Prove that:
Question: Prove that $\left|\begin{array}{ccc}-b c b^{2}+b c c^{2}+b c \\ a^{2}+a c -a c c^{2}+a c \\ a^{2}+a b b^{2}+a b -a b\end{array}\right|=(a b+b c+c a)^{3}$ Solution: $\Delta=\left|\begin{array}{ccc}-b c b^{2}+b c c^{2}+b c \\ a^{2}+a c -a c c^{2}+a c \\ a^{2}+a b b^{2}+a b -a b\end{array}\right|$ $=\frac{1}{a b c}\left|\begin{array}{ccc}-a b c a b^{2}+a b c a c^{2}+a b c \\ a^{2} b+a b c -a b c c^{2} b+a b c \\ a^{2} c+a b c b^{2} c+a b c -a b c\end{array}\right| \quad\left[\right.$ Appl...
Read More →Prove that:
Question: Prove that $\left|\begin{array}{ccc}-b c b^{2}+b c c^{2}+b c \\ a^{2}+a c -a c c^{2}+a c \\ a^{2}+a b b^{2}+a b -a b\end{array}\right|=(a b+b c+c a)^{3}$ Solution: $\Delta=\left|\begin{array}{ccc}-b c b^{2}+b c c^{2}+b c \\ a^{2}+a c -a c c^{2}+a c \\ a^{2}+a b b^{2}+a b -a b\end{array}\right|$ $=\frac{1}{a b c}\left|\begin{array}{ccc}-a b c a b^{2}+a b c a c^{2}+a b c \\ a^{2} b+a b c -a b c c^{2} b+a b c \\ a^{2} c+a b c b^{2} c+a b c -a b c\end{array}\right| \quad\left[\right.$ Appl...
Read More →Prove the following
Question: If $\frac{x}{y}+\frac{y}{x}=-1$ (where $x, y \neq 0$ ), then the value of $x^{3}-y^{3}$ is (a) 1 - (b) $-1$ (c) 0 (d) $\frac{1}{2}$ Solution: (c) Given, $\frac{x}{y}+\frac{y}{x}=-1$ $\Rightarrow$ $\frac{x^{2}+y^{2}}{x y}=-1$ $\Rightarrow$ $x^{2}+y^{2}=-x y$ $\Rightarrow$ $x^{2}+y^{2}+x y=0$ $\ldots($ i) Now, $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$ [using identity, $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$ $=(x-y) \times 0=0$ [from Eq. (i)]...
Read More →Making use of the cube root table, find the cube root 7800
Question: Making use of the cube root table, find the cube root7800 Solution: We have: $7800=78 \times 100$ $\therefore \sqrt[3]{7800}=\sqrt[3]{78 \times 100}=\sqrt[3]{78} \times \sqrt[3]{100}$ By the cube root table, we have: $\sqrt[3]{78}=4.273$ and $\sqrt[3]{100}=4.642$ $\sqrt[3]{7800}=\sqrt[3]{78} \times \sqrt[3]{100}=4.273 \times 4.642=19.835$ (upto three decimal places) Thus, the answer is 19.835...
Read More →Making use of the cube root table, find the cube root 780
Question: Making use of the cube root table, find the cube root780 Solution: We have: $780=78 \times 10$ $\therefore$ Cube root of 780 would be in the column of $\sqrt[3]{10 x}$ against 78 . By the cube root table, we have: $\sqrt[3]{780}=9.205$ Thus, the answer is 9.205....
Read More →Prove that:
Question: Prove that $\left|\begin{array}{ccc}-b c b^{2}+b c c^{2}+b c \\ a^{2}+a c -a c c^{2}+a c \\ a^{2}+a b b^{2}+a b -a b\end{array}\right|=(a b+b c+c a)^{3}$ Solution: $\Delta=\left|\begin{array}{ccc}-b c b^{2}+b c c^{2}+b c \\ a^{2}+a c -a c c^{2}+a c \\ a^{2}+a b b^{2}+a b -a b\end{array}\right|$ $=\frac{1}{a b c}\left|\begin{array}{ccc}-a b c a b^{2}+a b c a c^{2}+a b c \\ a^{2} b+a b c -a b c c^{2} b+a b c \\ a^{2} c+a b c b^{2} c+a b c -a b c\end{array}\right| \quad\left[\right.$ Appl...
Read More →Making use of the cube root table,
Question: Making use of the cube root table, find the cube root1100 Solution: We have: $1100=11 \times 100$ $\therefore \sqrt[3]{1100}=\sqrt[3]{11 \times 100}=\sqrt[3]{11} \times \sqrt[3]{100}$ By the cube root table, we have: $\sqrt[3]{11}=2.224$ and $\sqrt[3]{100}=4.642$ $\therefore \sqrt[3]{1100}=\sqrt[3]{11} \times \sqrt[3]{100}=2.224 \times 4.642=10.323(\mathrm{Up}$ to three decimal places) Thus, the answer is 10.323....
Read More →Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide.
Question: Two poles of equal heights are standing opposite to each other on eitherside of the road which is 80 m wide. From a point P between them onthe road, the angle of elevation of the top of one pole is 60 and theangle of depression from the top of another pole at P is 30. Find theheight of each pole and distances of the point P from the poles. Solution: Let AB and CD be the equal poles; and BD be the width of the road.We have, $\angle \mathrm{AOB}=60^{\circ}$ and $\angle \mathrm{COD}=60^{\...
Read More →The coefficient of x in the expansion of
Question: The coefficient of $x$ in the expansion of $(x+3)^{3}$ is (a) 1 (b) 9 (c) 18 (d) 27 Solution: (d) Now, (x + 3)3= x3+ 33+ 3x (3)(x + 3) [using identity, (a + b)3= a3+ b3+ 3ab (a + b)] = x3+ 27 + 9x (x + 3) = x3+27 + 9x2+27x Hence, the coefficient of x in (x + 3)3is 27....
Read More →Making use of the cube root table,
Question: Making use of the cube root table, find the cube root7000 Solution: We have: $\therefore \sqrt[3]{7000}=\sqrt[3]{7 \times 1000}=\sqrt[3]{7} \times \sqrt[3]{1000}$ By the cube root table, we have: $\sqrt[3]{7}=1.913$ and $\sqrt[3]{1000}=10$ $\therefore \sqrt[3]{7000}=\sqrt[3]{7} \times \sqrt[3]{1000}=1.913 \times 10=19.13$...
Read More →Which of the following is a factor of
Question: Which of the following is a factor of (x+ y)3 (x3+ y3)? (a) x2+ y2+ 2 xy (b) x2+ y2 xy (c) xy2 (d) 3xy Solution: (d) Now, (x+ y)3 (x3+ y3) = (x + y) (x + y)(x2 xy + y2) [using identity, a3+ b3= (a + b)(a2-ab+ b2)] = (x+ y)[(x+ y)2-(x2-xy+ y2)] = (x+ y)(x2+ y2+ 2xy- x2+ xy- y2) [using identity, (a + b)2= a2+ b2+ 2 ab)] = (x + y) (3xy) Hence, one of the factor of given polynomial is 3xy....
Read More →Prove that:
Question: Prove that $\left|\begin{array}{ccc}\frac{a^{2}+b^{2}}{c} c c \\ a \frac{b^{2}+c^{2}}{a} a \\ b b \frac{c^{2}+a^{2}}{b}\end{array}\right|=4 a b c$ Solution: $\Delta=\left|\begin{array}{ccc}\frac{a^{2}+b^{2}}{c} c c \\ a \frac{b^{2}+c^{2}}{a} a \\ b b \frac{c^{2}+a^{2}}{b}\end{array}\right|$ $=\frac{1}{a b c}\left|\begin{array}{ccc}a^{2}+b^{2} c^{2} c^{2} \\ a^{2} b^{2}+c^{2} a^{2} \\ b^{2} b^{2} c^{2}+a^{2}\end{array}\right|$ [Multiplying $R_{1}, R_{2}$ and $R_{3}$ by $c, a$ and $b$ an...
Read More →Making use of the cube root table,
Question: Making use of the cube root table, find the cube root7000 Solution: We have: $700=70 \times 10$ $\therefore$ Cube root of 700 will be in the column of $\sqrt[3]{10 x}$ against 70 . By the cube root table, we have: $\sqrt[3]{700}=8.879$ Thus, the answer is 8.879....
Read More →The factorization of
Question: The factorization of 4x2+ 8x+ 3 is (a) (x +1) (x + 3) (b) (2x+1) (2x + 3) (c) (2x + 2) (2x + 5) (d) (2x -1) (2x 3) Solution: (b) Now, 4x2+ 8x + 3= 4x2+ 6x + 2x + 3 [by splitting middle term] = 2x(2x+ 3) + 1 (2x+ 3) = (2x + 3) (2x + 1)...
Read More →The value of
Question: The value of 2492 2482is (a) 12 (b) 477 (c) 487 (d) 497 Solution: (d) Now, 2492 2482= (249 + 248) (249 248) [using identity, a2 b2= (a b)(a + b)] = 497 x 1 = 497....
Read More →Making use of the cube root table,
Question: Making use of the cube root table, find the cube root700 Solution: We have: $700=70 \times 10$ $\therefore$ Cube root of 700 will be in the column of $\sqrt[3]{10 x}$ against 70 . By the cube root table, we have $\sqrt[3]{700}=8.879$ Thus, the answer is 8.879....
Read More →On a horizontal plane there is a vertical tower with a flagpole on the top of the tower.
Question: On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top of bottom of the flagpole are 60 and 30 respectively. Find the height of the tower and the flagpole mounted on it.$[$ Use $\sqrt{3}=1.73]$ Solution: Let $O X$ be the horizontal plane, $A D$ be the tower and $C D$ be the vertical flagpole. We have: $A B=9 \mathrm{~m}, \angle D B A=30^{\circ}$ and $\angle C B A=60...
Read More →One of the factors of
Question: One of the factors of $\left(25 x^{2}-1\right)+(1+5 x)^{2}$ is (a) $5+x$ (b) $5-x$ (c) $5 x-1$ (d) $10 x$ Solution: (d) Now, (25x2-1) + (1 + 5x)2 = 25x2-1 + 1 + 25x2+ 10x [using identity, (a + b)2= a2+ b2+ 2ab] = 50x2+ 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x....
Read More →Making use of the cube root table,
Question: Making use of the cube root table, find the cube root70 Solution: Because 70 lies between 1 and 100, we will look at the row containing 70 in the column ofx.By the cube root table, we have: $\sqrt[3]{70}=4.121$...
Read More →Solve this
Question: $\left|\begin{array}{ccc}0 b^{2} a c^{2} a \\ a^{2} b 0 c^{2} b \\ a^{2} c b^{2} c 0\end{array}\right|=2 a^{3} b^{3} c^{3}$ Solution: $\Delta=\left|\begin{array}{ccc}0 b^{2} a c^{2} a \\ a^{2} b 0 c^{2} b \\ a^{2} c b^{2} c 0\end{array}\right|$ $=\frac{1}{a b c}\left|\begin{array}{ccc}0 b^{3} a c^{3} a \\ a^{3} b 0 c^{3} b \\ a^{3} c b^{3} c 0\end{array}\right|$ [Multiplying the three columns by $a, b$ and $c$ ] $=\frac{a b c}{a b c}\left|\begin{array}{ccc}0 b^{3} c^{3} \\ a^{3} 0 c^{3...
Read More →Making use of the cube root table,
Question: Making use of the cube root table, find the cube roots 7 Solution: Because 7 lies between 1 and 100, we will look at the row containing 7 in the column ofx.By the cube root table, we have: $\sqrt[3]{7}=1.913$ Thus, the answer is 1.913....
Read More →x+1 is a factor of the polynomial
Question: $x+1$ is a factor of the polynomial (a) $x^{3}+x^{2}-x+1$ (b) $x^{3}+x^{2}+x+1$ (c) $x^{4}+x^{3}+x^{2}+1$ (d) $x^{4}+3 x^{3}+3 x^{2}+x+1$ Solution: (b) Let assume (x + 1) is a factor of x3+ x2+ x+1. So, x = -1 is zero of x3+ x2+ x+1 (-1)3+ (-1)2+ (-1) + 1 = 0 = -1+1-1 + 1 = 0 = 0 = 0 Hence, our assumption is true....
Read More →Find the tens digit of the cube root of each of the numbers in Q. No. 15.
Question: Find the tens digit of the cube root of each of the numbers in Q. No. 15. Solution: (i) Let us consider the number 226981. The unit digit is 1; therefore,the unit digit of the cube root of 226981 is 1. After striking out the units, tens and hundreds digits of the given number, we are left with 226. Now, 6 is the largest number, whose cube is less than or equal to $226\left(6^{3}2267^{3}\right)$. Therefore, the tens digit of the cube root of 226981 is 6. (ii) Let us consider the number ...
Read More →If x+1 is a factor of the polynomial
Question: If $x+1$ is a factor of the polynomial $2 x^{2}+k x$, then the value of $k$ is (a) $-3$ (b) 4 (c) 2 (d)-2 Solution: (c) Let p(x) = 2x2+ kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 = 2-k = 0 = k= 2 Hence, the value of k is 2....
Read More →Solve this
Question: $\left|\begin{array}{ccc}b^{2}+c^{2} a b a c \\ b a c^{2}+a^{2} b c \\ c a c b a^{2}+b^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$ Solution: $\Delta=\left|\begin{array}{ccc}b^{2}+c^{2} a b a c \\ b a c^{2}+a^{2} b c \\ c a c b a^{2}+b^{2}\end{array}\right|$ $\left|\begin{array}{ccc}a\left(b^{2}+c^{2}\right) a^{2} b a^{2} c \\ b^{2} a b\left(c^{2}+a^{2}\right) b^{2} c \\ c^{2} a c^{2} b c\left(a^{2}+b^{2}\right)\end{array}\right| \quad[$ Multiplying the three rows by $a, b$ and $c]$ $=\f...
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