} = t

2 log_{3} (t – 5) = log_{3} 2 + log_{3} (t – 7/2)

→ log_{3} (t – 5)^{2} = log_{3} (2t – 7)

→ (t – 5)^{2} = 2t – 7

→ t^{2} – 10t + 25 = 2t – 7

→ t^{2} – 12t + 32 = 0

→ t = 4.8

2^{x} = 4.8

∴ x = 2, 3.

QUESTION: 21

A ray of light along the line gets reflected on the x-axis to become a ray along the line

(2014)

Solution:

Take –y in place of y

Then equation is

QUESTION: 22

If where p ≤ n, then the maximum value of X for n = 8 is :

(2014)

Solution:

X = (log_{10} 1 + log_{10} 2 + ... + log_{10} n) – (log_{10} 1 + log_{10} 2 + .... + log_{10} p) – (log_{10} 1 + log_{10} 2 + .... + log_{10} (n – p))

⇒ X = log_{10} n! – log_{10} p! – log_{10} (n – p)!

log (m × n) = log m + log n

X is maximum when is maximum.

QUESTION: 23

If x + y = 1, then what is the value of (x^{3} + y^{3} + 3xy)?

(2012)

Solution:

Given that x + y = 1

⇒ x + y – 1 = 0

⇒ x^{3} + y^{3} – 1 = – 3xy

(a^{3 }+ b^{3} + c^{3} = 3abc if a + b + c = 0)

⇒ x^{3} + y^{3} + 3xy = 1

QUESTION: 24

If log_{16}5 = m and log_{5}3 = n, then what is the value of log_{3}6 in terms of ‘m’ and ‘n’?

(2011)

Solution:

...(i)

...(ii)

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

Let log_{3}6 be equal to k; therefore,

QUESTION: 25

If a = b^{2} = c^{3} = d^{4} then the value of log_{a} (abcd) would be :

(2010)

Solution:

Let a = b^{2} = c^{3} = d^{4}

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