If a = 1 + i, then
Question: If $a=1+i$, then $a^{2}$ equals (a) 1 i (b) 2i (c) (1 +i) (1 i) (d)i 1 Solution: (b) 2i a= 1 +i On squaring both the sides, we get, $a^{2}=(1+i)^{2}$ $\Rightarrow a^{2}=1+i^{2}+2 i$ $\Rightarrow a^{2}=1-1+2 i \quad\left(\because i^{2}=-1\right)$ $\Rightarrow a^{2}=2 i$...
Read More →If z is a non-zero complex number, then
Question: If $z$ is a non-zero complex number, then $\left|\frac{|\bar{z}|^{2}}{z \bar{z}}\right|$ is equal to Solution: (a) $\left|\frac{\bar{z}}{z}\right|$ $\left|\frac{|\bar{z}|^{2}}{z \bar{z}}\right|=\left|\frac{|\bar{z}|^{2}}{|z|^{2}}\right| \quad\left(\because z \bar{z}=|z|^{2}\right)$ Let $z=a+i b$ $\Rightarrow|z|=\sqrt{a^{2}+b^{2}}$ Let $\bar{z}=a-i b$ $\Rightarrow|\bar{z}|=\sqrt{a^{2}+b^{2}}$ $\therefore\left|\frac{|\bar{z}|^{2}}{z \bar{z}}\right|=\left|\frac{|\bar{z}|^{2}}{|z|^{2}}\rig...
Read More →If the mean of 6, 7, x, 8, y, 14 is 9, then
Question: If the mean of 6, 7,x, 8,y, 14 is 9, then (a)x+y= 21(b)x+y= 19(c)xy= 19(d)xy= 21 Solution: The given observations are 6, 7,x, 8,y, 14. Mean = 9 (Given) $\Rightarrow \frac{6+7+x+8+y+14}{6}=9$ $\Rightarrow 35+x+y=54$ $\Rightarrow x+y=54-35=19$ Hence, the correct option is (b)....
Read More →If the mean of frequency distribution is 8.1
Question: If the mean of frequency distribution is $8.1$ and $\Sigma f_{i} x_{i}=132+5 k, \Sigma f_{i}=20$, then $k=$ (a) 3(b) 4(c) 5(d) 6 Solution: Given: $\sum f_{i} x_{i}=132+5 k, \Sigma f_{i}=20$ and mean $=8.1$ Then, Mean $=\frac{\sum f_{i} x_{i}}{\sum f_{i}}$ $8.1=\frac{132+5 k}{20}$ $162=132+5 k$ $5 k=30$ $k=6$ Hence, the correct option is (d)....
Read More →The least positive integer n such that
Question: The least positive integer $n$ such that $\left(\frac{2 i}{1+i}\right)^{n}$ is a positive integer, is (a) 16 (b) 8 (c) 4 (d) 2 Solution: (b) 8 Let $z=\left(\frac{2 i}{1+i}\right)$ $\Rightarrow z=\frac{2 i}{1+i} \times \frac{1-i}{1-i}$ $\Rightarrow z=\frac{2 i(1-i)}{1-i^{2}}$ $\Rightarrow z=\frac{2 i(1-i)}{1+1} \quad\left[\because i^{2}=-1\right]$ $\Rightarrow z=\frac{2 i(1-i)}{2}$ $\Rightarrow z=i-i^{2}$ $\Rightarrow z=i+1$ Now, $z^{n}=(1+i)^{n}$ For $n=2$, $z^{2}=(1+i)^{2}$ $=1+i^{2}+...
Read More →The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3,
Question: The mean of 1, 3, 4, 5, 7, 4 ism. The numbers 3, 2, 2, 4, 3, 3,phave meanm 1 and medianq. Then,p+q=(a) 4(b) 5(c) 6(d) 7 Solution: $1,3,4,5,7,4$ Mean $=\frac{1+3+4+5+7+4}{6}$ $=\frac{24}{6}$ $=4$ $M=4$ Consider the numbers 3, 2, 2, 4, 3, 3,p. Mean $=\frac{3+2+3+4+3+3+p}{7}$ $\Rightarrow 7 \times(4-1)=17+p$ $\Rightarrow 21=17+p$ $\Rightarrow p=4$ Arranging the numbers 3, 2, 2, 4, 3, 3, 4 in ascending order, we have 2, 2, 3, 3, 3, 4, 4 Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ te...
Read More →The least positive integer n such that
Question: The least positive integer $n$ such that $\left(\frac{2 i}{1+i}\right)^{n}$ is a positive integer, is (a) 16 (b) 8 (c) 4 (d) 2 Solution: (b) 8 Let $z=\left(\frac{2 i}{1+i}\right)$ $\Rightarrow z=\frac{2 i}{1+i} \times \frac{1-i}{1-i}$ $\Rightarrow z=\frac{2 i(1-i)}{1-i^{2}}$ $\Rightarrow z=\frac{2 i(1-i)}{1+1} \quad\left[\because i^{2}=-1\right]$ $\Rightarrow z=\frac{2 i(1-i)}{2}$ $\Rightarrow z=i-i^{2}$ $\Rightarrow z=i+1$ Now, $z^{n}=(1+i)^{n}$ For $n=2$, $z^{2}=(1+i)^{2}$ $=1+i^{2}+...
Read More →If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
Question: If the mode of the data: 16, 15, 17, 16, 15,x, 19, 17, 14 is 15, thenx= (a) 15(b) 16(c) 17(d) 19 Solution: It is given that the mode of the data is 15. So, it is the observation with the maximum frequency.This is possible only whenx= 15. In this case, the frequency of 15 would be 3. Hence, the correct answer is (a)....
Read More →If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
Question: If the mode of the data: 64, 60, 48,x, 43, 48, 43, 34 is 43, thenx+ 3 = (a) 44(b) 45(c) 46(d) 48 Solution: It is given that the mode of the given date is 43. So, it is the value with the maximum frequency.Now, this is possible only whenx= 43. In this case, the frequency of the observation 43 would be 3.Hence,x+ 3 = 46Hence, the correct option is (c)....
Read More →The median of first 10 prime numbers is
Question: The median of first 10 prime numbers is (a) 11(b) 12(c) 13(d) 14 Solution: First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. n= 10 (even) $\operatorname{Median}=\frac{\left(\frac{n}{2}\right)^{\text {th }} \operatorname{term}+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }}{2}$ $=\frac{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }}{2}$ $=\frac{11+13}{2}$ $=\frac{24}{2}$ $=12$ Hence, the correct option is (b)....
Read More →The principal value of the amplitude of (1 + i) is
Question: The principal value of the amplitude of (1 +i) is (a) $\frac{\pi}{4}$ (b) $\frac{\pi}{12}$ (c) $\frac{3 \pi}{4}$ (d) $\pi$ Solution: (a) $\frac{\pi}{4}$ Let $z=(1+i)$ $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ = 1 $\Rightarrow \alpha=\frac{\pi}{4}$ Since, $z$ lies in the first quadrant. Therefore, $\arg (z)=\frac{\pi}{4}$...
Read More →The principal value of the amplitude of (1 + i) is
Question: The principal value of the amplitude of (1 +i) is (a) $\frac{\pi}{4}$ (b) $\frac{\pi}{12}$ (c) $\frac{3 \pi}{4}$ (d) $\pi$ Solution: (a) $\frac{\pi}{4}$ Let $z=(1+i)$ $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ = 1 $\Rightarrow \alpha=\frac{\pi}{4}$ Since, $z$ lies in the first quadrant. Therefore, $\arg (z)=\frac{\pi}{4}$...
Read More →If the median of the data: 6, 7, x − 2, x, 17, 20, written in ascending order, is 16. Then x =
Question: If the median of the data: 6, 7,x 2,x, 17, 20, written in ascending order, is 16. Thenx= (a) 15(b) 16(c) 17(d) 18 Solution: The given observations arranged in ascending order are $6,7, x-2, x, 17,20$ $n=6($ even $)$, median $=16$ Median $=\frac{\left(\frac{n}{2}\right)^{\text {th }} \operatorname{term}+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }}{2}$ $=\frac{3 \text { rd term }+4 \text { th term }}{2}$ $=\frac{x-2+x}{2}$ $=\frac{2 x-2}{2}$ $16=\frac{2 x-2}{2}$ $2 x-2=32$ $x...
Read More →Solve the following
Question: If $\frac{\left(a^{2}+1\right)^{2}}{2 a-i}=x+i y$, then $x^{2}+y^{2}$ is equal to (a) $\frac{\left(a^{2}+1\right)^{4}}{4 a^{2}+1}$ (b) $\frac{(a+1)^{2}}{4 a^{2}+1}$ (c) $\frac{\left(a^{2}-1\right)^{2}}{\left(4 a^{2}-1\right)^{2}}$ (d) none of these Solution: (a) $\frac{\left(a^{2}+1\right)^{4}}{4 a^{2}+1}$ $x+i y=\frac{\left(a^{2}+1\right)^{2}}{2 a-i}$ Taking modulus on both the sides, we get: $\sqrt{x^{2}+y^{2}}=\frac{\left(a^{2}+1\right)^{2}}{\sqrt{4 a^{2}+1}}$ Squaring both sides, w...
Read More →If the median of the data: 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =
Question: If the median of the data: 24, 25, 26,x+ 2,x+ 3, 30, 31, 34 is 27.5, thenx=(a) 27(b) 25(c) 28(d) 30 Solution: The given observations are 24, 25, 26,x+ 2,x+ 3, 30, 31, 34. Median = 27.5 Here,n= 8 Median $=\frac{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }}{2}$ $27.5=\frac{4 \text { th term }+5 \text { th term }}{2}$ $27.5=\frac{(x+2)+(x+3)}{2}$ $27.5=\frac{2 x+5}{2}$ $2 x+5=55$ $2 x=50$ $x=25$ Hence, the correct option is ...
Read More →If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib,
Question: If $(1+i)(1+2 i)(1+3 i) \ldots(1+n i)=a+i b$, then $2.5 .10 .17 \ldots \ldots . .\left(1+n^{2}\right)=$ (a)aib (b)a2b2 (c)a2+b2 (d) none of these Solution: (c) $a^{2}+b^{2}$ (1 +i)(1 + 2i)(1 + 3i) ......(1 +ni) =a+ib Taking modulus on both the sides, we get, $|(1+i)(1+2 i)(1+3 i) \ldots \ldots(1+n i)|=a+i b$ $|(1+i)(1+2 i)(1+3 i) \ldots \ldots(1+n i)|$ can be wriiten as $|(1+i)||(1+2 i)||(1+3 i)| \ldots|(1+n i)|$ $\therefore \sqrt{1^{2}+1^{2}} \times \sqrt{1^{2}+2^{2}} \times \sqrt{1^{...
Read More →If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, the x =
Question: If the arithmetic mean ofx,x+ 3,x+ 6,x+ 9, andx+ 12 is 10, thex= (a) 1(b) 2(c) 6(d) 4 Solution: The given observations arex,x+ 3,x+ 6,x+ 9, andx+ 12. $\therefore \sum x=5 x+30, n=5, \bar{X}=10$ Now, $\bar{x}=\frac{\sum x}{n}$ $10=\frac{5 x+30}{5}$ $50=5 x+30$ $5 x=20$ $x=4$ Hence, the correct option is (d)....
Read More →If a = cos θ + i sin θ, then
Question: If $a=\cos \theta+i \sin \theta$, then $\frac{1+a}{1-a}=$ (a) $\cot \frac{\theta}{2}$ (b) $\cot \theta$ (c) $i \cot \frac{\theta}{2}$ (d) $i \tan \frac{\theta}{2}$ Solution: (c) $i \cot \frac{\theta}{2}$ $a=\cos \theta+i \sin \theta \quad$ (given ) $\Rightarrow \frac{1+a}{1-a}=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}$ $\Rightarrow \frac{1+a}{1-a}=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta} \times \frac{1-\cos \theta+i \sin \theta}{1-\cos \thet...
Read More →The mean of a discrete frequency distribution xi / fi, i = 1, 2, ......, n is given by
Question: The mean of a discrete frequency distribution $x_{i} / f_{i}, i=1,2$, $n$ is given by (a) $\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$ (b) $\frac{1}{n} \sum_{n}^{n} f_{i=1} x_{i}$ (C) $\frac{\sum_{i=1} f_{i} x_{i}}{\sum_{i=1}^{n} x_{i}}$ (d) $\frac{\sum_{i=1}^{n} f_{i} x_{i}}{\sum_{1=1}^{n} i}$ Solution: The mean of discrete frequency distribution is $\bar{x}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}$ Hence, the correct option is (a)....
Read More →Solve the following
Question: If $z=\frac{-2}{1+i \sqrt{3}}$, then the value of arg ( $\mathrm{z}$ ) is (a) $\pi$ (b) $\frac{\pi}{3}$ (c) $\frac{2 \pi}{3}$ (d) $\frac{\pi}{4}$ Solution: (c) $\frac{2 \pi}{3}$ $z=\frac{-2}{1+i \sqrt{3}}$ Rationalisingz, we get, $z=\frac{-2}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$ $\Rightarrow z=\frac{-2+i 2 \sqrt{3}}{1+3}$ $\Rightarrow z=\frac{-1+i \sqrt{3}}{2}$ $\Rightarrow z=\frac{-1}{2}+\frac{i \sqrt{3}}{2}$ $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorn...
Read More →The relationship between mean, median and mode for a moderately skewed distribution is
Question: The relationship between mean, median and mode for a moderately skewed distribution is (a) Mode = 2 Median 3 Mean(b) Mode = Median 2 Mean(c) Mode = 2 Median Mean(d) Mode = 3 Median 2 Mean Solution: Mode = 3Median 2Mean Hence, the correct option is (d)....
Read More →If the mean of the following distribution is 2.6,
Question: If the mean of the following distribution is 2.6, then the value ofyis (a) 3(b) 8(c) 13(d) 24 Solution: Consider the following table. Now, mean $=\frac{\sum f x}{\sum f}$ $\frac{2.6}{1}=\frac{28+3 y}{12+y}$ $31.2+2.6 y=28+3 y$ $3 y-2.6 y=31.2-28$ $0.4 y=3.2$ $y=\frac{3.2}{04}$ $y=8$ Hence, the correct option is (b)....
Read More →Solve the following
Question: If $P^{2}=-1$, then the sum $i+\beta^{2}+\beta^{3}+\ldots$ upto 1000 terms is equal to (a) 1 (b) 1 (c)i (d) 0 Solution: (d) 0 $i+i^{2}+i^{3}+i^{4} \ldots i^{1000}$ $i+i^{2}+i^{3}+i^{4} \quad\left[\because i^{2}=-1, i^{3}=-i\right.$ and $\left.i^{4}=1\right]$ $=i-1-i+1$ $=0$ Similarly, the sum of the next four terms of the series will be equal to 0 . This is because the powers of $i$ follow a cyclicity of 4 . Hence, the sum of all terms, till 1000 , will be zero. $i+i^{2}+i^{3}+i^{4} \l...
Read More →Solve the following
Question: If $P^{2}=-1$, then the sum $i+\beta^{2}+\beta^{3}+\ldots$ upto 1000 terms is equal to (a) 1 (b) 1 (c)i (d) 0 Solution: (d) 0 $i+i^{2}+i^{3}+i^{4} \ldots i^{1000}$ $i+i^{2}+i^{3}+i^{4} \quad\left[\because i^{2}=-1, i^{3}=-i\right.$ and $\left.i^{4}=1\right]$ $=i-1-i+1$ $=0$ Similarly, the sum of the next four terms of the series will be equal to 0 . This is because the powers of $i$ follow a cyclicity of 4 . Hence, the sum of all terms, till 1000 , will be zero. $i+i^{2}+i^{3}+i^{4} \l...
Read More →One of the methods of determining mode is
Question: One of the methods of determining mode is(a) Mode = 2 Median 3 Mean(b) Mode = 2 Median + 3 Mean(c) Mode = 3 Median 2 Mean(d) Mode = 3 Median + 2 Mean Solution: We have, Mode = 3 Median 2 Mean Hence, the correct option is (c)....
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