This is the solution for the test Class 11 Complex Numbers and Linear Inequality Test – 1. Let’s break down each problem step-by-step. Designed for both students and teachers, these answers make exam revision easier with clear explanations, formula highlights, and downloadable PDF options.
Class 11 Complex Numbers and Linear Inequality Test-1 2027 ⇒
Section 1: 2 Marks Questions
Q1. Find the conjugate of \frac{1}{(4+2i)^{2}}
Solution
Expand the denominator using the identity (a+b)^2 = a^2 + 2ab + b^2.
z = \frac{1}{16 + 16i + 4i^2}
Since i^2 = -1, substitute and simplify the expression.
z = \frac{1}{16 + 16i - 4} = \frac{1}{12 + 16i}
Factor out the common term and rationalize by multiplying the numerator and denominator by the conjugate of the denominator, (3-4i).
z = \frac{1}{4(3+4i)} \times \frac{3-4i}{3-4i}
z = \frac{3-4i}{4(9 - 16i^2)} = \frac{3-4i}{4(9 + 16)} = \frac{3-4i}{100}
Write in standard form a + ib and identify the conjugate \bar{z}.
z = \frac{3}{100} - \frac{4}{100}i
Final Answer: \bar{z} = \frac{3}{100} + \frac{4}{100}i
Q2. Solve \vert{}x-1\vert{}-\vert{}x-2\vert{}>10
Solution:
Recall the triangle inequality for absolute values: \vert{}\vert{}a\vert{} - \vert{}b\vert{}\vert{} \le \vert{}a - b\vert{}.
Apply this property to the left side of our expression where a = x-1 and b = x-2.
\vert{}\vert{}x-1\vert{} - \vert{}x-2\vert{}\vert{} \le \vert{}(x-1) - (x-2)\vert{} = \vert{}1\vert{} = 1
This implies that the absolute maximum value for the expression \vert{}x-1\vert{} - \vert{}x-2\vert{} is 1.
The question asks where this expression is greater than 10. Since the maximum possible value is 1, this condition can never be mathematically met.
Final Answer: No real solution, or x \in \emptyset.
Section 2: 3 Marks Questions
Q3. If (\frac{1+i}{1-i})^{3}-(\frac{1-i}{1+i})^{3}=x+iy, then find x, y.
Solution:
Simplify the fractions inside the parentheses first by rationalizing them.
\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1^2 - i^2} = \frac{2i}{2} = i
\frac{1-i}{1+i} = \frac{(1-i)^2}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1^2 - i^2} = \frac{-2i}{2} = -i
Substitute these simplified values back into the original cubic equation.
(i)^3 - (-i)^3 = x + iy
Evaluate the cubes (remembering that i^3 = -i).
-i - (-(-i)) = -i - i = -2i
Compare the real and imaginary parts of both sides.
0 - 2i = x + iy
Final Answer: x = 0, y = -2
Q4. If a=cos~B+i~sin~B then find the value of (\frac{1+a}{1-a})
Solution:
Substitute the given value of a into the numerator and denominator.
Numerator: 1+a = 1 + \cos B + i \sin B
Denominator: 1-a = 1 - (\cos B + i \sin B) = 1 - \cos B - i \sin B
Use half-angle identities to simplify:
1 + \cos B = 2\cos^2(B/2),
1 - \cos B = 2\sin^2(B/2), and
\sin B = 2\sin(B/2)\cos(B/2).
Numerator: 2\cos^2(B/2) + 2i\sin(B/2)\cos(B/2) = 2\cos(B/2)[\cos(B/2) + i\sin(B/2)]
Denominator: 2\sin^2(B/2) - 2i\sin(B/2)\cos(B/2) = 2\sin(B/2)[\sin(B/2) - i\cos(B/2)]
Notice a relationship in the denominator: \sin(B/2) - i\cos(B/2) = -i(\cos(B/2) + i\sin(B/2)).
Therefore, the denominator becomes: -2i\sin(B/2)[\cos(B/2) + i\sin(B/2)]
Divide the updated numerator by the denominator (the complex terms cancel out).
\frac{1+a}{1-a} = \frac{2\cos(B/2)}{-2i\sin(B/2)} = \frac{\cos(B/2)}{-i\sin(B/2)}
Multiply numerator and denominator by i to remove it from the denominator.
\frac{i\cos(B/2)}{-i^2\sin(B/2)} = i \cot(B/2)
Final Answer: i \cot(B/2)
Q5. Solve -5\le\frac{2-3x}{4}\le9
Solution:
Multiply the entire compound inequality by 4 to remove the fraction.
-20 \le 2 - 3x \le 36
Subtract 2 from all three segments.
-22 \le -3x \le 34
Divide by -3. Crucial point: flip the inequality signs when dividing by a negative number!
\frac{22}{3} \ge x \ge -\frac{34}{3}
Final Answer: x \in \left[-\frac{34}{3}, \frac{22}{3}\right]
Q6. If \frac{z-1}{z+1} is a purely imaginary and z\ne-1 then find the value of \vert{}z\vert{}.
Solution:
Let w = \frac{z-1}{z+1}.
For any purely imaginary number w, the sum of the number and its conjugate is exactly zero (w + \bar{w} = 0).
Apply this algebraic property to our fraction.
\frac{z-1}{z+1} + \overline{\left(\frac{z-1}{z+1}\right)} = 0
\frac{z-1}{z+1} + \frac{\bar{z}-1}{\bar{z}+1} = 0
Cross-multiply to combine them into a single fraction.
\frac{(z-1)(\bar{z}+1) + (\bar{z}-1)(z+1)}{(z+1)(\bar{z}+1)} = 0
Set the numerator to zero (since z \ne -1) and expand.
(z\bar{z} + z - \bar{z} - 1) + (z\bar{z} + \bar{z} - z - 1) = 0
2z\bar{z} - 2 = 0
Simplify and substitute the standard property z\bar{z} = \vert{}z\vert{}^2.
z\bar{z} = 1 \implies \vert{}z\vert{}^2 = 1 \implies \vert{}z\vert{} = 1
Final Answer: \vert{}z\vert{} = 1
Q7. A company manufactures cassettes. Its cost and revenue functions are C(x)=26000+30x and R(x)=43x respectively. How many cassettes must be sold by the company to realise some profit?
Solution:
Profit P(x) is calculated as Revenue minus Cost.
For the company to realize some profit, P(x) must be strictly greater than 0.
P(x) = R(x) - C(x) > 0
Substitute the given functions.
43x - (26000 + 30x) > 0
Solve the linear inequality for x.
13x - 26000 > 0
13x > 26000 \implies x > 2000
Final Answer: The company must sell at least 2001 cassettes to realize a profit.
Section 3: 4 Marks Questions
Q8. The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.
Solution:
Define all three sides in terms of a single variable.
Let the shortest side be x cm.
Longest side = 2x cm.
Third side = (x + 2) cm.
Set up the perimeter inequality based on the word problem.
x + 2x + (x + 2) > 166
Combine like terms and solve for x.
4x + 2 > 166
4x > 164 \implies x > 41
Final Answer: The minimum length of the shortest side must be strictly greater than 41 cm.
Q9. Solve the inequalities: \frac{2x+1}{7x-1}>5 and \frac{x+7}{x-8}>2
Solution:
(i)
\frac{2x+1}{7x-1} > 5
Bring 5 to the left side to maintain zero on the right (never cross-multiply variable expressions in inequalities).
\frac{2x+1}{7x-1} - 5 > 0
Find a common denominator and simplify.
\frac{2x+1 - 5(7x-1)}{7x-1} > 0 \implies \frac{2x + 1 - 35x + 5}{7x-1} > 0 \implies \frac{6 - 33x}{7x-1} > 0
The critical points are x = \frac{6}{33} = \frac{2}{11} and x = \frac{1}{7}.
Using the wavy curve/sign analysis method, the inequality holds for x \in \left(\frac{1}{7}, \frac{2}{11}\right).
(ii)\frac{x+7}{x-8} > 2
Follow the exact same zero-side process.
\frac{x+7}{x-8} - 2 > 0 \implies \frac{x+7 - 2(x-8)}{x-8} > 0 \implies \frac{23 - x}{x-8} > 0
The critical points are x = 23 and x = 8.
This expression is strictly positive when x \in (8, 23).
Q10. Find the complex number satisfying the equation z+\sqrt{2}\vert{}z+1\vert{}+i=0
Solution:
Let z = x + iy.
Substitute this standard form into the equation.
x + iy + \sqrt{2}\sqrt{(x+1)^2 + y^2} + i = 0
Group the real and imaginary parts together.
\left(x + \sqrt{2}\sqrt{(x+1)^2 + y^2}\right) + i(y + 1) = 0
For a complex number to equal zero, both its real and imaginary parts must be zero.
Imaginary part: y + 1 = 0 \implies y = -1
Real part: x + \sqrt{2}\sqrt{(x+1)^2 + y^2} = 0 \implies \sqrt{2}\sqrt{(x+1)^2 + y^2} = -x
Substitute y = -1 into the real part equation.
\sqrt{2}\sqrt{(x+1)^2 + 1} = -x
(Important logic check: A principal square root is non-negative. Therefore, for the equation to hold, the right side must also be non-negative, meaning x \le 0.)
Square both sides to eliminate the radical and solve for x.
2((x+1)^2 + 1) = x^2
2(x^2 + 2x + 2) = x^2 \implies 2x^2 + 4x + 4 = x^2
x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2
Verify against the domain constraint x \le 0.
Since -2 \le 0 is true, the solution is valid.
Solution: z = -2 - i
Q11. Let x and y be real numbers such that 50(\frac{2x}{1+3i}-\frac{y}{1-2i})=31+17i then find the value of 10(x-3y).
Soution:
Rationalize the individual fractions inside the parentheses.
\frac{2x}{1+3i} = \frac{2x(1-3i)}{1^2 + 3^2} = \frac{2x(1-3i)}{10} = \frac{x(1-3i)}{5}
\frac{y}{1-2i} = \frac{y(1+2i)}{1^2 + 2^2} = \frac{y(1+2i)}{5}
Substitute these simplified fractions back into the parent equation.
50 \left[ \frac{x - 3ix - y - 2iy}{5} \right] = 31 + 17i
10 [ (x-y) - i(3x+2y) ] = 31 + 17i
Equate the real parts and imaginary parts to create a system of equations.
Real: 10(x-y) = 31 \implies x - y = 3.1
Imaginary: -10(3x+2y) = 17 \implies 3x + 2y = -1.7
Solve the linear system.
Multiply the first equation by 2.
2x - 2y = 6.2
Add this result to the second equation
(3x + 2y = -1.7).
5x = 4.5 \implies x = 0.9
Substitute x back into the first equation to find y.
0.9 - y = 3.1 \implies y = -2.2
Calculate the final target expression 10(x - 3y).
10(0.9 - 3(-2.2)) = 10(0.9 + 6.6) = 10(7.5) = 75
Final Answer: 75
Class 11 Complex Numbers and Linear Inequality Test-1 2027 ⇒