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QUESTION: 1

If a_{1} + a_{2} + a_{3} + ... + a_{n} = 3(2^{n+1} – 2), for every n ≥ 1, then a_{11} equals

(2019)

Solution:

If n = 1, then a_{1} = 3(2^{1 + 1} – 2) = 6 = 3 × 2^{1}

If n = 2, then a_{1} + a_{2} = 3(2^{2 + 1} – 2) = 18 ⇒ a_{2}

= 18 – a1 = 18 – 6 = 12 = 3 × 2^{2}

If n = 3, then a_{1} + a_{2} + a_{3} = 3(2^{3 + 1} – 2) = 42 ⇒ a_{3}

= 42 – (a_{1} + a_{2}) = 24 = 3 × 2^{3}

Similarly, a_{n} = 3 × 2^{n}

Therefore, a_{11} = 3 × 2^{11} = 6144

QUESTION: 2

The number of solutions to the equation |x|(6x^{2} + 1) = 5x^{2} is

(2019)

Solution:

**Case I:** x = 0.

Clearly, x = 0 satisfy the equation

**Case II:** x > 0 |x| (6x^{2} + 1) = 5x^{2} ⇒ x(6x^{2} + 1) = 5x^{2}

⇒ x(6x^{2} + 1 – 5x) = 0

∴ x ≠ 0

∴ 6x^{2} + 1 – 5x = 0

On solving the quadratic equation, we get

**Case III:** x < 0

|x| (6x^{2} + 1) = 5x^{2 }⇒ –x(6x^{2 }+ 1) = 5x2

⇒ –x(6x^{2} + 1 + 5x) = 0

⇒ x(6x^{2} + 5x + 1) = 0

∵ x ≠ 0, ∴ 6x^{2} + 5x + 1 = 0

On solving the quadratic equation, we get

Hence there are 5 solutions.

QUESTION: 3

The product of the distinct roots of |x^{2} – x – 6| = x + 2 is

(2019)

Solution:

x^{2} – x – 6 = (x + 2)(x – 3)

**Case 1: **x^{2} – x – 6 < 0 ⇒ (x + 2)(x – 3) < 0

⇒ –2 < x < 3 and |x^{2} – x – 6| = –(x^{2} – x – 6)

∴ |x^{2} – x – 6| = x + 2 ⇒ –(x + 2)(x – 3) = x + 2

⇒ (x – 3) = –1 ⇒ x = 2

**Case 2:** x^{2} – x – 6 ≥ 0 ⇒ (x + 2) (x – 3) ≥ 0

⇒ x ≤ –2 or x ≥ 3

Checking for boundary conditions:

For x = –2, |x^{2} – x – 6| = x + 2, therefore,x = -2 is also the root.

But for x = 3, |x^{2} – x – 6| ≠ x + 2

Hence x = 3 is NOT the root.

And for the interval x < –2 or x > 3 the expression |x^{2} – x – 6|

= x^{2} – x – 6

∴ |x^{2} – 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

QUESTION: 4

The number of the real roots of the equation 2cos (x(x + 1)) = 2^{x} + 2^{–x} is

(2019)

Solution:

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.

QUESTION: 5

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) – f(m) = 2, then m equals

(2019)

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

⇒ m^{2} – 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(m^{2} + 3m + 2) – m – 3 = 2

⇒ 8m^{2} + 24m + 16 – m – 3 = 2

⇒ 8m^{2} + 23m + 11 = 0,

Which is not possible and hence no solution.

QUESTION: 6

Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2^{n} – 1) then a is equal to

(2019)

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

QUESTION: 7

The real root of the equation 2^{6x} + 2^{3x+2} – 21 = 0 is

(2019)

Solution:

2^{6x} + 2^{3x+2} – 21 = 0

⇒ 2^{6x} + 2^{3x} × 2^{2} – 21 = 0

Put 2^{3x} = y, then y^{2} + 4y – 21 = 0

⇒ (y – 3) (y + 7) = 0 ⇒ y = 3 or y = –7

⇒ 2^{3x} = 3 or 2^{3x} = –7 (No solution)

⇒ 3x = log_{2}3 ⇒ x = log_{2} 3 / 3

QUESTION: 8

What is the largest positive integer n such that is also a positive integer?

(2019)

Solution:

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

QUESTION: 9

If 5^{x} – 3^{y} = 13438 and 5^{x – 1} + 3^{y + 1} = 9686, then x + y equals

(2019)

Solution:

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 3^{4} = 81, and 3^{8} = 6561

Also, 9686 – 81 = 9605 and 9686 – 6561 = 3125

We observe that 3125 = 5^{5 }

∴ 5^{x–1} = 5^{5} ⇒ x = 6

and 3^{y+1} = 3^{8} ⇒ y = 7

x = 6 and y = 7 also satisfies the first given equation 5^{x} – 3^{y} = 13438

∴ x + y = 6 + 7 = 13

QUESTION: 10

The quadratic equation x^{2} + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b^{2} + c?

(2019)

Solution:

Sum of roots = 4a + 3a = 7a = –b

∴ b = –7a

Product of roots = 4a × 3a = c

∴ c = 12a^{2}

Now, b2 + c = (–7a)^{2} + 12a^{2} = 61a^{2}

Comparing the options

**Option (a) :** 61a^{2} = 3721 ⇒ a^{2} = 61, clearly a is not an integer.

**Option (b) :** 61a^{2} = 361 ⇒ a^{2} = 361 / 61, clearly a is not an integer.

**Option (c) : **61a^{2} = 549 ⇒ a^{2} = 9, we can have a = –3 or 3 (an integer)

**Option (d) :** 61a^{2} = 427 ⇒ a^{2} = 7, clearly a is not an integer

QUESTION: 11

Let a, b, x, y be real numbers such that a^{2} + b^{2} = 25, x^{2} + y^{2} = 169, and ax + by = 65. If k = ay – bx, then

(2019)

Solution:

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

QUESTION: 12

If u^{2} + (u − 2v − 1)^{2} = −4v(u + v), then what is the value of u + 3v?

(2018)

Solution:

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

QUESTION: 13

If f(x) = 5x + 2 / 3x - 5 and g(x) = x^{2} – 2x – 1, then the value of g (f(f(3))) is

(2017)

Solution:

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

QUESTION: 14

The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12,and z ≤ 12 is

(2017)

Solution:

x = 25 + y + z

Possible value table below.

27 ≤ x ≤ 40

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

QUESTION: 15

Suppose, log_{3} x = log_{12} y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log_{6} G is equal to

(2017)

Solution:

log_{3}x = a, x = 3^{a }

log_{12}y = a, y = 12^{a}

Therefore, the geometric mean of x and y equals

Hence, 6^{a} or log_{6} G = a

QUESTION: 16

If the sum to infinity of the series 2 + (2 – d) 2/3 + (2 + d) 4/9 + (2 + 3d) 8/27 + .... is 5/2, what is the value of d?

(2016)

Solution:

Subtracting,

QUESTION: 17

If S = 2/10 + 6/10^{2} + 12/10^{3} + 20/10^{4} + 30/10^{5} + 42/10^{6 }+ ......., find the value of S?

(2016)

Solution:

∵ 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.

QUESTION: 18

If the roots of the equation (x + 1) (x + 9) + 8 = 0 are a and b, then the roots of the equation (x + a) (x + b) – 8 = 0 are

(2016)

Solution:

Here, (x + 1) (x + 9) + 8 = 0

x^{2} + 10x + 17 = 0

The roots of the equation are a and b

∴ a + b = – 10

ab = 17

(x + a) (x + b) – 8 = 0

x^{2} + (a + b) x + ab – 8 = 0

x^{2} – 10x + 9 = 0

Therefore, roots of (x + a) (x + b) – 8 = 0 are 1 and 9.

QUESTION: 19

If x^{2} + (x + 1) (x + 2) (x + 3) (x + 6) = 0, where x is a real number, then one value of x that satisfies this equation is

(2015)

Solution:

x^{2} + ( x + 1) (x + 2) (x + 3) (x + 6) = 0

(x + 1) (x + 2) (x + 3) (x + 6) = –x^{2}

(x^{2} + 7x + 6) (x^{2} + 5x + 6) = –x^{2}

then, (y + 7) (y + 5) = –1

⇒ y^{2} + 12y + 35 = –1

y^{2} + 12y + 36 = 0

⇒ (y + 6)^{2 } = 0

⇒ y = – 6

QUESTION: 20

Find the solution set for [x] + [2x] + [3x] = 8, where x is a real number and [x] is the greatest integer less than or equal to x.

(2015)

Solution:

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

QUESTION: 21

If 3x + y + 4 = 2xy, where x and y are natural numbers, then find the ratio of the sum of all possible values of x to the sum of all possible values of y.

(2015)

Solution:

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

QUESTION: 22

Let x, y, z and t be the positive numbers which satisfy the following conditions:

I. If x > y, then z > t and

II. If x > z, then y < t

Which of the following is necessarily true?

(2015)

Solution:

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.

QUESTION: 23

Both the roots of the quadratic equation x^{2} + rx + s = 0 are real and greater than 1. If then which of the following is definitely true?

(2015)

Solution:

Take roots as 2, 2

⇒ r = – 4 & s = 4

QUESTION: 24

If x(x – 3) = –1, then the value of x^{3}(x^{3} – 18) is

(2014)

Solution:

According to question,

x (x – 3) = –1

cubing on both side.

⇒ x^{3} (x – 3)^{3} = (–1)^{3}

⇒ x^{3 }(x^{3}– 27 – 9x^{2} + 27x) = – 1

⇒ x^{3} (x^{3} – 18) + x^{3} (–9 – 9x^{2} + 27x) = –1

⇒ x^{3} (x^{3} – 18) – 9x^{3} (x^{2} – 3x + 1) = –1

⇒ x^{3} (x^{3 }– 18) – 9x^{3} (–1 + 1) = –1

⇒ x^{3} (x^{3} – 18) = –1

QUESTION: 25

Find the value of P.

(2014)

Solution:

nth term of the series can be written as

QUESTION: 26

What is the sum of the roots of all the quadratic equations that can be formed such that both the roots of the quadratic equation are common with the roots of equation (x – a) (x – b) (x – c) = 0?

(2014)

Solution:

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).

QUESTION: 27

The sum of the coefficients of the polynomial (x – 1)^{9} (x – 2)^{4 }(x – 4) is

(2014)

Solution:

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.

QUESTION: 28

An equation with all positive roots is written as X^{n} + a_{n}X^{n-1} + a_{n-1}X^{n-2} + .... a_{1} = 0. Which of the following in necessarily true?

(2014)

Solution:

Here, (x – 1) (x – 2) (x – 3) = 0

⇒ (x^{2} – 3x + 2) (x – 3) = 0

⇒ x^{3} – 6x^{2} + 11x – 6 = 0

According to question,

x = 3, a_{n} = –6, a_{1} = –6

Substitute the above values in the option,

Option (a) : (–6)^{3} ≥ 3^{3} × – 6

⇒ – 216 ≥ – 168

This is incorrect, thus option (a) is incorrect.

Opiton (b) : 3^{3} ≥ (–6)^{3} × –6 27 ≥ 6^{4}

This is incorrect, thus option (b) is incorrect.

Option (c) : (–6)^{3} ≥ 3^{3} × – 6

⇒ –216 ≥ – 168

This is incorrect, thus option (c) is incorrect.

Thus, none of the options is necessarily true.

QUESTION: 29

The coefficient of a^{12}b^{8} in the expansion of (a^{2} + b)^{13} is

(2013)

Solution:

General term expansion of

(a^{2} + b)^{13} = ^{n}C_{r} x^{n-r }y^{r}

To get coefficient of a^{12 } b^{8}

(a^{2})^{6} b^{8} where 6 + 8 ≠ 13.

So the term is not possible

QUESTION: 30

Let R(x) = mx^{3} – 100x^{2 }+ 3n, where m and n are positive integers. For how many ordered pairs (m, n) will (x – 2) be a factor of R(x)?

(2013)

Solution:

(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.

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