Solve the following
Question: If $\frac{1-i x}{1+i x}=a+i b$, then $a^{2}+b^{2}=$ (a) 1 (b) 1 (c) 0 (d) none of these Solution: (a) 1 $\frac{1-i x}{1+i x}=a+i b$ Taking modulus on both the sides, we get: $\left|\frac{1-i x}{1+i x}\right|=|a+i b|$ $\Rightarrow \frac{\sqrt{1^{2}+x^{2}}}{\sqrt{1^{2}+x^{2}}}=\sqrt{a^{2}+b^{2}}$ $\Rightarrow \sqrt{a^{2}+b^{2}}=1$ Squaring both the sides, we get: $a^{2}+b^{2}=1$...
Read More →If mode of a series exceeds its mean by 12,
Question: If mode of a series exceeds its mean by 12, then mode exceeds the median by(a) 4(b) 8(c) 6(d) 10 Solution: Given: Mode Mean = 12We know thatMode = 3Median 2Mean Mode Mean = 3(Median Mean)⇒ 12 = 3(Median Mean)⇒Median Mean = 4 .....(1)Again,Mode = 3Median 2Mean⇒ 2Mode = 6Median 4Mean⇒ Mode Mean + Mode = 6Median 5Mean⇒ 12 + (Mode Median) = 5(Median Mean)⇒12 + (Mode Median) = 20 [Using (1)]⇒ Mode Median = 20 12 = 8Hence, the correct option is (b)....
Read More →If the arithmetic mean, 7, 8, x, 11, 14 is x, then x =
Question: If the arithmetic mean, 7, 8,x, 11, 14 isx, thenx= (a) 9(b) 9.5(c) 10(d) 10.5 Solution: The given observations are 7, 8,x, 11, 14.Mean =x(Given)Now, Mean $=\frac{7+8+x+11+14}{5}$ $\Rightarrow x=\frac{40+x}{5}$ $\Rightarrow 5 x=40+x$ $\Rightarrow 4 x=40$ $\Rightarrow x=10$ Hence, the correct option is (c)....
Read More →Solve the following
Question: If $x+i y=\frac{3+5 i}{7-6 i}$, then $y=$ (a) 9/85 (b) 9/85 (c) 53/85 (d) none of these Solution: (c) $\frac{53}{85}$ $x+i y=\frac{3+5 i}{7-6 i}$ $\Rightarrow x+i y=\frac{3+5 i}{7-6 i} \times \frac{7+6 i}{7+6 i}$ $\Rightarrow x+i y=\frac{21+53 i+30 i^{2}}{49-36 i^{2}}$ $\Rightarrow x+i y=\frac{21-30+53 i}{49+36}$ $\Rightarrow x+i y=\frac{-9}{85}+i \frac{53}{85}$ On comparing both the sides : $y=\frac{53}{85}$...
Read More →If the difference of mode and median of a data is 24,
Question: If the difference of mode and median of a data is 24, then the difference of median and mean is(a) 12(b) 24(c) 8(d) 36 Solution: Given: Mode Median = 24We know thatMode = 3Median 2MeanNow,Mode Median = 2(Median Mean)⇒ 24 = 2(Median Mean)⇒ Median Mean = 12Hence, the correct option is (a)....
Read More →Solve the following
Question: If $z=\frac{1}{1-\cos \theta-i \sin \theta}$, then $\operatorname{Re}(z)=$ (a) 0 (b) $\frac{1}{2}$ (c) $\cot \frac{\theta}{2}$ (d) $\frac{1}{2} \cot \frac{\theta}{2}$ Solution: (b) $\frac{1}{2}$ $z=\frac{1}{1-\cos \theta-i \sin \theta}$ $z=\frac{1}{1-\cos \theta-i \sin \theta} \times \frac{1-\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}$ $\Rightarrow z=\frac{1-\cos \theta+i \sin \theta}{(1-\cos \theta)^{2}-(i \sin \theta)^{2}}$ $\Rightarrow z=\frac{1-\cos \theta+i \sin \theta...
Read More →Find the values of a and b so that the polynomial
Question: Find the values of $a$ and $b$ so that the polynomial $\left(x^{4}+a x^{3}-7 x^{2}-8 x+b\right)$ is exactly divisible by $(x+2)$ as well as $(x+3)$. Solution: Let: $f(x)=x^{4}+a x^{3}-7 x^{2}-8 x+b$ Now, $x+2=0 \Rightarrow x=-2$ By the factor theorem, we can say: $f(x)$ will be exactly divisible by $(x+2)$ if $f(-2)=0$. Thus, we have: $f(-2)=\left[(-2)^{4}+a \times(-2)^{3}-7 \times(-2)^{2}-8 \times(-2)+b\right]$ $=(16-8 a-28+16+b)$ $=(4-8 a+b)$ $\therefore f(-2)=0 \Rightarrow 8 a-b=4 \...
Read More →The mean of first n odd natural numbers is
Question: The mean of first $n$ odd natural numbers is $\frac{n^{2}}{81}$, then $n=$ (a) 9(b) 81(c) 27(d) 18 Solution: The firstnodd natural numbers are 1, 3, 5, ... , (2n 1). $\therefore$ Mean of first $n$ odd natural numbers $=\frac{1+3+5+\ldots+(2 n-1)}{n}$ $=\frac{\frac{n}{2}(1+2 n-1)}{n} \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$ $=\frac{2 n}{2}$ $=n$ Now, Mean of first $n$ odd natural numbers $=\frac{n^{2}}{81}$ (Given) $\therefore n=\frac{n^{2}}{81}$ $\Rightarrow n=81$ Hence, the correct o...
Read More →If x+iy=(1+i)(1+2i)(1+3i),
Question: If $x+i y=(1+i)(1+2 i)(1+3 i)$, then $x^{2}+y^{2}=$ (a) 0 (b) 1 (c) 100 (d) none of these Solution: (c) 100 $\because x+i y=(1+i)(1+2 i)(1+3 i)$ Taking modulus on both the sides: $|x+i y|=|(1+i)(1+2 i)(1+3 i)|$ $\Rightarrow|x+i y|=|1+i| \times|1+2 i| \times|1+3 i|$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{1^{2}+1^{2}} \sqrt{1^{2}+2^{2}} \sqrt{1^{2}+3^{2}}$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{2} \sqrt{5} \sqrt{10}$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{100}$ Squaring both the sides, $\Ri...
Read More →If x+iy=(1+i)(1+2i)(1+3i),
Question: If $x+i y=(1+i)(1+2 i)(1+3 i)$, then $x^{2}+y^{2}=$ (a) 0 (b) 1 (c) 100 (d) none of these Solution: (c) 100 $\because x+i y=(1+i)(1+2 i)(1+3 i)$ Taking modulus on both the sides: $|x+i y|=|(1+i)(1+2 i)(1+3 i)|$ $\Rightarrow|x+i y|=|1+i| \times|1+2 i| \times|1+3 i|$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{1^{2}+1^{2}} \sqrt{1^{2}+2^{2}} \sqrt{1^{2}+3^{2}}$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{2} \sqrt{5} \sqrt{10}$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{100}$ Squaring both the sides, $\Ri...
Read More →The mean of first n odd natural number is
Question: The mean of firstnodd natural number is (a) $\frac{n+1}{2}$ (b) $\frac{n}{2}$ (c) $n$ (d) $n^{2}$ Solution: The firstnodd natural numbers are 1, 3, 5, ... , (2n 1). $\therefore$ Mean of first $n$ odd natural numbers $=\frac{1+3+5+\ldots+(2 n-1)}{n}$ $=\frac{\frac{n}{2}(1+2 n-1)}{n} \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$ $=\frac{2 n}{2}$ $=n$ Hence, the correct answer is (c)....
Read More →If z=1−cosθ+i sinθ,
Question: If $z=1-\cos \theta+i \sin \theta$, then $|z|=$ (a) $2 \sin \frac{\theta}{2}$ (b) $2 \cos \frac{\theta}{2}$ (c) $2\left|\sin \frac{\theta}{2}\right|$ (d) $2\left|\cos \frac{\theta}{2}\right|$ Solution: (c) $2\left|\sin \frac{\theta}{2}\right|$ $\because z=1-\cos \theta+i \sin \theta$ $\Rightarrow|z|=\sqrt{(1-\cos \theta)^{2}+\sin ^{2} \theta}$ $\Rightarrow|z|=\sqrt{1+\cos ^{2} \theta-2 \cos \theta+\sin ^{2} \theta}$ $\Rightarrow|z|=\sqrt{1+1-2 \cos \theta}$ $\Rightarrow|z|=\sqrt{2(1-\c...
Read More →The arithmetic mean and mode of a data
Question: The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is(a) 25(b) 18(c) 20(d) 22 Solution: Given: Mean = 24 and Mode = 12We know thatMode = 3Median 2Mean⇒ 12 = 3Median 2 24⇒ 3Median = 12 + 48 = 60⇒ Median = 20Hence, the correct option is (c)....
Read More →If the mean of first n natural numbers is
Question: If the mean of first $n$ natural numbers is $\frac{5 n}{9}$, then $n=$ (a) 5(b) 4(c) 9(d) 10 Solution: Given: Mean of first $n$ natural number $=\frac{5 n}{9}$ $\Rightarrow \frac{1+2+3+\ldots+n}{n}=\frac{5 n}{9}$ $\Rightarrow \frac{\frac{n(n+1)}{2}}{n}=\frac{5 n}{9}$ $\Rightarrow \frac{n+1}{2}=\frac{5 n}{9}$ $\Rightarrow 9 n+9=10 n$ $\Rightarrow n=9$ Hence, the correct option is (c)....
Read More →Solve the following
Question: If $z=\frac{1}{(1-i)(2+3 i)}$, than $|z|=$ (a) 1 (b) $1 / \sqrt{26}$ (c) $5 / \sqrt{26}$ (d) none of these Solution: (b) $1 / \sqrt{26}$ Let $z=\frac{1}{(1-i)(2+3 i)}$ $\Rightarrow z=\frac{1}{2+i-3 i^{2}}$ $\Rightarrow z=\frac{1}{2+i+3}$ $\Rightarrow z=\frac{1}{5+i} \times \frac{5-i}{5-i}$ $\Rightarrow z=\frac{5-i}{25-i^{2}}$ $\Rightarrow z=\frac{5-i}{25+1}$ $\Rightarrow z=\frac{5-i}{26}$ $\Rightarrow z=\frac{5}{26}-\frac{i}{26}$ $\Rightarrow|z|=\sqrt{\frac{25}{676}+\frac{1}{676}}$ $\R...
Read More →Solve the following
Question: If $z=\frac{1}{(2+3 i)^{2}}$, than $|z|=$ (a) $\frac{1}{13}$ (b) $\frac{1}{5}$ (c) $\frac{1}{12}$ (d) none of these Solution: (a) $\frac{1}{13}$ Let $z=\frac{1}{(2+3 i)^{2}}$ $\Rightarrow z=\frac{1}{4+9 i^{2}+12 i}$ $\Rightarrow z=\frac{1}{4-9+12 i}$ $\Rightarrow z=\frac{1}{-5+12 i}$ $\Rightarrow z=\frac{1}{-5+12 i} \times \frac{-5-12 i}{-5-12 i}$ $\Rightarrow z=\frac{-5-12 i}{25+144}$ $\Rightarrow z=\frac{-5}{169}-\frac{12 i}{169}$ $\Rightarrow|z|=\sqrt{\frac{25}{169^{2}}+\frac{144}{1...
Read More →Solve the following
Question: If $z=\frac{1+2 i}{1-(1-i)^{2}}$, then arg $(z)$ equal (a) 0 (b) $\frac{\pi}{2}$ (c) (d) none of these. Solution: (a) 0 Let $z=\frac{1+2 i}{1-(1-i)^{2}}$ $\Rightarrow z=\frac{1+2 i}{1-\left(1+i^{2}-2 i\right)}$ $\Rightarrow z=\frac{1+2 i}{1-(1-1-2 i)}$ $\Rightarrow z=\frac{1+2 i}{1+2 i}$ $\Rightarrow z=1$ Since point $(1,0)$ lies on the positive direction of real axis, we have: $\arg (z)=0$...
Read More →Solve the following
Question: If $z=\frac{1+2 i}{1-(1-i)^{2}}$, then arg $(z)$ equal (a) 0 (b) $\frac{\pi}{2}$ (c) (d) none of these. Solution: (a) 0 Let $z=\frac{1+2 i}{1-(1-i)^{2}}$ $\Rightarrow z=\frac{1+2 i}{1-\left(1+i^{2}-2 i\right)}$ $\Rightarrow z=\frac{1+2 i}{1-(1-1-2 i)}$ $\Rightarrow z=\frac{1+2 i}{1+2 i}$ $\Rightarrow z=1$ Since point $(1,0)$ lies on the positive direction of real axis, we have: $\arg (z)=0$...
Read More →Solve the following
Question: If $z=\left(\frac{1+i}{1-i}\right)$, then $z^{4}$ equals (a) 1 (b) 1 (c) 0 (d) none of these Solution: (a) 1 Let $z=\frac{1+i}{1-i}$ Rationalising the denominator: $z=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$ $\Rightarrow z=\frac{1+i^{2}+2 i}{1-i^{2}}$ $\Rightarrow z=\frac{1-1+2 i}{1+1}$ $\Rightarrow z=\frac{2 i}{2}$ $\Rightarrow z=i$ $\Rightarrow z^{4}=i^{4}$ Since $i^{2}=-1$, we have: $\Rightarrow z^{4}=i^{2} \times i^{2}$ $\Rightarrow z^{4}=1$...
Read More →The argument of
Question: The argument of $\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$ is (a) 60 (b) 120 (c) 210 (d) 240 Solution: (d) 240 $\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$ Rationalising the denominator, $\frac{1-i \sqrt{3}}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$ $=\frac{1+3 i^{2}-2 \sqrt{3} i}{1-3 i^{2}}$ $=\frac{-2-2 \sqrt{3} i}{4}$$\left(\because i^{2}=-1\right)$ $=\frac{-1}{2}-i \frac{\sqrt{3}}{2}$ $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ Then, $\tan \alpha=\lef...
Read More →Solve the following
Question: $(\sqrt{-2})(\sqrt{-3})$ is equal to (a) $\sqrt{6}$ (b) $-\sqrt{6}$ (c) $i \sqrt{6}$ (d) none of these. Solution: (b) $-\sqrt{6}$ $\sqrt{-2} \times \sqrt{-3}$ $=\sqrt{2} \times \sqrt{3} \times \sqrt{-1} \times \sqrt{-1}$ $=\sqrt{6} \times i \times i$ $=\sqrt{6} \times i^{2}$ $=-\sqrt{6}$ $\left[\because i^{2}=-1\right]$...
Read More →If (x + iy)1/3 = a + ib, then
Question: If $(x+i y) 1 / 3=a+i b$, then $\frac{x}{a}+\frac{y}{b}=$ (a) 0 (b) 1 (c) 1 (d) none of these Solution: (d) none of these $(x+i y)^{\frac{1}{3}}=a+i b$ Cubing on both the sides, we get: $x+i y=(a+i b)^{3}$ $\Rightarrow x+i y=a^{3}+(i b)^{3}+3 a^{2} b i+3 a(i b)^{2}$ $\Rightarrow x+i y=a^{3}+i^{3} b^{3}+3 a^{2} i b+3 i^{2} a b^{2}$ $\Rightarrow x+i y=a^{3}-i b^{3}+3 a^{2} i b-3 a b^{2} \quad\left(\because i^{2}=-1, i^{3}=-i\right)$ $\Rightarrow x+i y=a^{3}-3 a b^{2}+i\left(-b^{3}+3 a^{2...
Read More →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 →Find the values of a and b so that the polynomial
Question: Find the values of $a$ and $b$ so that the polynomial $\left(x^{3}-10 x^{2}+a x+b\right)$ is exactly divisible by $(x-1)$ as well as $(x-2)$. Solution: Let: $f(x)=x^{3}-10 x^{2}+a x+b$ Now, $x-1=0 \Rightarrow x=1$ By the factor theorem, we can say: $f(x)$ will be exactly divisible by $(x-1)$ if $f(1)=0$. Thus, we have: $f(1)=1^{3}-10 \times 1^{2}+a \times 1+b$ $=(1-10+a+b)$ $=-9+a+b$ $\therefore f(1)=0 \Rightarrow a+b=9 \quad \ldots(1)$ Also, $x-2=0 \Rightarrow x=2$ By the factor theor...
Read More →The mean of n observation is x.
Question: The mean of $n$ observation is $\bar{x}$. If the first observation is increased by 1 , the second by 2 , the third by 3 , and so on, then the new mean is (a) $\bar{x}+(2 n+1)$ (b) $\bar{x}+\frac{n+1}{2}$ (c) $\bar{x}+(n+1)$ (d) $\bar{x}-\frac{n+1}{2}$ Solution: Let $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ be the $n$ observations. Mean $=\bar{x}=\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}$ $\Rightarrow x_{1}+x_{2}+x_{3}+\ldots+x_{n}=n \bar{x}$ If the first item is increased by 1, the second by 2, th...
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