Prove the following
Question: If $x^{51}+51$ is divided by $x+1$, then the remainder is (a) 0 (b) 1 (c) 49 (d) 50 Solution: (d) Let p(x) = x51+ 51 . (i) When we divide p(x) by x+1, we get the remainder p(-1) On putting x= -1 in Eq. (i), we get p(-1) = (-1)51+ 51 = -1 + 51 = 50 Hence, the remainder is 50....
Read More →Find the units digit of the cube root of the following numbers:
Question: Find the units digit of the cube root of the following numbers: (i) 226981 (ii) 13824 (iii) 571787 (iv) 175616 Solution: (i) Cube root using units digit:Let us consider the number 226981. The unit digit is 1; therefore, the unit digit of the cube root of 226981 is 1.(ii) Cube root using units digit:Let us consider the number 13824. The unit digit is 4; therefore,the unit digit of the cube root of 13824 is 4.(iii) Cube root using units digit:Let us consider the number 571787. The unit d...
Read More →The angle of elevation of the top of an unfinished tower at a distance of 75 m form its base is 30°.
Question: The angle of elevation of the top of an unfinished tower at a distance of 75 m form its base is 30. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60? Solution: Let $A B$ be the unfinished tower, $A C$ be the raised tower and $\mathrm{O}$ be the point of observation. We have: $O A=75 \mathrm{~m} \cdot \angle A O B=30^{\circ}$ and $\angle A O C=60^{\circ}$ Let $A C=H \mathrm{~m}$ such that $B C=(H-h) \mathrm{m}$. In ∆AOB, we h...
Read More →Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that
Question: Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that (i) 3048625 = 3375 729 (ii) 20346417 = 9261 2197 (iii) 210644875 = 42875 4913 (iv) 57066625 = 166375 343 Solution: (i)To find the cube root, we use the following property: $\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for two integers $a$ and $b$ now $\sqrt[3]{3048625}$ $=\sqrt[3]{3375} \times 729$ $=\sqrt[3]{3375} \times \sqrt[3]{729}$ (By the above property) $=\sqrt[3]{3 \times 3 \times 3 \...
Read More →Solve this
Question: $\left|\begin{array}{ccc}b+c a a \\ b c+a b \\ c c a+b\end{array}\right|=4 a b c$ Solution: $\Delta=\left|\begin{array}{ccc}b+c a a \\ b c+a b \\ c c a+b\end{array}\right|$ $=\left|\begin{array}{ccc}0 -2 c -2 b \\ b c+a b \\ c c a+b\end{array}\right| \quad$ [Applying $\left.R_{1} \rightarrow R_{1}-\left(R_{2}+R_{3}\right)\right]$ $=\left|\begin{array}{ccc}0 -2 c -2 b \\ b c+a-b 0 \\ c 0 a+b-c\end{array}\right| \quad$ [Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $\left.C_{3} \rightarro...
Read More →A statue 1.46 m tall, stands on the top of a pedestal.
Question: A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60 and from the same point, the angle of elevation of the top of the pedestal is 45. Find the height of the pedestal.$[$ Use $\sqrt{3}=1.73]$ Solution: Let $A C$ be the pedestal and $B C$ be the statue such that $B C=1.46 \mathrm{~m}$. We have: $\angle A D C=45^{\circ}$ and $\angle A D B=60^{\circ}$ Let: $A C=h \mathrm{~m}$ and $A D=x \mathrm{~m}$ In the...
Read More →Solve this
Question: $\left|\begin{array}{ccc}a+b+c -c -b \\ -c a+b+c -a \\ -b -a a+b+c\end{array}\right|=2(a+b)(b+c)(c+a)$ Solution: Let LHS $=\Delta=\mid a+b+c \quad-c \quad-b$ $\begin{array}{lll}-c a+b+c -a \\ -b -a a+b+c \mid\end{array}$ $=\mid \mathrm{a} \quad-\mathrm{c} \quad-\mathrm{b}$ $\begin{array}{ccc}b a+b+c -a \\ c -a a+b+c \mid\end{array}$ $=\mid \mathrm{a} \quad-\mathrm{c} \quad-\mathrm{b}$ $\begin{array}{ccc}b a+b+c -a \\ c -a a+b+c \mid\end{array}$[Applying $\mathrm{C}_{1} \rightarrow \mat...
Read More →Evaluate:
Question: Evaluate: (i) $\sqrt[3]{36} \times \sqrt[3]{384}$ (ii) $\sqrt[3]{96} \times \sqrt[3]{144}$ (iii) $\sqrt[3]{100} \times \sqrt[3]{270}$ (iv) $\sqrt[3]{121} \times \sqrt[3]{297}$ Solution: (i)36 and 384 are not perfect cubes; therefore, we use the following property: $\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for any two integers $a$ and $b$ $\therefore \sqrt[3]{36} \times \sqrt[3]{384}$ $=\sqrt[3]{36 \times 384}$ $=\sqrt[3]{(2 \times 2 \times 3 \times 3) \times(2 \times 2 \times 2 \t...
Read More →A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m.
Question: A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30 and that of the top of the flagstaff is 60. Find the height of the tower.[Use $\sqrt{3}=1.732]$ Solution: Let AB be the tower and BC be the flagstaff.We have, $\mathrm{BC}=6 \mathrm{~m}, \angle \mathrm{AOB}=30^{\circ}$ and $\angle \mathrm{AOC}=60^{\circ}$ Let $\mathrm{AB}=h$ In $\triangle \mathrm{AOB}...
Read More →One of the zeroes of the polynomial
Question: One of the zeroes of the polynomial $2 x^{2}+7 x-4$ is (a) 2 (b) $1 / 2$ (c)-1 (d)- 2 Thinking Process (i) Firstly, determine the factor by using splitting method. (ii) Further, put the factors equals to zero, then determine the values of $x$. Solution: (b) Let p (x) = 2x2+ 7x-4 = 2x2+ 8x-x-4 [by splitting middle term] = 2x(x+ 4)-1(x+ 4) = (2x-1)(x+ 4) For zeroes of p(x), put p(x) = 0= (2x -1) (x + 4) = 0 = 2x-1 = 0 and x+4 = 0 = x = and x = -4 Hence, one of the zeroes of the polynomia...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{ccc}1 a a^{2} \\ a^{2} 1 a \\ a a^{2} 1\end{array}\right|=\left(a^{3}-1\right)^{2}$ Solution: Let LHS $=\Delta=\mid \begin{array}{lll}1 a a^{2}\end{array}$ $\begin{array}{lll}a^{2} 1 a \\ a a^{2} 1\end{array}$ $\Delta=\mid 1+a^{2}+a \quad 1+a^{2}+a \quad 1+a^{2}+a$ $\begin{array}{ccccc}a^{2} 1 a \\ a a^{2} 1 \text { [Applyng } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{2} \text { ] } \\ = \left(1+a^{2}+a\right) \mid 1 1 1 \end{...
Read More →Prove that:
Question: $\left|\begin{array}{ccc}1 a a^{2} \\ a^{2} 1 a \\ a a^{2} 1\end{array}\right|=\left(a^{3}-1\right)^{2}$ Solution: Let LHS $=\Delta=\mid \begin{array}{lll}1 a a^{2}\end{array}$ $\begin{array}{lll}a^{2} 1 a \\ a a^{2} 1\end{array}$ $\Delta=\mid 1+a^{2}+a \quad 1+a^{2}+a \quad 1+a^{2}+a$ $\begin{array}{ccccc}a^{2} 1 a \\ a a^{2} 1 \text { [Applyng } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{2} \text { ] } \\ = \left(1+a^{2}+a\right) \mid 1 1 1 \end{array}$ $a^{...
Read More →Zero of the polynomial
Question: Zero of the polynomial $p(x)=2 x+5$ is (a) $-2 / 5$ (b) $-5 / 2$ (c) $2 / 5$ (d) $5 / 2$ Solution: (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 2x + 5 = 0 = -5/2 Hence, zero of the polynomial p(x) is -5/2....
Read More →Find the side of a cube whose volume is $
Question: Find the side of a cube whose volume is $\frac{24389}{216} \mathrm{~m}^{3}$. Solution: Volume of a cube with sidesis given by: $V=s^{3}$ $\therefore s=\sqrt[3]{\bar{V}}$ $=\sqrt[3]{\frac{24389}{216}}$ $=\frac{\sqrt[3]{24389}}{\sqrt[3]{216}}$ $=\frac{\sqrt[3]{29 \times 29 \times 29}}{\sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3}}$ (By prime factorisation) $=\frac{29}{2 \times 3}$ $=\frac{29}{6}$ Thus, the length of the side is $\frac{29}{6} \mathrm{~m}$....
Read More →From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°.
Question: From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30. The angle of elevation of the top of a water tank (on the top of the tower) is 45. Find (i) the height of the tower, (ii) the depth of the tank. Solution: Let BC be the tower and CD be the water tank. We have, $\mathrm{AB}=40 \mathrm{~m}, \angle \mathrm{BAC}=30^{\circ}$ and $\angle \mathrm{BAD}=45^{\circ}$ In $\Delta$ ABD $\tan 45^{\circ}=\frac{\mathrm{BD}}{\mathrm{AB}}$...
Read More →Zero of the zero polynomial is
Question: Zero of the zero polynomial is (a) 0 (b) 1 (c) any real number (d) not defined Solution: (c) Zero of the zero polynomial is any real number. e.g., Let us consider zero polynomial be 0(x-k), where k is a real number For determining the zero, putx-k = 0=x = k Hence, zero of the zero polynomial be any real number....
Read More →Prove the following
Question: If $p(x)=x+3$, then $p(x)+p(-x)$ is equal to (a) 3 (b) $2 x$ (c) 0 (d) 6 Solution: (d) Given $p(x)=x+3$, put $x=-x$ in the given equation, we get $p(-x)=-x+3$ Now, $p(x)+p(-x)=x+3+(-x)+3=6$...
Read More →Three numbers are to one another 2 : 3 : 4.
Question: Three numbers are to one another 2 : 3 : 4. The sum of their cubes is 0.334125. Find the numbers. Solution: Let the numbers be2x, 3xand 4x. According to the question: $(2 x)^{3}+(3 x)^{3}+(4 x)^{3}=0.334125$ $\Rightarrow 8 x^{3}+27 x^{3}+64 x^{3}=0.334125$ $\Rightarrow 8 x^{3}+27 x^{3}+64 x^{3}=0.334125$\ $\Rightarrow 99 x^{3}=0.334125$ $\Rightarrow x=\sqrt[3]{\frac{3375}{1000000}}$ $\Rightarrow x=\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}$ $\Rightarrow x=\frac{15}{100}=0.15$ Thus, the ...
Read More →The value of the polynomial
Question: The value of the polynomial $5 x-4 x^{2}+3$, when $x=-1$ is (a)-6 (b) 6 (c) 2 (d) $-2$ Solution: (a) Let $p(x)=5 x-4 x^{2}+3 \ldots$ (i) On putting $x=-1$ in Eq. (i), we get $p(-1)=5(-1)-4(-1)^{2}+3=-5-4+3=-6$...
Read More →Prove the following
Question: If $p(x)=x 2-2 \sqrt{2} x+1$, then $p(2 \sqrt{2})$ is equal to (a) 0 (b) 1 (c) $4 \sqrt{2}$ (d) $8 \sqrt{2}+1$ Solution: (b) Given, $p(x)=x^{2}-2 \sqrt{2} x+1 \ldots$ (i) On putting $x=2 \sqrt{2}$ in Eq. (i), we get $P(2 \sqrt{2})=(2 \sqrt{2})^{2}-(2 \sqrt{2})(2 \sqrt{2})+1=8-8+1=1$...
Read More →The volume of a cubical box is 474.552 cubic metres.
Question: The volume of a cubical box is 474.552 cubic metres. Find the length of each side of the box. Solution: Volume of a cube is given by: $V=s^{3}$, where $s=$ side of the cube Now $s^{3}=474.552$ cubic metres $\Rightarrow s=\sqrt[3]{474.552}=\sqrt[3]{\frac{474552}{1000}}=\frac{\sqrt[3]{474552}}{\sqrt[3]{1000}}$ To find the cube root of 474552, we need to proceed as follows: On factorising 474552 into prime factors, we get: $474552=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 13 \...
Read More →Degree of the zero polynomial is
Question: Degree of the zero polynomial is (a) 0 (b) 1 (c) any natural number (d) not defined Solution: (d) The degree of zero polynomial is not defined, because in zero polynomial, the coefficient of any variable is zero i.e., $\mathrm{Ox}^{2}$ or $\mathrm{Ox}^{5}$, etc. Hence, we cannot exactly determine the degree of variable....
Read More →Prove that:
Question: Prove that $\left|\begin{array}{ccc}a^{2}+1 a b a c \\ a b b^{2}+1 b c \\ c a c b c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$ Solution: Let LHS $=\Delta=\mid a^{2}+1 \quad a b \quad a c$ $a b \quad b^{2}+1 \quad b c$ $c a \quad c b \quad c^{2}+1 \mid$ $=(a b c) \mid a+\frac{1}{a} \quad b \quad c$ $a \quad b+\frac{1}{b} \quad c$ $\begin{array}{llll}a b c+\frac{1}{c}\end{array} \quad$ [Taking out $a, b$ and $c$ common from $R_{1}, R_{2}$ and $R_{3}$ ] $=(a b c) \mid a+\frac{1}{a} \qua...
Read More →The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°.
Question: The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60, then find the height of the flagstaff[Use $\sqrt{3}=1.732]$ Solution: Let BC and CD be the heights of the tower and the flagstaff, respectively.We have, $\mathrm{AB}=120 \mathrm{~m}, \angle \mathrm{BAC}=45^{\circ}, \angle \mathrm{BAD}=60^{\circ}$ Let $\mathrm{CD}=x$ In $\Delta \mathrm{ABC}$...
Read More →Degree of the polynomial
Question: Degree of the polynomial $4 x^{4}+0 x^{3}+0 x^{5}+5 x+7$ is (a) 4 (b) 5 (c) 3 (d) 7 Solution: (a) Degree of $4 x^{4}+0 x^{3}+0 x^{5}+5 x+7$ is equal to the highest power of variable $x$. Here, the highest power of $x$ is 4 , Hence, the degree of a polynomial is 4 ....
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