Number System

Miscellaneous

Medium

The expression 2x³ + ax² + bx + 3, where a and b are constants, has a factor of x-1 and leaves a remainder of 15 when divided by x + 2. Then, (a, b) =

(-3,8)

(3,-8)

(-3,-8)

(3, 8)

✅ Correct Option: 2

The expression: $2x^3 + ax^2 + bx + 3$

Since $x-1$ is a factor, the expression should be equal to $0$ when $x=1$

When $x=1$ :

$2(1)^3 + a(1)^2 + b(1) + 3 = 0$

Simplifies to: $2 + a + b + 3 = 0$ $2 + a + b + 3 = 0$
- Therefore: $a + b = -5$ ...(1)

When divided by $x+2$ , remainder is $15$

When $x=-2$ :

$2(-2)^3 + a(-2)^2 + b(-2) + 3 = 15$

$-16 + 4a - 2b + 3 = 15$ $- 16 + 4 a - 2 b + 3 = 15$
- $4a - 2b = 28$ ...(2)
- $2a - b = 14$

We now have two equations:

$a + b = -5$ $a + b = - 5$ ...(1)
- $2a - b = 14$ ...(2)

From (1): $b = -5 - a$

Substitute in (2):

$2a - (-5-a) = 14$ $2 a - (- 5 - a) = 14$
- $2a + 5 + a = 14$
- $3a = 9$
- $a = 3$

Put back to find $b$ :

$b = -5 - 3 = -8$

Therefore, $(a,b) = (3,-8)$

JIPMAT 2023

Given below are two statements:

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Statement (II): The last digit in the decimal representation of 2⁹¹ is 6.

In light of the above statements, choose the most appropriate answer from the options given below.

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Amicable numbers are a pair of distinct natural numbers (a, b) such that the sum of the proper divisors of a equals b and the sum of the proper divisors of b equals a. Given that (220, y) is a pair of amicable numbers, y equals:

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Given below are two statements:

Statement I : The sum of exponents of prime factors in the prime factorization of 392 is 5.

Statement II : The decimal representation $\frac{13}{2^3 \times 5}$ will terminate after 2 decimal places.

In the light of the above statements, choose the correct answer from the options given below:

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Arrange the following fractions in descending order.

(A) $\frac{5}{9}$

(B) $\frac{7}{11}$

(D) $\frac{11}{17}$

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After adding 5 to both numerator and denominator of the following fractions, arrange them in ascending order of the magnitude of change in their values.

A. $\frac{2}{3}$

B. $\frac{3}{4}$

C. $\frac{4}{5}$

D. $\frac{5}{6}$

The expression 2x3 + ax2 + bx + 3, where a and b are constants, has a factor of x-1 and leaves a remainder of 15 when divided by x + 2. Then, (a, b) =

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JIPMAT Past Year Questions (Topic-Wise)

The expression 2x³ + ax² + bx + 3, where a and b are constants, has a factor of x-1 and leaves a remainder of 15 when divided by x + 2. Then, (a, b) =