Let P_n = alpha^n + beta^n, n in mathbbN. If P_10 = 123, P_9 = 76, P_8 = 47 and P_1 = 1, then the quadratic equation having roots frac1alpha and frac1beta is:

Solution & Explanation

### Related Formula Newton's Sums for the roots of a quadratic equation ax^2 + bx + c = 0: a P_n + b P_n-1 + c P_n-2 = 0 ### Core Logic Observe the recurrence relation from the given numerical values of P_n to construct the base quadratic equation satisfied by alpha and beta, then invert the roots. ### Step 1: Identify the Linear Recurrence Relation Compare the provided sequence values: P_8 + P_9 = 47 + 76 = 123 = P_10 This fits the general sequence relation: P_n = P_n-1 + P_n-2 implies P_n - P_n-1 - P_n-2 = 0 ### Step 2: Construct the Base Quadratic Equation The characteristic equation corresponding to this recurrence relation is: x^2 - x - 1 = 0 Thus, alpha and \(\beta\) are roots of x^2 - x - 1 = 0, giving sum alpha+beta = 1 and product alphabeta = -1 (which matches P_1 = alpha+beta = 1). ### Step 3: Construct Equation with Reciprocal Roots To find the equation with roots frac1alpha and frac1beta, apply the transformations: textSum of new roots = frac1alpha + frac1beta = fracalpha + betaalphabeta = frac1-1 = -1 textProduct of new roots = frac1alphabeta = frac1-1 = -1 The new quadratic equation is: x^2 - (textSum)x + (textProduct) = 0 implies x^2 - (-1)x + (-1) = 0 implies x^2 + x - 1 = 0 ### Pattern Recognition The recurrence pattern P_n = P_n-1 + P_n-2 is the Fibonacci sequence recurrence line. Its roots generate the Golden Ratio layout from x^2-x-1=0. Inverting roots swaps the coefficients of x^2 and the constant term, yielding x^2+x-1=0 instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 11 Mathematics: Sequences and Series

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 7

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument
If alpha denotes the number of solutions of |1 - i|^x = 2^x and beta = left(frac|z|arg(z)right), where z = fracpi4 (1 + i)^4 left(frac1 - sqrtpi isqrtpi + i + fracsqrtpi - i1 + sqrtpi iright), i = sqrt-1, then the distance of the point (alpha, beta) from the line 4x - 3y = 7 is
Numerical Answer. Answer: 3 to 3

Solution

### Core Logic |1 - i|^x = 2^x implies (sqrt2)^x = 2^x implies 2^x/2 = 2^x This implies fracx2 = x implies x = 0. There is exactly 1 solution, so alpha = 1. ### Step 1: Simplify complex number z (1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4 Thus, z = -pi left( frac(1-sqrtpii)(sqrtpi-i)pi + 1 + frac(sqrtpi-i)(1-sqrtpii)1 + pi right) ### Step 2: Simplify Bracket Let's expand the terms directly: z = fracpi4(-4) left[ fracsqrtpi - pi i - i - sqrtpipi + 1 + fracsqrtpi - i - pi i - sqrtpi1 + pi right] = -pi left[ frac-i(pi+1)pi+1 + frac-i(pi+1)pi+1 right] = -pi [ -i - i ] = 2pi i ### Step 3: Find beta For z = 2pi i: |z| = 2pi and arg(z) = fracpi2. beta = frac|z|arg(z) = frac2pipi/2 = 4 ### Step 4: Distance from Line Distance of point (alpha, beta) = (1, 4) from the line 4x - 3y - 7 = 0: D = frac|4(1) - 3(4) - 7|sqrt4^2 + (-3)^2 = frac|4 - 12 - 7|5 = frac|-15|5 = 3 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines
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