# Complex Numbers and Quadratic Equations - Class 11 - Mathematics

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## Extra Questions - Complex Numbers and Quadratic Equations | NCERT | Mathematics | Class 11

**Question:**

Calculate the value of $\frac{2^5}{2^8} \times 2^{-7}$.

A) $2^{10}$

B) $2^6$

C) $2^{-10}$

D) $2^{-4}$

To find the value of $ \frac{2^5}{2^8} \times 2^{-7} $, we multiply the given numbers:

$$ \frac{2^5}{2^8} \times 2^{-7} = \frac{2^{-2}}{2^8} = 2^{-10} $$

**Hence, the correct answer is option C.**

If $|z - 3| \leq 2$, then the maximum of $|z + 1|$ is

A) 4

B) 6

C) 2

D) 5

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$\frac{y}{y-2} + \frac{1}{y+2} + \frac{3y}{4-y^{2}} =$

A) $\frac{y^{2}-2}{y^{2}-4}$

B) $\frac{y^{2}-2}{y^{2}+2}$

C) $\frac{y+2}{y-2}$

D) $\frac{1}{y+2}$

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If $z+z^{3}=0$, then which of the following must be true on the complex plane?

(A) $\operatorname{Re}(z)<0$ (B) $\operatorname{Re}(z)=0$ (C) $\operatorname{Im}(z)=0$ (D) $z^{4}=1$

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The value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5\pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$ is equal to

(A) $1 / 4$

(B) $1 / 6$

(C) $1 / 8$

(D) $1 / 2$

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The value of $\left((1+i)^{2} \times (1-i)^{2}\right)$ is

(A) -8

(B) $8i$ (C) 8

(D) 32

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Find the range of $f(x) = \frac{x^{2}-x+1}{x^{2}+x+1}$

A $(-\infty, \frac{1}{3})$

B $(3, \infty)$

C $[\frac{1}{3}, 3]$

D $[-1,1]$