Let mathrmA = left\theta in [0,2pi ]:1 + 10operatorname Releft(frac2costheta + mathrmisinthetacostheta - 3mathrmisinthetaright) = 0right\. Then sum_theta in mathrmAtheta^2 is equal to

Solution & Explanation

### Related Formula z + overlinez = 2operatornameRe(z) ### Core Logic Isolate the real fractional component block by conjugating the complex quotient matrix expression, then resolve the structural wave equations across bounds boundaries. ### Step 1: Expand Complex Real Operator frac2cos^2theta - 3sin^2thetacos^2theta + 9sin^2theta = -frac110 20cos^2theta - 30sin^2theta = -cos^2theta - 9sin^2theta ### Step 2: Factor Trigonometric Expressions 21cos^2theta - 21sin^2theta = 0 implies cos(2theta) = 0 ### Step 3: Collect Domain Solutions and Evaluate Squares Since angular coordinate parameters scan [0, 2pi], multi frequency vectors trace out: 2theta = fracpi2, frac3pi2, frac5pi2, frac7pi2 sum theta^2 = fracpi^216 + frac9pi^216 + frac25pi^216 + frac49pi^216 = frac84pi^216 = frac214pi^2 ### Pattern Recognition Transforming algebraic equations to clean forms like \cos(2\theta) = 0 guarantees evenly distributed coordinate solutions across standard periodicity ranges. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Trigonometric Functions

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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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