Find the value
Question: Find the value $x^{2}+6 \sqrt{2} x+10$ Solution: Splitting the middle term, $=x^{2}+5 \sqrt{2} x+\sqrt{2} x+10$ $[\therefore 6 \sqrt{2}=5 \sqrt{2}+\sqrt{2}$ and $5 \sqrt{2} \times \sqrt{2}=10]$ $=x(x+5 \sqrt{2})+\sqrt{2}(x+5 \sqrt{2})$ $=(x+5 \sqrt{2})(x+\sqrt{2})$ $\therefore x^{2}+6 \sqrt{2} x+10$ $=(x+5 \sqrt{2})(x+\sqrt{2})$...
Read More →The area of the triangle formed by the lines
Question: The area of the triangle formed by the linesx= 3,y= 4 andx=yis (a) $1 / 2$ sq. unit (b) 1 sq. unit (c) 2 sq. unit (d) None of these Solution: Given $x=3, y=4$ and $x=y$ We have plotting points as $(3,4)(3,3)(4,4)$ when $x=y$ Therefore, area of $\triangle A B C=\frac{1}{2}($ Base $\times$ Height $)=\frac{1}{2}(A B \times A C)=\frac{1}{2}(1 \times 1)=\frac{1}{2}$ Area of triangle $A B C$ is $\frac{1}{2}$ square units Hence, the correct choice is a...
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Question: Find the value $x^{2}-y^{2}-4 x z+4 z^{2}$ Solution: On rearranging the terms $=x^{2}-4 x z+4 z^{2}-y^{2}$ $=(x)^{2}-2 \times x \times 2 z+(2 z)^{2}-y^{2}$ Using the identity $x^{2}-2 x y+y^{2}=(x-y)^{2}$ $=(x-2 z)^{2}-y^{2}$ Using the identity $p^{2}-q^{2}=(p+q)(p-q)$ $=(x-2 z+y)(x-2 z-y)$ $\therefore x^{2}-y^{2}-4 x z+4 z^{2}=(x-2 z+y)(x-2 z-y)$...
Read More →If the system of equations kx − 5y = 2, 6x + 2y = 7
Question: If the system of equationskx 5y= 2, 6x+ 2y= 7 has no solution, thenk= (a) $-10$ (b) $-5$ (c) $-6$ (d) $-15$ Solution: The given systems of equations are $k x-5 y=2$ $6 x+2 y=7$ If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ Here $a_{1}=k, a_{2}=6, b_{1}=-5, b_{2}=2$ $\frac{k}{6}=\frac{-5}{2}$ $k=\frac{-30}{2}$ $k=-15$ Hence, the correct choice is $d$....
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Question: Find the value $x^{4}+x^{2} y^{2}+y^{4}$ Solution: Adding $x^{2} y^{2}$ and subtracting $x^{2} y^{2}$ to the given equation $=x^{4}+x^{2} y^{2}+y^{4}+x^{2} y^{2}-x^{2} y^{2}$ $=x^{4}+2 x^{2} y^{2}+y^{4}-x^{2} y^{2}$ $=\left(x^{2}\right)^{2}+2 x x^{2}+y^{2}+\left(y^{2}\right)^{2}-(x y)^{2}$ Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$ $=\left(x^{2}+y^{2}\right)^{2}-(x y)^{2}$ Using the identity $p^{2}-q^{2}=(p+q)(p-q)$ $=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$ $\th...
Read More →If the system of equations 2x + 3y = 5, 4x + ky = 10
Question: If the system of equations 2x+ 3y= 5, 4x+ky= 10 has infinitely many solutions, thenk= (a) 1 (b) $1 / 2$ (c) 3 (d) 6 Solution: The given system of equations $2 x+3 y=5$ $4 x+k y=10$ $\frac{a_{1}}{a_{2}}=\frac{2}{4}, \frac{b_{1}}{b_{2}}=\frac{3}{k}, \frac{c_{1}}{c_{2}}=\frac{5}{10}$ For the equations to have infinite number of solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ $\frac{2}{4}=\frac{3}{k}=\frac{5}{10}$ If we take $\frac{2}{4}=\frac{3}{4}$ $2 k=12$ $k=\fr...
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Question: Find the value $x y^{9}-y x^{9}$ Solution: $=x y\left(y^{8}-x^{8}\right)$ $=x y\left(\left(y^{4}\right)^{2}-\left(x^{4}\right)^{2}\right)$ Using the identity $p^{2}-q^{2}=(p+q)(p-q)$ $=x y\left(y^{4}+x^{4}\right)\left(y^{4}-x^{4}\right)$ $=x y\left(y^{4}+x^{4}\right)\left(\left(y^{2}\right)^{2}-\left(x^{2}\right)^{2}\right)$ Using the identity $p^{2}-q^{2}=(p+q)(p-q)$ $=x y\left(y^{4}+x^{4}\right)\left(y^{2}+x^{2}\right)\left(y^{2}-x^{2}\right)$ $=x y\left(y^{4}+x^{4}\right)\left(y^{2}...
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Question: Find the value $a^{2}+2 a b+b^{2}-c^{2}$ Solution: Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$ $=(a+b)^{2}-c^{2}$ Using the identity $p^{2}-q^{2}=(p+q)(p-q)$ $=(a+b+c)(a+b-c)$ $\therefore a^{2}+2 a b+b^{2}-c^{2}=(a+b+c)(a+b-c)$...
Read More →The area of the triangle formed by the lines y = x, x = 6 and y = 0 is
Question: The area of the triangle formed by the linesy=x,x= 6 andy= 0 is (a) 36 sq. units (b) 18 sq. units (c) 9 sq. units (d) 72 sq. units Solution: Given $x=6, y=0$ and $x=y$ We have plotting points as $(6,0)(0,0)(6,6)$ when $x=y$ Therefore, area of $\triangle A B C=\frac{1}{2}($ Base $\times$ Height $)=\frac{1}{2}(C A \times A B)=\frac{1}{2}(6 \times 6)=\frac{1}{2} \times 36=18$ Area of triangle $A B C$ is 18 square units Hence, the correct choice is $b$...
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Question: Find the value $a^{2}+2 a b+b^{2}-c^{2}$ Solution: Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$ $=(a+b)^{2}-c^{2}$ Using the identity $p^{2}-q^{2}=(p+q)(p-q)$ $=(a+b+c)(a+b-c)$ $\therefore a^{2}+2 a b+b^{2}-c^{2}=(a+b+c)(a+b-c)$...
Read More →The area of the triangle formed by the line
Question: The area of the triangle formed by the line $\frac{x}{a}+\frac{y}{b}=1$ with the coordinate axes is (a) $a b$ (b) $2 a b$ (c) $\frac{1}{2} a b$ (d) $\frac{1}{4} a b$ Solution: Given the area of the triangle formed by the line $\frac{x}{a}+\frac{y}{b}=1$ If in the equation $\frac{x}{a}+\frac{y}{b}=1$ either $\mathrm{A}$ and $\mathrm{B}$ approaches infinity, The line become parallel to either $\mathrm{x}$ axis or $\mathrm{y}$ axis respectively, Therefore $x=a$ $y=b$ Area of triangle $=\f...
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Question: Find the value $a^{2}-b^{2}+2 b c-c^{2}$ Solution: $a^{2}-\left(b^{2}-2 b c+c^{2}\right)$ Using the identity $(a-b)^{2}=a^{2}+b^{2}-2 a b$ $=a^{2}-(b-c)^{2}$ Using the identity $a^{2}-b^{2}=(a+b)(a-b)$ $=(a+b-c)(a-(b-c))$ $=(a+b-c)(a-b+c)$ $\therefore a^{2}-b^{2}+2 b c-c^{2}=(a+b-c)(a-b+c)$...
Read More →If a pair of linear equations in two variables is consistent, then the lines represented by two equations are
Question: If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (a) intersecting(b) parallel(c) always coincident(d) intersecting or coincident Solution: If a pair of linear equations in two variables is consistent, then its solution exists. $\therefore$ The lines represented by the equations are either intersecting or coincident. Hence, correct choice is $d$....
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Question: Find the value $4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}$ Solution: Let $(x-y)=x,(x+y)=y$ $=4 x^{2}-12 x y+9 y^{2}$ Splitting the middle term - 12 = 6 - 6 also 4 9 = 6 6 $=4 x^{2}-6 x y-6 x y+9 y^{2}$ $=2 x(2 x-3 y)-3 y(2 x-3 y)$ $=(2 x-3 y)(2 x-3 y)$ $=(2 x-3 y)^{2}$ Substituting x = x - y y = x + y $=[2(x-y)-3(x+y)]^{2}=[2 x-2 y-3 x-3 y]^{2}$ $=(2 x-3 x-2 y-3 y)^{2}$ $=[-x-5 y]^{2}$ $=[(-1)(x+5 y)]^{2}$ $=(x+5 y)^{2} \quad\left[?(-1)^{2}=1\right]$ $\therefore 4(x-y)^{2}-12(x-y)(x+y)+9(x+y)...
Read More →If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b
Question: If 2x 3y= 7 and (a+b)x (a+b 3)y= 4a+brepresent coincident lines, thenaandbsatisfy the equation (a) $a+5 b=0$ (b) $5 a+b=0$ (c) $a-5 b=0$ (d) $5 a-b=0$ Solution: The given system of equations are $2 x-3 y=7$ $(a+b) x-(a+b-3) y=4 a+b$ For coincident lines , infinite number of solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b 2}=\frac{c_{1}}{c_{2}}$ $\Rightarrow \frac{2}{(a+b)}=\frac{-3}{-(a+b-3)}=\frac{7}{4 a+b}$ $\Rightarrow 2(a+b-3)=3(a+b)$ $\Rightarrow 2 a+2 b-6=3 a+3 b$ $\Rightarrow 2 a+2...
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Question: Find the value $a^{2}+b^{2}+2(a b+b c+c a)$ Solution: $=a^{2}+b^{2}+2 a b+2 b c+2 c a$ Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$ We get, $=(a+b)^{2}+2 b c+2 c a$ $=(a+b)^{2}+2 c(b+a)$ Or $(a+b)^{2}+2 c(a+b)$ Taking (a + b) common $=(a+b)(a+b+2 c)$ $\therefore a^{2}+b^{2}+2(a b+b c+c a)=(a+b)(a+b+2 c)$...
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Question: Find the value $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)$ Solution: Let (a - b + c) = x and (b - c + a) = y $=x^{2}+y^{2}+2 x y$ Using the identity $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $=(x+y)^{2}$ Now, substituting x and y $(a-b+c+b-c+a)^{2}$ Cancelling -b, +b + c, -c $=(2 a)^{2}$ $=4 a^{2}$ $\therefore(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)=4 a^{2}$...
Read More →The value of k for which the system of equations has no solution is
Question: The value ofkfor which the system of equations has no solution is $x+2 y=5$ $3 x+k y+15=0$ (a) 6 (b) $-6$ (c) $3 / 2$ (d) None of these Solution: The given system of equation is $x+2 y=5$ $3 x+k y+15=0$ If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ then the equation have no solution. $\frac{1}{3}=\frac{2}{k}=\frac{-5}{15}$ By cross multiply we get $k \times 1=3 \times 2$ $k=6$ Hence, the correct choice is a...
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Question: Find the value $2 a^{2}+2 \sqrt{6} a b+3 b^{2}$ Solution: $=(\sqrt{2} a)^{2}+2 \times \sqrt{2} a \times \sqrt{3} b+(\sqrt{3} b)^{2}$ Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$ $=(\sqrt{2} a+\sqrt{3} b)^{2}$ $=(\sqrt{2} a+\sqrt{3} b)(\sqrt{2} a+\sqrt{3} b)$ $\therefore 2 a^{2}+2 \sqrt{6} a b+3 b^{2}$ $=(\sqrt{2} a+\sqrt{3} b)(\sqrt{2} a+\sqrt{3} b)$...
Read More →Find the value
Question: Find the value $(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x$ Solution: $(x+2)\left(x^{2}+25\right)-10 x(x+2)$ Taking (x + 2) common in both the terms $=(x+2)\left(x^{2}+25-10 x\right)$ $=(x+2)\left(x^{2}-10 x+25\right)$ Splitting the middle term of $\left(x^{2}-10 x+25\right)$ $=(x+2)\left(x^{2}-5 x-5 x+25\right)$ $=(x+2)\{x(x-5)-5(x-5)\}$ $=(x+2)(x-5)(x-5)$ $\therefore(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x=(x+2)(x-5)(x-5)$...
Read More →If the system of equations has infinitely many solutions, then
Question: If the system of equations has infinitely many solutions, then $2 x+3 y=7$ $2 a x+(a+b) y=28$ (a) $a=2 b$ (b) $b=2 a$ (c) $a+2 b=0$ (d) $2 a+b=0$ Solution: Given the system of equations are $2 x+3 y=7$ $2 a x+(a+b) y=28$ For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ $a_{1}=2, a_{2}=2 a, b_{1}=3, b_{2}=a+b$ $\frac{2}{2 a}=\frac{3}{a+b}=\frac{7}{28}$ By cross multiplication we have $\frac{2}{2 a}=\frac{3}{a+b}$ $2(a+...
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Question: Find the value $x(x-2)(x-4)+4 x-8$ Solution: = x(x - 2)(x - 4) + 4(x - 2) Taking (x - 2) common in both the terms =(x - 2){x(x - 4) + 4} $=(x-2)\left\{x^{2}-4 x+4\right\}$ Now splitting the middle term of $x^{2}-4 x+4$ $=(x-2)\left\{x^{2}-2 x-2 x+4\right\}$ $=(x-2)\{x(x-2)-2(x-2)\}$ $=(x-2)\{(x-2)(x-2)\}$ $=(x-2)(x-2)(x-2)$ $=(x-2)^{3}$ $\therefore x(x-2)(x-4)+4 x-8=(x-2)^{3}$...
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Question: Find the value $\left[x^{2}+1 / x^{2}\right]-4[x+1 / x]+6$ Solution: $=x^{2}+1 / x^{2}-4 x-4 / x+4+2$ $=x^{2}+1 / x^{2}+4+2-4 x-4 x$ $=\left(x^{2}\right)+(1 / x)^{2}+(-2)^{2}+2 \times x \times 1 / x+2 \times 1 / x \times(-2)+2(-2) x$ Using identity $x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x=(x+y+z)^{2}$ We get, $=[x+1 / x+(-2)]^{2}$ $=[x+1 / x-2]^{2}$ $=[x+1 / x-2][x+1 / x-2]$ $\therefore\left[x^{2}+1 / x^{2}\right]-4[x+1 / x]+6=[x+1 / x-2][x+1 / x-2]$...
Read More →If am ≠ bl, then the system of equations
Question: Ifambl, then the system of equations $a x+b y=c$ $l x+m y=n$ (a) has a unique solution (b) has no solution (c) has infinitely many solutions (d) may or may not have a solution Solution: Given $a m \neq b l$, the system of equations has $a x+b y=c$ $l x+m y=n$ We know that intersecting lines have unique solution $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ $a_{1} \times b_{2} \neq a_{2} \times b_{1}$ Here $a_{1}=a, a_{2}=l, b_{1}=b, b_{2}=m$ $\frac{a}{l} \neq \frac{b}{m}$ $a \times m ...
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Question: Find the value $6 a b-b^{2}+12 a c-2 b c$ Solution: Taking $b$ common in $\left(6 a b-b^{2}\right)$ and $2 c$ in $(12 a c-2 b c)$ $=b(6 a-b)+2 c(6 a-b)$ Taking $(6 a-b)$ common in the terms $=(6 a-b)(b+2 c)$ $\therefore 6 a b-b^{2}+12 a c-2 b c=(6 a-b)(b+2 c)$...
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