Let S = x³ + ax² + bx + c : a, b, c in N and a, b, c ≤ 20 be a set of polynomials. Then the number of polynomials in S, which are divisible by x² + 2, is

Solution & Explanation

Core Logic

For a polynomial P(x) = x³ + ax² + bx + c to be divisible by x² + 2, we can perform polynomial long division or use synthetic substitution (roots of x²+2=0). Alternatively, factorize P(x): Since the degree is 3 and the leading coefficient is 1, we must have:

x³ + ax² + bx + c = (x² + 2)(x + k)

where k is a constant.

Step 1: Equating Coefficients

Expand the right side:

(x² + 2)(x + k) = x³ + kx² + 2x + 2k

Comparing coefficients with x³ + ax² + bx + c: x² coefficient: a = k x coefficient: b = 2 Constant term: c = 2k

Step 2: Apply Constraints

From the coefficient matching, we have a = (c)/(2) and b = 2. Since a, b, c in N and a, b, c ≤ 20: b = 2 (this is fixed, always valid). c must be an even natural number such that c ≤ 20. The possible values for c are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. For each such c, a is uniquely determined as a = c/2 and a ≤ 10 (which easily satisfies a ≤ 20).

Step 3: Final Count

The number of valid (a, b, c) tuples corresponds to the number of valid c values. There are exactly 10 values for c. Thus, the number of polynomials in S is 10.

Chapter Mix

Class 10 Mathematics: Polynomials Class 11 Mathematics: Complex Numbers and Quadratic Equations

Reference Study Guides

Polynomials

More Polynomials Previous-Year Questions — Page 6

Q20 jee_main_2024_31_jan_morning Sign of Quadratic Expressions
Let S be the set of positive integral values of a for which (ax² + 2(a + 1)x + 9a + 4)/(x² - 8x + 32) < 0, x in R. Then, the number of elements in S is:
  • A. 1
  • B. 0
  • C. ∞
  • D. 3

Solution

Core Logic

For the denominator x² - 8x + 32, D = 64 - 128 < 0 and a = 1 > 0. Thus, x² - 8x + 32 > 0 x in R.

Step 1: Constraint on Numerator

Since the denominator is always positive, the numerator must be strictly negative for all x in R.

ax² + 2(a + 1)x + 9a + 4 < 0 x in R

This requires a < 0 and D < 0.

Step 2: Conclusion

Since a must be strictly less than 0, there are no positive integral values of a that satisfy the condition. Hence, S is an empty set. Number of elements is 0.

Chapter Mix

Class 11 Maths: Quadratic Equations

Rankbit System
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