Which of the following are perfect cubes?
Question: Which of the following are perfect cubes? (i) 64 (ii) 216 (iii) 243 (iv) 1000 (v) 1728 (vi) 3087 (vii) 4608 (viii) 106480 (ix) 166375 (x) 456533 Solution: (i)On factorising 64 into prime factors, we get $64=2 \times 2 \times 2 \times 2 \times 2 \times 2$ Group the factors in triples of equal factors as: $64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$ It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore,...
Read More →Which of the following are perfect cubes?
Question: Which of the following are perfect cubes? (i) 64 (ii) 216 (iii) 243 (iv) 1000 (v) 1728 (vi) 3087 (vii) 4608 (viii) 106480 (ix) 166375 (x) 456533 Solution: (i)On factorising 64 into prime factors, we get $64=2 \times 2 \times 2 \times 2 \times 2 \times 2$ Group the factors in triples of equal factors as: $64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$ It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore,...
Read More →The angle of depression from the top of a tower of a point A on the ground is 30°.
Question: The angle of depression from the top of a tower of a point A on the ground is 30. On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60. Find the height of the tower and its distance from the point A. Solution: Let PQ be the tower.We have, $\mathrm{AB}=20 \mathrm{~m}, \angle \mathrm{PAQ}=30^{\circ}$ and $\angle \mathrm{PBQ}=60^{\circ}$ Let $\mathrm{BQ}=x$ and $\mathrm{PQ}=h$ I...
Read More →If a, b, c are real numbers such that
Question: If $a, b, c$ are real numbers such that $\left|\begin{array}{lll}b+c c+a a+b \\ c+a a+b b+c \\ a+b b+c c+a\end{array}\right|=0$, then show that either $a+b+c=0$ or, $a=b=c$. Solution: Let $\Delta=\mid b+c c+a \quad a+b$ $c+a \quad a+b b+c$ $a+b b+c \quad c+a$ $=\mid 2(a+b+c) 2(a+b+c) 2(a+b+c)$ $\begin{array}{lll}c+a a+b b+c \\ a+b b+c c+a \mid\end{array}$ [Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ] $=2(a+b+c) \mid \begin{array}{lll}1 1 1\end{array}$ $c+a+b b+c$ $a+b b+c c+a$ $=2(...
Read More →Prove the following
Question: Factorise (i) x2+ 9x +18 (ii) 6x2+7x -3 (iii) 2x2-7x.-15 (iv) 84-2r-2r2 Solution: (i) $x^{2}+9 x+18=x^{2}+6 x+3 x+18$ [by splitting middle term] $=x(x+6)+3(x+6)=(x+3)(x+6)$ (ii) $6 x^{2}+7 x-3=6 x^{2}+9 x-2 x-3$ [by splitting middle term] $=3 x(2 x+3)-1(2 x+3)=(3 x-1)(2 x+3)$ (iii) $2 x^{2}-7 x-15=2 x^{2}-10 x+3 x-15$ [by splitting middle term] $=2 x(x-5)+3(x-5)=(2 x+3)(x-5)$ (iv) $84-2 r-2 r^{2}=-2\left(r^{2}+r-42\right)$ $=-2\left(r^{2}+7 r-6 r-42\right)$ $=-2[r(r+7)-6(r+7)]$ $=-2(r-...
Read More →Show that
Question: Show that $\left|\begin{array}{lll}x-3 x-4 x-\alpha \\ x-2 x-3 x-\beta \\ x-1 x-2 x-\gamma\end{array}\right|=0$, where $\alpha, \beta, \gamma$ are in AP. Solution: Given: $\alpha, \beta, \gamma$ are in A.P. Now, $2 \beta=\alpha+\gamma$ $\Delta=\left|\begin{array}{lll}x-3 x-4 x-\alpha \\ x-2 x-3 x-\beta \\ x-1 x-2 x-\gamma\end{array}\right|$ $\Delta=\frac{1}{2}\left|\begin{array}{ccc}x-3 x-4 x-\alpha \\ 2 x-4 2 x-6 2 x-2 \beta \\ x-1 x-2 x-\gamma\end{array}\right|$ [Applying $R_{2} \rig...
Read More →If x +1 is a factor of ax3 +x2 -2x+4a-9,
Question: If x +1 is a factor of ax3+x2-2x+4a-9, then find the value of a. Solution: Let $p(x)=a x^{3}+x^{2}-2 x+4 a-9$ Since, $x+1$ is a factor of $p(x)$, then put $p(-1)=0$ $\therefore \quad a(-1)^{3}+(-1)^{2}-2(-1)+4 a-9=0$ $\Rightarrow \quad-a+1+2+4 a-9=0$ $\Rightarrow \quad 3 a-6=0 \Rightarrow 3 a=6$ $\Rightarrow \quad a=\frac{6}{3}=2$ Hence, the value of $a$ is $2 .$...
Read More →Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
Question: Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following: 'The cube of a multiple of 7 is a multiple of 73'. Solution: Five multiples of 7 can be written by choosing different values of a natural numbernin the expression 7n. Let the five multiples be $7,14,21,28$ and 35 . The cubes of these numbers are: $7^{3}=343,14^{3}=2744,21^{3}=9261,28^{3}=21952$, and $35^{3}=42875$ Now, write the above cubes as a multiple of 73. Proceed as follows: $343=7^{3} \tim...
Read More →Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
Question: Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following: 'The cube of a multiple of 7 is a multiple of 73'. Solution: Five multiples of 7 can be written by choosing different values of a natural numbernin the expression 7n. Let the five multiples be $7,14,21,28$ and 35 . The cubes of these numbers are: $7^{3}=343,14^{3}=2744,21^{3}=9261,28^{3}=21952$, and $35^{3}=42875$ Now, write the above cubes as a multiple of 73. Proceed as follows: $343=7^{3} \tim...
Read More →Find the value of m,
Question: Find the value of m, so that 2x -1 be a factor of 8x4+4x3-16x2+10x+07. Solution: Let $p(x)=8 x^{4}+4 x^{3}-16 x^{2}+10 x+m$ Since, $2 x-1$ is a factor of $p(x)$, then put $p\left(\frac{1}{2}\right)=0$ $\therefore \quad 8\left(\frac{1}{2}\right)^{4}+4\left(\frac{1}{2}\right)^{3}-16\left(\frac{1}{2}\right)^{2}+10\left(\frac{1}{2}\right)+m=0$ $\Rightarrow \quad 8 \times \frac{1}{16}+4 \times \frac{1}{8}-16 \times \frac{1}{4}+10\left(\frac{1}{2}\right)+m=0$ $\Rightarrow \quad \frac{1}{2}+\...
Read More →Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11, ...) and verify the following:
Question: Write the cubes of 5 natural numbers of the form 3n+ 2 (i.e. 5, 8, 11, ...) and verify the following: 'The cube of a natural number of the form 3n+ 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'. Solution: Five natural numbers of the form $(3 n+2)$ could be written by choosing $n=1,2,3 \ldots$ etc. Let five such numbers be $5,8,11,14$, and 17 . The cubes of these five numbers are: $125=3 \times 41+2$, which is of the form $(3 n+2)$ for $n=41$ $...
Read More →If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3,
Question: If x + 2a is a factor of a5-4a2x3+2x + 2a +3, then find the value of a. Solution: Let p(x) =a5-4a2x3+2x + 2a +3 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 4a2(-2a)3+ 2(-2a) + 2a + 3 = 0= -32a5+ 32a5-4a + 2a+ 3 = 0 = -2a + 3=0 2a =3 a = 3/2. Hence, the value of a is 3/2....
Read More →Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, ...) and verify the following:
Question: Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g. 4, 7, 10, ...) and verify the following: 'The cube of a natural number of the form 3n+ 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'. Solution: Five natural numbers of the form $(3 n+1)$ could be written by choosing $n=1,2,3 \ldots$ etc. Let five such numbers be $4,7,10,13$, and 16 . The cubes of these five numbers are: $4^{3}=64,7^{3}=343,10^{3}=1000,13^{3}=2197$ and $16^{...
Read More →For what value of m is x3 -2mx2 +16 divisible by x + 2?
Question: For what value of m is x3-2mx2+16 divisible by x + 2? Solution: Let p(x) = x3-2mx2+16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 = (-2)3-2m(-2)2+ 16=0 = -8-8m+16=0 = 8=8 m m = 1 Hence, the value of m is 1 ....
Read More →Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
Question: Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings: 'The cube of a natural number which is a multiple of 3 is a multiple of 27' Solution: Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15. Cubes of these five numbers are: $3^{3}=3 \times 3 \times 3=27$ $6^{3}=6 \times 6 \times 6=216$ $9^{3}=9 \times 9 \times 9=729$ $12^{3}=12 \times 12 \times 12=1728$ $15^{3}=15 \times 15 \times 15=3375$ Now, let us write the cubes as a multiple...
Read More →Show that p-1 is a factor of p10 -1 and also of p11 -1.
Question: Show that p-1 is a factor of p10-1 and also of p11 -1. Solution: Let g (p) = p10-1 and h(p) = p11-1. On putting p=1 in Eq. (i), we get g(1)=110-1= 1-1=0 Hence, p-1 is a factor of g(p). Again, putting p = 1 in Eq. (ii), we get h (1) = (1)11 1 = 1 1 = 0 Hence, p -1 is a factor of h(p)....
Read More →Determine which of the following polynomial has
Question: Determine which of the following polynomial has $x-2$ a factor (i) $3 x^{2}+6 x-24$ (ii) $4 x^{2}+x-2$ Solution: Let $p(x)=x^{5}-4 a^{2} x^{3}+2 x+2 a+3$ Since, $x+2 a$ is a factor of $p(x)$, then put $p(-2 a)=0$ $\therefore \quad(-2 a)^{5}-4 a^{2}(-2 a)^{3}+2(-2 a)+2 a+3=0$ $\Rightarrow \quad-32 a^{5}+32 a^{5}-4 a+2 a+3=0$ $\Rightarrow \quad-2 a+3=0$ $\Rightarrow \quad 2 a=3$ $\therefore$ $a=\frac{3}{2}$ Hence, the value of $a$ is $\frac{3}{2}$....
Read More →Observe the following pattern:
Question: Observe the following pattern:13= 113+ 23= (1 + 2)213+ 23+ 33= (1 + 2 + 3)2Write the next three rows and calculate the value of 13+ 23+ 33+ ... + 93+ 103by the above pattern. Solution: Extend the pattern as follows: Now, from the above pattern, the required value is given by: $1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}+9^{3}+10^{3}=(1+2+3+4+5+6+7+8+9+10)^{2}=55^{2}$ = 3025 Thus, the required value is 3025 ....
Read More →Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
Question: Write the cubes of all natural numbers between 1 and 10 and verify the following statements: (i) Cubes of all odd natural numbers are odd. (ii) Cubes of all even natural numbers are even. Solution: The cubes of natural numbers between 1 and 10 are listed and classified in the following table. We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2. If the number i...
Read More →Find the cubes of the following numbers:
Question: Find the cubes of the following numbers: (i) 7 (ii) 12 (iii) 16 (iv) 21 (v) 40 (vi) 55 (vii) 100 (viii) 302 (ix) 301 Solution: Cube of a number is given by the number raised to the power three. (i) Cube of $7=7^{3}=7 \times 7 \times 7=343$ (ii) Cube of $12=12^{3}=12 \times 12 \times 12=1728$ (iii) Cube of $16=16^{3}=16 \times 16 \times 16=4096$ (iv) Cube of $21=21^{3}=21 \times 21 \times 21=9261$ (v) Cube of $40=40^{3}=40 \times 40 \times 40=64000$ (vi) Cube of $55=55^{3}=55 \times 55 ...
Read More →Show that,
Question: Show that, (i) x + 3 is a factor of 69 + 11c x2+ x3 (ii) 2x 3 is a factor of x + 2x3-9x2+12 Solution: (i) Let $p(x)=x^{3}-x^{2}+11 x+69$ We have to show that, $x+3$ is a factor of $p(x)$. ie. $\quad D(-3)=0$ Now, $p(-3)=(-3)^{3}-(-3)^{2}+11(-3)+69$ $=-27-9-33+69=-69+69=0$ Hence, $(x+3)$ is a factor of $p(x)$. (ii) Let $p(x)=2 x^{3}-9 x^{2}+x+12$ We have to show that, $2 x-3$ is a factor of $p(x)$. i.e., $p\left(\frac{3}{2}\right)=0$ Now, $p\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\ri...
Read More →Check whether p(x) is a multiple of g(x) or not
Question: Check whether p(x) is a multiple of g(x) or not (i) p(x) = x3 5x2+ 4x 3, g(x) = x 2. (ii) p(x) = 2x3 11x2 4x+ 5, g(x) = 2x + l Thinking Process (i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it. (ii) lf p(a) = Q then p(x) is a multiple of g(x) and f(p(a) # Q then p(x) is not a multiple of g(x) where a is a zero of g (x)). Solution: (i) $\because$ $g(x)=x-2$ [given] Then, zero of $g(x)$ is 2 . Now, $p(2)=(2)^{3}-5(2)^{2}+4(2)-3$ $\left[\because p(x)=x^...
Read More →Show that
Question: Show that $\left|\begin{array}{lll}x+1 x+2 x+a \\ x+2 x+3 x+b \\ x+3 x+4 x+c\end{array}\right|=0$ where $a, b, c$ are in A.P. Solution: Given: $a, b, c$ are in A.P. $2 b=a+c$ $\Delta=\left|\begin{array}{lll}x+1 x+2 x+a \\ x+2 x+3 x+b \\ x+3 x+4 x+c\end{array}\right|$ [Applying $R_{2}=2 R_{2}$ ] $\Delta=\frac{1}{2}\left|\begin{array}{ccc}x+1 x+2 x+a \\ 2 x+4 2 x+6 2 x+2 b \\ x+3 x+4 x+c\end{array}\right|$ $\Delta=\frac{1}{2}\left|\begin{array}{ccc}x+1 x+2 x+a \\ 0 0 0 \\ x+3 x+4 x+c\end...
Read More →What is the length of the side of a cube whose volume is 275 cm
Question: What is the length of the side of a cube whose volume is 275 cm3. Make use of the table for the cube root. Solution: Volume of a cube is given by: $V=a^{3}$, where $a=$ side of the cube $\therefore$ Side of a cube $=a=\sqrt[3]{V}$ If the volume of a cube is $275 \mathrm{~cm}^{3}$, the side of the cube will be $\sqrt[3]{275}$. We have: $270275280 \Rightarrow \sqrt[3]{270}\sqrt[3]{275}\sqrt[3]{280}$ From the cube root table, we have: $\sqrt[3]{270}=6.463$ and $\sqrt[3]{280}=6.542$ For th...
Read More →By remainder theorem,
Question: By remainder theorem, find the remainder when p(x) is divided by g(x) (i) p(x) = x3-2x2-4x-1, g(x)=x + 1 (ii) p(x) = x3-3x2+ 4x + 50, g(x)= x 3 (iii) p(x) = x3 12x2+ 14x -3, g(x)= 2x 1 1 (iv) p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2) x Solution: (i) Given, $p(x)=x^{3}-2 x^{2}-4 x-1$ and $g(x)=x+1$ Here, zero of $g(x)$ is $-1$. When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(-1)$ $\therefore \quad p(-1)=(-1)^{3}-2(-1)^{2}-4(-1)-1$ $=-1-2+4-1$ $=4-4=0$ Hence, r...
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