Let S = x³ + ax² + bx + c : a, b, c in N and a, b, c ≤ 20$S = \{ x^3 + ax^2 + bx + c : a, b, c \in \mathbb{N} \text{ and } a, b, c \leq 20 \}$ be a set of polynomials. Then the number of polynomials in S$S$, which are divisible by x² + 2$x^2 + 2$, is

Related Formula

X² = |X|²$X^2 = |X|^2$ Quadratic factorization: t² - 5t + 6 = (t-2)(t-3)$t^2 - 5t + 6 = (t-2)(t-3)$

Core Logic

Rewrite the equation taking |x - 1| = t$|x - 1| = t$, where t ≥ 0$t \geq 0$. Since (x-1)² = |x-1|²$(x-1)^2 = |x-1|^2$, the equation becomes:

Step 1: Solve for modulus

Since both roots are positive, both provide valid solutions for the modulus. |x - 1| = 2 and |x - 1| = 3$|x - 1| = 2 \quad \text{and} \quad |x - 1| = 3$

Step 2: Unpack x values

From |x - 1| = 2$|x - 1| = 2$:

From |x - 1| = 3$|x - 1| = 3$:

The roots are 3, -1, 4, -2$3, -1, 4, -2$.

Step 3: Sum of Roots

Pattern Recognition

In symmetric modulus equations f(|x-a|) = 0$f(|x-a|) = 0$, every valid root t$t$ generates twin solutions (a+t)$(a+t)$ and (a-t)$(a-t)$. The sum of each pair is perfectly 2a$2a$. If there are n$n$ distinct valid positive roots for t$t$, the sum of all x$x$-roots is exactly n × 2a$n \times 2a$. Here, n=2, a=1 ⇒ 2 × 2(1) = 4$n=2, a=1 \Rightarrow 2 \times 2(1) = 4$.

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Class 11 Maths: Quadratic Equations

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Core Logic

Rewrite the series in summation form:

Step 1: Simplify the Equation

Substitute the standard summation formulas:

Divide the entire equation by n$n$:

Let the roots be α$\alpha$ and β$\beta$.

Step 2: Apply the Condition for Roots

Since the roots are two consecutive even integers, their difference is 2$2$.

Since n$n$ is a positive integer, n = 3$n = 3$.

Pattern Recognition

When dealing with equations where roots have a specific difference k$k$, instantly use D = a²k²$D = a^2k^2$. Here k=2$k=2$, so D=4a²$D=4a^2$.

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Class 11 Maths: Quadratic Equations Class 11 Maths: Sequence and Series

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Core Logic

Given f(x) = x² + 2ax + (3a + 10)$f(x) = x^2 + 2ax + (3a + 10)$. Since the coefficient of x²$x^2$ is 1 > 0$1 > 0$, the parabola opens upward. For 1$1$ to lie between the roots α$\alpha$ and β$\beta$, the value of the function at x = 1$x = 1$ must be strictly less than 0$0$. f(1) < 0$f(1) < 0$

Step 1: Evaluate Inequality

5a + 11 < 0$5a + 11 < 0$

So, a in (-∞, -(11)/(5))$a \in \left(-\infty, -\frac{11}{5}\right)$.

Pattern Recognition

When a specified value k$k$ lies between roots, a · f(k) < 0$a \cdot f(k) < 0$. If a>0$a>0$, this simplifies to f(k) < 0$f(k) < 0$. No need to check discriminant Δ > 0$\Delta > 0$ manually because a f(k) < 0$a f(k) < 0$ guarantees real distinct roots.

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Class 11 Maths: Quadratic Equations

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Core Logic

Break the domain into intervals based on the critical points of the absolute values, which are x = -4$x = -4$ and x = -2$x = -2$. Intervals to test: Case I: x < -4$x < -4$ Case II: -4 ≤ x < -2$-4 \leq x < -2$ Case III: x ≥ -2$x \geq -2$

Step 1: Analyzing Case I

For x < -4$x < -4$: Both (x+4)$(x+4)$ and (x+2)$(x+2)$ are negative. The equation becomes:

Roots are x = -7 ± √(49 + 16)2 = -7 ± √(65)2$x = \frac{-7 \pm \sqrt{49 + 16}}{2} = \frac{-7 \pm \sqrt{65}}{2}$.

Check if they fall in the interval x < -4$x < -4$: √(65) ≈ 8.06$\sqrt{65} \approx 8.06$ x₁ = (-7 + 8.06)/(2) ≈ 0.53$x_1 = \frac{-7 + 8.06}{2} \approx 0.53$ (Rejected, not <-4$<-4$) x₂ = (-7 - 8.06)/(2) ≈ -7.53$x_2 = \frac{-7 - 8.06}{2} \approx -7.53$ (Accepted) So, 1 valid solution here.

Step 2: Analyzing Case II

For -4 ≤ x < -2$-4 \leq x < -2$: (x+4) ≥ 0$(x+4) \geq 0$ and (x+2) < 0$(x+2) < 0$. The equation becomes:

x² + x + 4 = 0$x^2 + x + 4 = 0$

Discriminant D = 1² - 4(1)(4) = 1 - 16 = -15 < 0$D = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0$. No real roots in this interval.

Step 3: Analyzing Case III

For x ≥ -2$x \geq -2$: Both (x+4)$(x+4)$ and (x+2)$(x+2)$ are positive. The equation becomes:

Discriminant D = 49 - 64 = -15 < 0$D = 49 - 64 = -15 < 0$. No real roots in this interval.

Step 4: Final Count

Combining the results from all cases, there is exactly 1 distinct real solution.

Pattern Recognition

For sum-of-absolute-value equations, systematically partition the number line using the roots of the arguments. Discard roots generated by the quadratics that fall outside their respective assumed interval.

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Class 11 Maths: Quadratic Equations

Related Formula

Difference of roots formula:

For 12x² - 20x + 3λ = 0$12x^2 - 20x + 3\lambda = 0$: α + β = (20)/(12) = (5)/(3)$\alpha + \beta = \frac{20}{12} = \frac{5}{3}$, αβ = (3λ)/(12) = (λ)/(4)$\alpha\beta = \frac{3\lambda}{12} = \frac{\lambda}{4}$.

Core Logic

Square the inequality (1)/(2) ≤ |α - β| ≤ (3)/(2)$\frac{1}{2} \le |\alpha - \beta| \le \frac{3}{2}$:

Step 1: Integer Values of Lambda

Since λ in Z$\lambda \in \mathbb{Z}$, the valid integer values are λ = 1, 2$\lambda = 1, 2$.

Pattern Recognition

Convert root difference inequality into quadratic discriminant bounds for quick integer extraction.

Chapter Mix

Class 11 Maths: Quadratic Equations