Solve each of the following in equations and represent the solution set on
Question: Solve each of the following in equations and represent the solution set on the number line. $3-2 x \geq 4 x-9$, where $x \in R$ Solution: Given: $3-2 x \geq 4 x-9$, where $x \in R$ $3-2 x \geq 4 x-9$ Subtracting 3 from both the sides in the above equation, $3-2 x-3 \geq 4 x-9-3$ $-2 x \geq 4 x-12$ Now, subtracting 4x from both the sides in the above equation, $-2 x-4 x \geq 4 x-12-4 x$ $-6 x \geq-12$ Now, dividing both the sides by 6 in the above equation $\frac{-6 x}{6} \geq \frac{-12...
Read More →Solve each of the following in equations and represent the solution set on
Question: Solve each of the following in equations and represent the solution set on the number line. $3 x-4x+6$, where $x \in R$ Solution: Given: $3 x-4x+6$, where $x \in R$ $3 x-4x+6$ Adding 4 to both sides in above equation $3 x-4+4x+6+4$ $3 xx+10$ Now, subtracting x from both the sides in above equation $3 x-xx+10-x$ $2 x10$ Now, dividing both the sides by 2 in above equation $\frac{2 x}{2}\frac{10}{2}$ $x5$ Therefore, $x \in(5, \infty)$...
Read More →Write the Pythagorean triplet
Question: Write the Pythagorean triplet whose one of the numbers is $4 .$ Solution: We know that, for any natural number greater than $1,\left(2 m, m^{2}-1, m^{2}+1\right)$ is a pythagorean triplet. So, one number is $2 m$, then other two numbers are $m^{2}+1$ and $m^{2}-1$. Hence, one number is 4 , then pythagorean triplet, $2 m=4 \Rightarrow m=2$ $\therefore \quad m^{2}+1=2^{2}+1=4+1=5$ and $\quad m^{2}-1=2^{2}-1=4-1=3$ Now, $\quad 3^{2}+4^{2}=5^{2}$ $\Rightarrow \quad 9+16=25 \Rightarrow 25=2...
Read More →Can a right triangle with sides 6cm,
Question: Can a right triangle with sides 6cm, 10cm and 8cm be formed? Give reason. Solution: From the question it is given that, Sides of the right angle triangle are 6 cm, 10 cm and 8 cm We know that, the square one side must be equal to the sum of square of other two sides of right angle triangle. i.e. a2= b2+ c2 Where, a = 10 cm, b = 8 cm, c = 6 cm 102= 82+ 62 100 = 64 + 36 100 = 100 Right angle triangle is formed by the given sides 6cm, 10cm and 8cm....
Read More →Solve each of the following in equations and represent the solution set on
Question: Solve each of the following in equations and represent the solution set on the number line. $5 x+217$, where (i) $x \in Z$, (ii) $x \in R$. Solution: (i) $5 x+217, x \in Z$ Subtracting 2 from both the sides in the above equation, $5 x+2-217-2$ $5 x15$ Dividing both the sides by 5 in the above equation, $\frac{5 x}{5}\frac{15}{5}$ $x3$ Since x is an integer Therefore, possible values of x can be $x=\{\ldots,-2,-1,0,1,2\}$ (ii) $5 x+217, x \in R$ Subtracting 2 from both the sides in abov...
Read More →Using distributive law,
Question: Using distributive law, find the squares of (a) 101 (b) 72 Solution: (a) We have, 1012= 101 x 101 = 101(100+ 1)= 10100+ 101 = 10201 (b) We have, 722= 72 x 72 = 72 x (70 + 2) = 5040+ 144= 5184...
Read More →Using prime factorisatioji,
Question: Using prime factorisatioji, find which of the following are perfect cubes, (a) 128 (b) 343 (c) 729 (d) 1331 Solution: (a) We have, 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Since, 2 remains after grouping in triplets. So, 128 is not a perfect cube. (b) We have, 343 = 7 x 7 x 7 Since, the prime factors appear in triplets. So, 343 is a perfect cube. (c) We have, 729 = 3 x 3 x 3 x 3 x 3 x3 Since, the prime factors appear in triplets. So, 729 is a perfect cube. (d) We have, 1331 =11x11x11 Since, the...
Read More →Using prime factorisation,
Question: Using prime factorisation, find which of the following are perfect squares. (a) 484 (b) 11250 (c) 841 (d) 729 . Solution: (a) Prime factors of $484=(2 \times 2) \times(11 \times 11)$ As grouping, there is no unpaired factor left over. So, 484 is a perfect square. (b) 11250 First we have to find out the factors by using prime factorisation method. So, prime factors of 11250 = 2 3 3 5 5 5 5 Now, grouping the prime factors = 2 (3 3) (5 5) (5 5) = 2 32 52 52 Factor 2 has no pair. 11250 is ...
Read More →Solve each of the following in equations and represent the solution set on
Question: Solve each of the following in equations and represent the solution set on the number line. $3 x+82$, where (i) $x \in Z$, (ii) $x \in R$. Solution: (i) $3 x+82, x \in Z$ Subtracting 8 from both the sides in above equation $3 x+8-82-8$ $3 x-6$ Dividing both the sides by 3 in above equation $\frac{3 x}{3}\frac{-6}{3}$ Thus, $x-2$ Since x is an integer Therefore, possible values of x can be $x=\{-1,0,1,2,3, \ldots\}$ (ii) $3 x+82, x \in R$ Subtracting 8 from both the sides in above equat...
Read More →Express 81 as the sum of first nine
Question: Express 81 as the sum of first nine consecutive odd numbers. Solution: 81= (9)2=1+3+ 5+ 7 + 9+ 11 + 13+ 15+ 17 = Sum of first nine consecutive odd numbers...
Read More →Show that 500 is not a perfect square.
Question: Show that 500 is not a perfect square. Solution: Resolving 500 into prime factors, we have $500=2 \times 2 \times 5 \times 5 \times 5$ Grouping the factors into pairs of equal factors, we get $500=(2 \times 2) \times(5 \times 5 \times 5)$ Clearly, by grouping into pairs of equal factors, we are left with one factor 5 , which cannot be paired. Hence, 500 is not a perfect square....
Read More →Solve each of the following in equations and represent the solution set on
Question: Solve each of the following in equations and represent the solution set on the number line. $-2 x5$, where (i) $x \in Z$, (ii) $x \in R$. Solution: (i) $-2 x5, x \in Z$ Multiply both the sides by -1 in above equation, $-2 x(-1)5(-1)$ $2 x-5$ Dividing both the sides by 2 in above equation, $\frac{2 x}{2}\frac{-5}{2}$ $x\frac{-5}{2}$ $x2.5$ Since, x is an integer Therefore, possible values of x can be $x=\{\ldots,-2,-1,0,1,2\}$ (ii) $-2 x5, x \in R$ Multiply both the sides by -1 in above...
Read More →Write cubes of first three multiples of 3.
Question: Write cubes of first three multiples of 3. Solution: Since, the first three multiples of 3 are 3, 6 and 9. Hence, the cubes of first three multiples of 3 are (3)3, (6)3and (9)3, i.e. 27, 216 and 729....
Read More →Write the first five square numbers.
Question: Write the first five square numbers. Solution: First five square numbers are 12,22, 32, 42and 52, i.e. 1, 4, 9,16and 25....
Read More →Square of a number is positive,
Question: Square of a number is positive, so the cube of positive. Solution: False e.g. $(-2)^{2}$, i.e. 4 is a positive number and $(-2)^{3}$. i.e. $-8$ is a negative number....
Read More →There is no cube root
Question: There is no cube root of a negative integer. Solution: False e.g. Let $-8$ be a negative number. Then, cube root of $-8=\sqrt[3]{-8}=\sqrt[3]{(-2) \times(-2)(-2)}=-2$...
Read More →Solve each of the following in equations and represent the solution set on
Question: Solve each of the following in equations and represent the solution set on the number line. $6 x \leq 25$, where (i) $x \in N$, (ii) $x \in Z$. Solution: (i) $6 x \leq 25, x \in N$ Dividing both the sides by 6 in the above equation, $\frac{6 x}{6} \leq \frac{25}{6}$ $x \leq \frac{25}{6}$ x 4.166 Since x is a natural number, therefore the value of x can be less than or equal to 4 Therefore, x = {1,2,3,4} (ii) $6 x \leq 25, x \in Z$ Dividing both the sides by 6 in the above equation, $\f...
Read More →Fill in the blanks with correct inequality sign
Question: Fill in the blanks with correct inequality sign (, , , ) (i) $5 x20 \Rightarrow x$ 4 (ii) $-3 x9 \Rightarrow x \ldots \ldots \ldots .3$ (iii) $4 x-16 \Rightarrow x$ $-4$ (iv) $-6 x \leq-18 \Rightarrow x \ldots \ldots \ldots .3$ (v) $x-3 \Rightarrow-2 x \ldots \ldots \ldots .6$ (vi) $\mathrm{a}\mathrm{b}$ and $\mathrm{c}0 \Rightarrow \frac{\mathrm{a}}{\mathrm{c}} \ldots \ldots \ldots \frac{\mathrm{b}}{\mathrm{c}}$ (vii) $p-q=-3 \Rightarrow p \ldots \ldots \ldots . q$ (viii) $u-v=2 \Righ...
Read More →Solve this
Question: If $|z|=2$ and $\arg (z)=\frac{\pi}{4}$, find $z$ Solution: We have, $|z|=2$ and $\arg (z)=\frac{\pi}{4}$ Let z = r(cos + i sin) We know that, |z| = r = 2 And $\arg (z)=\theta=\frac{\pi}{4}$ Thus, $z=r(\cos \theta+i \sin \theta)=2\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$...
Read More →Solve this
Question: Write $z=(-1+i \sqrt{3})$ in polar form. Solution: We have, $z=(-1+i \sqrt{3})$ Let $-1=r \cos \theta$ and $\sqrt{3}=r \sin \theta$ By squaring and adding, we get $(-1)^{2}+(\sqrt{3})^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 1+3=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow 4=r^{2}$ $\Rightarrow r=2$ $\therefore \cos \theta=\frac{-1}{2}$ and $\sin \theta=\frac{\sqrt{3}}{2}$ Since, lies in second quadrant, we have $\theta=\pi-\frac{\pi}{3}=\frac{2 \p...
Read More →Write z = (1 – i) in polar form.
Question: Write z = (1 i) in polar form. Solution: We have, z = (1 i) Let 1 = rcos and -1 = rsin By squaring and adding, we get $(1)^{2}+(-1)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 1+1=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow 2=r^{2}$ $\Rightarrow r=\sqrt{2}$ $\therefore \cos \theta=\frac{1}{\sqrt{2}}$ and $\sin \theta=\frac{-1}{\sqrt{2}}$ Since, lies in fourth quadrant, we have $\theta=-\frac{\pi}{4}$ Thus, the required polar form is $\sqrt{2}\left(\c...
Read More →Write –3i in polar form.
Question: Write 3i in polar form. Solution: Let, z = -3i Let $0=r \cos \theta$ and $-3=r \sin \theta$ By squaring and adding, we get $(0)^{2}+(-3)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 0+9=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow 9=r^{2}$ $\Rightarrow r=3$ $\therefore \cos \theta=0$ and $\sin \theta=-1$ Since, lies in fourth quadrant, we have $\theta=\frac{3 \pi}{2}$ Thus, the required polar form is $3\left(\cos \left(\frac{3 \pi}{2}\right)+\operatorn...
Read More →Write 2i in polar form.
Question: Write 2i in polar form. Solution: Let, $z=2 i$ Let $0=r \cos \theta$ and $2=r \sin \theta$ By squaring and adding, we get $(0)^{2}+(2)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 0+4=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow 4=r^{2}$ $\Rightarrow r=2$ $\therefore \cos \theta=0$ and $\sin \theta=1$ Since, lies in first quadrant, we have $\theta=\frac{\pi}{2}$ Thus, the required polar form is $2\left(\cos \left(\frac{\pi}{2}\right)+i \sin \left(\frac...
Read More →Write –9 in polar form.
Question: Write 9 in polar form. Solution: We have, z = 9 Let $-9=r \cos \theta$ and $0=r \sin \theta$ By squaring and adding, we get $(-9)^{2}+(0)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 81=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow 81=r^{2}$ $\Rightarrow r=9$ $\therefore \cos \theta=-1$ and $\sin \theta=0$ $\Rightarrow \theta=\pi$ Thus, the required polar form is $9(\cos \pi+i \sin \pi)$...
Read More →Write the principal argument of
Question: Write the principal argument of $(1+i \sqrt{3})^{2}$ Solution: Let, $\mathrm{z}=(1+\mathrm{i} \sqrt{3})^{2}$ $=(1)^{2}+(i \sqrt{3})^{2}+2 \sqrt{3} i$ $=1-1+2 \sqrt{3} i$ $z=0+2 \sqrt{3} i$ Let $0=r \cos \theta$ and $2 \sqrt{3}=r \sin \theta$ By squaring and adding, we get $(0)^{2}+(2 \sqrt{3})^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 0+(2 \sqrt{3})^{2}=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow(2 \sqrt{3})^{2}=r^{2}$ $\Rightarrow r=2 \sqrt{3}$ co...
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