If z=a+ib lies in third quadrant, then
Question: If $z=a+i b$ lies in third quadrant, then $\frac{\bar{z}}{z}$ also lies in third quadrant if (a) $ab0$ (b) $ab0$ (c) $ba0$ (d) $ba0$ Solution: Since, $z=a+i b$ lies in third quadrant. $\Rightarrow a0$ and $b0$ Now, $\frac{\bar{z}}{z}=\frac{\overline{a+i b}}{a+i b}$ $=\frac{a-i b}{a+i b}$ $=\frac{a-i b}{a+i b} \times \frac{a-i b}{a-i b}$ $=\frac{a^{2}+i^{2} b^{2}-2 a b i}{a^{2}-i^{2} b^{2}}$ $=\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}$ Since, $\frac{\bar{z}}{z}$ also lies in third quadran...
Read More →If z=a+ib lies in third quadrant, then
Question: If $z=a+i b$ lies in third quadrant, then $\frac{\bar{z}}{z}$ also lies in third quadrant if (a) $ab0$ (b) $ab0$ (c) $ba0$ (d) $ba0$ Solution: Since, $z=a+i b$ lies in third quadrant. $\Rightarrow a0$ and $b0$ Now, $\frac{\bar{z}}{z}=\frac{\overline{a+i b}}{a+i b}$ $=\frac{a-i b}{a+i b}$ $=\frac{a-i b}{a+i b} \times \frac{a-i b}{a-i b}$ $=\frac{a^{2}+i^{2} b^{2}-2 a b i}{a^{2}-i^{2} b^{2}}$ $=\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}$ Since, $\frac{\bar{z}}{z}$ also lies in third quadran...
Read More →Solve this
Question: (a) $\sqrt[3]{y}+4$ (b) $\sqrt{y}-3$ (c) $y$ (d) $\frac{1}{\sqrt{y}}+7$ Solution: (c) $y$ y is a polynomial because it has a non-negative integral power 1....
Read More →Which of the following expression is a polynomial?
Question: Which of the following expression is a polynomial? (a) $\sqrt{x}-1$ (b) $\frac{x-1}{x+1}$\ (c) $x^{2}-\frac{2}{x^{2}}+5$ (d) $x^{2}+\frac{2 x^{3 / 2}}{\sqrt{x}}+6$ Solution: (d) $x^{2}+\frac{2 x^{3 / 2}}{\sqrt{x}}+6$ We have: $x^{2}+\frac{2 x^{\frac{3}{2}}}{\sqrt{x}}+6=x^{2}+2 x^{\frac{3}{2}} x^{-\frac{1}{2}}+6$ $=x^{2}+2 x+6$ It a polynomial because it has only non-negative integral powers ofx....
Read More →If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40,
Question: If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by(a) 2(b) 1.5(c) 1(d) 0.5 Solution: The given data set is $30,34,35,36,37,38,39,40$ $n=8$ (even) median $=\frac{\left(\frac{n}{2}\right)^{\text {th }} \operatorname{term}+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }}{2}$ $=\frac{4^{\text {th }} \text { term }+5^{\text {th }} \text { term }}{2}$ $=\frac{36+37}{2}$ $=\frac{73}{2}$ median $=36.5$ If 35 is removed, then the new data set is...
Read More →The value of
Question: The value of $(1+i)^{4}+(1-i)^{4}$ is (a) 8 (b) 4 (c) 8 (d) 4 Solution: (c) 8 Using $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}$ $(1+i)^{4}+(1-i)^{4}$ $=\left((1+i)^{2}+(1-i)^{2}\right)^{2}-2(1+i)^{2}(1-i)^{2}$ $=\left(1+i^{2}+2 i+1+i^{2}-2 i\right)^{2}-2\left(1+i^{2}+2 i\right)\left(1+i^{2}-2 i\right)$ $=(1-1+2 i+1-1-2 i)^{2}-2(1-1+2 i)(1-1-2 i)$ $=(0)-2(2 i)(-2 i) \quad\left(\because i^{2}=-1\right)$ $=8 i^{2}$ $=-8$...
Read More →The value of
Question: The value of $(1+i)^{4}+(1-i)^{4}$ is (a) 8 (b) 4 (c) 8 (d) 4 Solution: (c) 8 Using $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}$ $(1+i)^{4}+(1-i)^{4}$ $=\left((1+i)^{2}+(1-i)^{2}\right)^{2}-2(1+i)^{2}(1-i)^{2}$ $=\left(1+i^{2}+2 i+1+i^{2}-2 i\right)^{2}-2\left(1+i^{2}+2 i\right)\left(1+i^{2}-2 i\right)$ $=(1-1+2 i+1-1-2 i)^{2}-2(1-1+2 i)(1-1-2 i)$ $=(0)-2(2 i)(-2 i) \quad\left(\because i^{2}=-1\right)$ $=8 i^{2}$ $=-8$...
Read More →The value of
Question: The value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1$ is (a) 1 (b) 2 (c) 3 (d) 4 Solution: (b) 2 $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1$ $=\frac{i^{4 \times 148}+i^{4 \times 147+2}+i^{4 \times 147}+i^{4 \times 146+2}+i^{4 \times 16} 16}{i^{4} \times 15+2}+i^{\times 1 \times 45}+i^{4 \times 144+2}+i^{4 \times 144}+i^{4 \times 143+2}-1 \quad\left[\because i^{4}=1\right.$ and $\left.i^{2...
Read More →Which of the following expressions is a polynomial in one variable?
Question: Which of the following expressions is a polynomial in one variable? (a) $x+\frac{2}{x}+3$ (b) $3 \sqrt{x}+\frac{2}{\sqrt{x}}+5$ (c) $\sqrt{2} x^{2}-\sqrt{3} x+6$ (d) $x^{10}+y^{5}+8$ Solution: (c) $\sqrt{2} x^{2}-\sqrt{3} x+6$ Clearly, $\sqrt{2} x^{2}-\sqrt{3} x+6$ is a polynomial in one variable because it has only non-negative integral powers of $x$....
Read More →The value of
Question: The value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1$ is (a) 1 (b) 2 (c) 3 (d) 4 Solution: (b) 2 $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1$ $=\frac{i^{4 \times 148}+i^{4 \times 147+2}+i^{4 \times 147}+i^{4 \times 146+2}+i^{4 \times 16} 16}{i^{4} \times 15+2}+i^{\times 1 \times 45}+i^{4 \times 144+2}+i^{4 \times 144}+i^{4 \times 143+2}-1 \quad\left[\because i^{4}=1\right.$ and $\left.i^{2...
Read More →Solve the following
Question: $\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$ equals (a)i (b) 1 (c) i (d) 4 Solution: (c) i Let $z=\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$ $\Rightarrow z=\frac{1+2 i-3}{1-2 i-3}$ $\Rightarrow z=\frac{-2+2 i}{-2-2 i} \times \frac{-2+2 i}{-2+2 i}$ $\Rightarrow z=\frac{(-2+2 i)^{2}}{(-2)^{2}-(2 i)^{2}}$ $\Rightarrow z=\frac{4+4 i^{2}-8 i}{4+4}$ $\Rightarrow z=\frac{4-4-8 i}{8}$ $\Rightarrow z=\frac{-8 i}{8}$ $\Rightarrow z=-i$...
Read More →Use factor theorem to prove that
Question: Use factor theorem to prove that $(x+a)$ is a factor of $\left(x^{n}+a^{n}\right)$ for any odd positive integer. Solution: Let $f(x)=x^{n}+a^{n}$ Putting $x=-a$ in $f(x)$, we get $f(-a)=(-a)^{n}+a^{n}$ Ifnis any odd positive integer, then $f(-a)=(-a)^{n}+a^{n}=-a^{n}+a^{n}=0$ Therefore, by factor theorem, $(x+a)$ is a factor of $\left(x^{n}+a^{n}\right)$ for any odd positive integer....
Read More →Solve the following
Question: $\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$ equals (a)i (b) 1 (c) i (d) 4 Solution: (c) i Let $z=\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$ $\Rightarrow z=\frac{1+2 i-3}{1-2 i-3}$ $\Rightarrow z=\frac{-2+2 i}{-2-2 i} \times \frac{-2+2 i}{-2+2 i}$ $\Rightarrow z=\frac{(-2+2 i)^{2}}{(-2)^{2}-(2 i)^{2}}$ $\Rightarrow z=\frac{4+4 i^{2}-8 i}{4+4}$ $\Rightarrow z=\frac{4-4-8 i}{8}$ $\Rightarrow z=\frac{-8 i}{8}$ $\Rightarrow z=-i$...
Read More →What must be subtracted from
Question: What must be subtracted from $\left(x^{4}+2 x^{3}-2 x^{2}+4 x+6\right)$ so that the result is exactly divisible by $\left(x^{2}+2 x-3\right) ?$ Solution: Dividing $\left(x^{4}+2 x^{3}-2 x^{2}+4 x+6\right)$ by $\left(x^{2}+2 x-3\right)$ using long division method, we have Here, theremainder obtained is (2x+ 9). Thus, the remainder $(2 x+9)$ must be subtracted from $\left(x^{4}+2 x^{3}-2 x^{2}+4 x+6\right)$ so that the result is exactly divisible by $\left(x^{2}+2 x-3\right)$....
Read More →Solve the following
Question: The value of $\left(i^{5}+i^{6}+i^{7}+i^{8}+i^{9}\right) /(1+i)$ is (a) $\frac{1}{2}(1+i)$ (b) $\frac{1}{a}(1-i)$ (c) 1 (d) $\frac{1}{2}$ Solution: (a) $\frac{1}{2}(1+i)$ $\frac{i^{5}+i^{6}+i^{7}+i^{8}+i^{9}}{1+i}$ $\left[\right.$ As, $\left.i^{5}=i, i^{6}=-1, i^{7}=-i, i^{8}=1, i^{9}=i\right]$ $=\frac{i-1-i+1+i}{1+i}$ $=\frac{i}{i+1}$ $=\frac{i}{i+1} \times \frac{i-1}{i-1}$ $=\frac{i(i-1)}{i^{2}-1}$ $=\frac{i^{2}-i}{-2}$ $=\frac{1}{2}(1+i)$...
Read More →The amplitude of
Question: The amplitude of $\frac{1+i \sqrt{3}}{\sqrt{3}+i}$ is (a) $\frac{\pi}{3}$ (b) $-\frac{\pi}{3}$ (a) $\frac{\pi}{3}$ (b) $-\frac{\pi}{3}$ (c) $\frac{\pi}{6}$ (d) $-\frac{\pi}{6}$ Solution: (c) $\frac{\pi}{6}$ Let $z=\frac{1+i \sqrt{3}}{\sqrt{3}+i}$ $\Rightarrow z=\frac{1+i \sqrt{3}}{\sqrt{3}+i} \times \frac{\sqrt{3}-i}{\sqrt{3}-i}$ $\Rightarrow z=\frac{\sqrt{3}+2 i-\sqrt{3} i^{2}}{3-i^{2}}$ $\Rightarrow z=\frac{\sqrt{3}+\sqrt{3}+2 i}{4}$ $\Rightarrow z=\frac{2 \sqrt{3}+2 i}{4}$ $\Rightar...
Read More →Mean of a certain number of observation is x.
Question: Mean of a certain number of observation is $\bar{x}$. If each observation is divided by $m(m \neq 0)$ and increased by $n$, then the mean of new observation is (a) $\frac{\bar{x}}{\frac{m}{x}}+n$ (b) $\frac{\frac{m}{x}}{n}+m$ (c) $\frac{n}{x}+\frac{n}{m}$ (d) $\bar{x}+\frac{m}{n}$ Solution: Let $y_{1}, y_{2}, y_{3}, \ldots, y_{k}$ be $k$ observations. Mean of the observations $=\bar{x}$ $\Rightarrow \frac{y_{1}+y_{2}+y_{3}+\ldots+y_{k}}{k}=\bar{x}$ $\Rightarrow y_{1}+y_{2}+y_{3}+\ldots...
Read More →What must be added to
Question: What must be added to $2 x^{4}-5 x^{3}+2 x^{2}-x-3$ so that the result is exactly divisible by $(x-2) ?$ Solution: Let $k$ be added to $2 x^{4}-5 x^{3}+2 x^{2}-x-3$ so that the result is exactly divisible by $(x-2)$. Here, $k$ is a constant. $\therefore f(x)=2 x^{4}-5 x^{3}+2 x^{2}-x-3+k$ is exactly divisible by $(x-2)$. Using factor theorem, we have $f(2)=0$ $\Rightarrow 2 \times 2^{4}-5 \times 2^{3}+2 \times 2^{2}-2-3+k=0$ $\Rightarrow 32-40+8-5+k=0$ $\Rightarrow-5+k=0$ $\Rightarrow ...
Read More →The argument of
Question: The argument of $\frac{1-i}{1+i}$ is (a) $-\frac{\pi}{2}$ (b) $\frac{\pi}{2}$ (c) $\frac{3 \pi}{2}$ (d) $\frac{5 \pi}{2}$ Solution: (a) $-\frac{\pi}{2}$ Let $z=\frac{1-i}{1+i}$ $\Rightarrow z=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$ $\Rightarrow z=\frac{1+i^{2}-2 i}{1-i^{2}}$ $\Rightarrow z=\frac{1-1-2 i}{1+1}$ $\Rightarrow z=\frac{-2 i}{2}$ $\Rightarrow z=-i$ Since, $z$ lies on negative direction of imaginary axis. Therefore, $\arg (z)=\frac{-\pi}{2}$...
Read More →The amplitude of
Question: The amplitude of $\frac{1}{i}$ is equal to (a) 0 (b) $\frac{\pi}{2}$ (c) $-\frac{\pi}{2}$ (d) $\pi$ Solution: (c) $-\frac{\pi}{2}$ Let $z=\frac{1}{i}$ $\Rightarrow z=\frac{1}{i} \times \frac{i}{i}$ $\Rightarrow z=\frac{i}{i^{2}}$ $\Rightarrow z=-i$ Since, $z(0,-1)$ lies on the negative imaginary axis. Therefore, $\arg (z)=\frac{-\pi}{2}$...
Read More →Without actual division, prove that
Question: Without actual division, prove that $2 x^{4}-5 x^{3}+2 x^{2}-x+2$ is divisible by $x^{2}-3 x+2$ Solution: Let $f(x)=2 x^{4}-5 x^{3}+2 x^{2}-x+2$ and $g(x)=x^{2}-3 x+2$ $x^{2}-3 x+2$ $=x^{2}-2 x-x+2$ $=x(x-2)-1(x-2)$ $=(x-1)(x-2)$ Now,f(x) will be divisible byg(x) iff(x) is exactly divisible by both (x 1) and(x 2).Puttingx= 1 inf(x), we get $f(1)=2 \times 1^{4}-5 \times 1^{3}+2 \times 1^{2}-1+2=2-5+2-1+2=0$ By factor theorem,(x2)is a factor off(x). So,f(x) is exactly divisible by (x2).T...
Read More →Solve the following
Question: If $z=\frac{1+7 i}{(2-i)^{2}}$, then (a) $|z|=2$ (b) $|z|=\frac{1}{2}$ (c) $\operatorname{amp}(z)=\frac{\pi}{4}$ (d) $\operatorname{amp}(z)=\frac{3 \pi}{4}$ Solution: (d) $\operatorname{amp}(z)=\frac{3 \pi}{4}$ $z=\frac{1+7 i}{(2-i)^{2}}$ $\Rightarrow z=\frac{1+7 i}{4+i^{2}-4 i}$ $\Rightarrow z=\frac{1+7 i}{4-1-4 i}$ $\left[\because i^{2}=-1\right]$ $\Rightarrow z=\frac{1+7 i}{3-4 i}$ $\Rightarrow z=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}$ $\Rightarrow z=\frac{3+4 i+21 i+28 i^{2...
Read More →If the mean of observation x1, x2,
Question: If the mean of observation $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$, then the mean of $x_{1}+a, x_{2}+a_{1} \ldots \ldots, x_{n}+a$ is (a) $a \bar{x}$ (b) $\bar{x}-a$ (c) $\bar{x}+a$ (d) $\frac{\bar{x}}{a}$ Solution: The mean of $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. $\therefore \frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\bar{x}$ $\Rightarrow x_{1}+x_{2}+x_{3}+\ldots+x_{n}=n \bar{x}$ Mean of $x_{1}+a_{1} x_{2}+a, \ldots, x_{n}+a$ $=\frac{\left(x_{1}+a\right)+\left(x_{2}+a\right)+\...
Read More →If θ is the amplitude of
Question: If $\theta$ is the amplitude of $\frac{a+i b}{a-i b}$, than $\tan \theta=$ (a) $\frac{2 a}{a^{2}+b^{2}}$ (b) $\frac{2 a b}{a^{2}-b^{2}}$ (c) $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$ (d) none of these Solution: (b) $\frac{2 a b}{a^{2}-b^{2}}$ $z=\frac{a+i b}{a-i b} \times \frac{a+i b}{a+i b}$ $\Rightarrow z=\frac{a^{2}+i^{2} b^{2}+2 a b i}{a^{2}-i^{2} b^{2}}$ $\Rightarrow z=\frac{a^{2}-b^{2}+2 a b i}{a^{2}+b^{2}}$ $\Rightarrow z=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}+i \frac{2 a b}{a^{2}+b^{2}}$ $\Ri...
Read More →If the mean of first n natural number is 15, then n =
Question: If the mean of firstnnatural number is 15, thenn=(a) 15(b) 30(c) 14(d) 29 Solution: Given:Mean of firstnnatural numbers = 15 $\Rightarrow \frac{1+2+3+\ldots+n}{n}=15$ $\Rightarrow \frac{\frac{n(n+1)}{2}}{n}=15$ $\Rightarrow \frac{n+1}{2}=15$ $\Rightarrow n+1=30$ $\Rightarrow n=29$ Hence, the correct option is (d)....
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