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Question: $\sqrt{-11-60 i}$ Solution: Let, $(a+i b)^{2}=-11-60 i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=-11-60 i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=-11-60 i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=-11 \ldots \ldots \ldots \ldots \ldots$ eq. 1 $\Rightarrow 2 a b=-60 \ldots \ldots . .$ eq. 2 $\Rightarrow \mathrm{a}=-\frac{30}{b}$ Now, using the value of a in eq.1, we get $\Rightarrow\left(-\frac{30}{b}\right)^...
Read More →Cube roots of 8 are + 2 and – 2 .
Question: Cube roots of 8 are + 2 and 2 . Solution: False Cube root of 8 is 2 only and cube root of 8 is 2....
Read More →363 x 81 is a perfect cube.
Question: 363 x 81 is a perfect cube. Solution: False $\because \quad 363 \times 81=3 \times 11 \times 11 \times 3 \times 3 \times 3 \times 3$ $=3 \times 3 \times 3 \times 3 \times 3 \times 11 \times 11$ Grouping the factors in triplets of equal factors, we have $363 \times 81=(3 \times 3 \times 3) \times 3 \times 3 \times 11 \times 11$ Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors $3 \times 3$ and $11 \times 11$. Hence, $363 \times 81$ is not a per...
Read More →999 is a perfect cube.
Question: 999 is a perfect cube. Solution: False Resolving 999 into prime factors, we get $999=3 \times 3 \times 3 \times 37$ Grouping the factors in triplets of equal factors, we get $999=(3 \times 3 \times 3) \times 37$ Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor 37 . Therefore, 999 is not a perfect cube....
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Question: $\sqrt{-15-8 i}$ Solution: Let, $(a+i b)^{2}=-15-8 i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=-15-8 i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=-15-8 i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=-15$ ..eq.1 $\Rightarrow 2 a b=-8$ .. eq.2 $\Rightarrow \mathrm{a}=-\frac{4}{b}$ Now, using the value of a in eq.1, we get $\Rightarrow\left(-\frac{4}{b}\right)^{2}-\mathrm{b}^{2}=-15$ $\Rightarrow 16-\mathrm{b}^{4}=-1...
Read More →Cube of an odd number is odd.
Question: Cube of an odd number is odd. Solution: True We know that, the cube of an odd number is always an odd number, e.g. 9 is an odd number. Then,93 =9 x 9 x 9 = 729 Clearly, 729 is also an odd number....
Read More →Cube of an even number is even.
Question: Cube of an even number is even. Solution: True We know that, the cube of an even number is always an even number, e.g. 4 is an even number. Then, 43 = 4 x 4 x 4 = 64 Clearly, 64 is also an even number....
Read More →Cube of an odd number is even.
Question: Cube of an odd number is even. Solution: False We know that, the cube of an odd number is always an odd number, e.g. 3 is an odd number. Then, 33 = 3 x 3 x 3 = 27 Clearly, 27 is not an even number....
Read More →Cube of an even number is odd.
Question: Cube of an even number is odd. Solution: False We know that, the cube of an even number is always an even number, e.g. 2 is an even number. Then, 23= 2 x 2 x 2 = 8 Clearly, 8 is also an even number....
Read More →The cube of a one-digit number cannot
Question: The cube of a one-digit number cannot be a two-digit number. Solution: False e.g. 3 is a one-digit number and $(3)^{3}$, i.e. 27 is a two-digit number....
Read More →A number having 7 at its one’s place will have
Question: A number having 7 at its ones place will have 3 at the ones place of its cube. Solution: True Cube of 7 = 7 x 7 x 7 =343 Cube of 17 = 17 x 17 x 17=4913 Cube of 27 = 27 x 27 x 27 = 19683 and so on....
Read More →A number having 7 at its one’s place will have
Question: A number having 7 at its ones place will have 3 at the units place of its square. Solution: False Square of 7 = 7 x 7 = 49 Square of 17 = 17 x 17 = 289 Square of 27 = 27 x 27 = 729 and so on....
Read More →Square root of a number x
Question: Square root of a number x is denoted by 4x. Solution: True Square root of a number x is denoted by 4x....
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Question: $\sqrt{-4-3 i}$ Solution: Let, $(a+i b)^{2}=-4-3 i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=-4-3 i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=-4-3 i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=-4 \ldots \ldots \ldots \ldots \ldots$ eq. 1 $\Rightarrow 2 \mathrm{ab}=-3 \ldots \ldots .$ eq. 2 $\Rightarrow \mathrm{a}=-\frac{3}{2 b}$ Now, using the value of a in eq.1, we get $\Rightarrow\left(-\frac{3}{2 b}\right)^{2...
Read More →The square root of a perfect square of
Question: The square root of a perfect square of $n$ digits will have $\left(\frac{n+1}{2}\right)$ digits, if $n$ is odd. Solution: True If the square has 3 digits, then its square root has 2 i.e. $\left(\frac{3+1}{2}\right)$ digits. Similarly, if the square has 5 digits, then its square root has 3 i.e. $\left(\frac{5+1}{2}\right)$ digits....
Read More →If a2 ends in 9,
Question: If a2ends in 9, thena3ends in 7. Solution: False $\because(7)^{2}=49$ ends in 9 and $(7)^{3}$, i.e. 343 does not end in 7 ....
Read More →If a2 ends in 5,
Question: If a2ends in 5, then a3ends in 25: Solution: False $\because(35)^{2}=1225$ ends in 5 and $(35)^{3}$, i.e. 42875 does not end in 25 ....
Read More →Let x and y be natural numbers.
Question: Let x and y be natural numbers. If x divides y, then x3divides y3. Solution: True If $x$ divides $y$, then $\frac{y}{x}$ is a natural number. $\Rightarrow \quad\left(\frac{y}{x}\right)^{3}$ is also a natural number. $\Rightarrow \quad \frac{y^{3}}{x^{3}}$ is a natural number. $\Rightarrow \quad x^{3}$ divides $y^{3}$...
Read More →If x and y are integers
Question: If $x$ and $y$ are integers such that $x^{2}y^{2}$, then $x^{3}y^{3}$. Solution: False Suppose, $-1$ and $-2$ are integers Then, $(-2)^{2^{\prime}}(-1)^{2}=41$ and $(-2)^{3}(-1)^{3}=-8-1$...
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Question: $\sqrt{16-30 i}$ Solution: Let, $(a+i b)^{2}=16-30 i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=16-30 i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=16-30 i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=16$ ..eq.1 $\Rightarrow 2 a b=-30$ .. eq.2 $\Rightarrow \mathrm{a}=-\frac{15}{b}$ Now, using the value of a in eq.1, we get $\Rightarrow\left(-\frac{15}{b}\right)^{2}-b^{2}=16$ $\Rightarrow 225-b^{4}=16 b^{2}$ $\Righta...
Read More →For an integer a,
Question: For an integer a, a3is always greater than a2. Solution: False e.g. $-1$ is an integer and $(-1)^{3}$, i.e. $-1$ is not greater than $(-1)^{2}$, i.e. 1...
Read More →All numbers of a Pythagorean
Question: All numbers of a Pythagorean triplet are odd. Solution: False 3, 4 and 5 are the numbers of Pythagorean triplet as 52= 42+ 32where, 4 is not an odd number....
Read More →For every natural number
Question: For every natural number $m,\left(2 m-1,2 m^{3}-2 m, 2 m^{2}-2 m+1\right)$ is a pythagorean triplet. Solution: False $\because$ $(2 m-1)^{2} \neq\left(2 m^{3}-2 m\right)^{2}+\left(2 m^{2}-2 m+1\right)^{2}$ $\left(2 m^{3}-2 m\right)^{2} \neq(2 m-1)^{2}+\left(2 m^{2}-2 m+1\right)^{2}$ and $\left(2 m^{2}-2 m+1\right)^{2} \neq(2 m-1)^{2}+\left(2 m^{3}-2 m\right)^{2}$...
Read More →A perfect square can have 8
Question: A perfect square can have 8 as its units digit. Solution: False A perfect square can never have 8 as its units digit....
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Question: $\sqrt{3+4 \sqrt{-7}}$ Solution: Let, $(a+i b)^{2}=3+4^{\sqrt{7}} i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=3+4^{\sqrt{7}} i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=3+4^{\sqrt{7}} i$ now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=3 \ldots \ldots \ldots \ldots \ldots e q \cdot 1$ $\Rightarrow 2 \mathrm{ab}=4^{\sqrt{7}} \ldots \ldots \ldots$ eq.2 $\Rightarrow a=\frac{2 \sqrt{7}}{b}$ Now, using the value of a in eq...
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