SSC CGL 202021)If \(a+{1\over a}=5\) then \((a^3+{1\over a^3})\) is :
110
\(a+{1\over a}=5\) ; Cubing both sides, \((a+{1\over a})^3=5^3\); ⇒ \(a^3+{1\over a^3}+3a\times{1\over a}(a+{1\over a})=125\); ⇒ \(a^3+{1\over a^3} = 125-15 = 110\)
SSC CGL 202022)If the value of \((a+b-2)^2+(b+c-5)^2+(c+a-5)^2=0\), then the value of \(\sqrt{(b+c)^a+(c+a)^b-1}\) is :
3
\((a+b-2)^2+(b+c-5)^2+(c+a-5)^2=0\) ; If \(x^2+y^2+z^2=0, then\space x =0=y=z\) ; \((a+b-2)^2=0\); ⇒ a + b = 2; Similarly, b + c = 5 and c + a = 5; therefore b + c = c + a ;⇒ b = a; so a + b = 2; ⇒ a + a = 2; a = 1; a = b =1; Again, c + a = 5; ⇒ c = 4; \(\therefore \sqrt{(b+c)^a+(c+a)^b-1}=\sqrt{(1+4)^1+(4+1)^1-1}=3\)
SSC CGL 202023)If \(3^a=27^b=81^c\) and abc = 144 then the value of \(12({1\over a}+{1\over2b}+{1\over5c})\) is :
\(33\over10\)
\(3^a=27^b=81^c\); ⇒ \(3^a=3^{3b}=3^{4c}\); ⇒ a = 3b = 4c = k (let); (a = k, b = \(k\over3\), c = \(k\over4\)) and abc = 144; ⇒ \(\therefore k\times {k\over3}\times {k\over4}=144\) ; ⇒ k = 12; \(12({1\over a}+{1\over2b}+{1\over5c})= 12({1\over12}+{1\over2\times4}+{1\over5\times3})={33\over10}\)
SSC CGL 202024)If \(a+b+c=9\) and \(ab+bc+ca=-22\), then the value of \(a^3+b^3+c^3-3abc\) is:
1323
a + b + c = 9; ⇒ \((a+b+c)^2=9^2\); ⇒ \(a^2+b^2+c^2+2(ab+bc+ca)=81\); ⇒ \(a^2+b^2+c^2+2\times (-22)=81\); ⇒ \(a^2+b^2+c^2=125\) ; \(a^3+b^3+c^3-3abc= (a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]=9[125-(-22)]=1323\)
SSC CGL 202025)The coefficient of \(x^2\) in \((2x+y)^3\) is:
12y
\((2x+y)^3= (2x)^3+y^3+3(2x)y(2x+y)=8x^3+y^3+12x^2y+6xy^2\); Co-efficient of \(x^2\) = 12y
SSC CGL 202026)\((a+b+c-d)^2-(a-b-c+d)^2=?\)
4a (b + c - d)
Put a=b= c=d=1 in \((a+b+c-d)^2-(a-b-c+d)^2\)= 4..
Now check all the given options value 4 will come in option C only i.e.
= 4a(b + c - d)
SSC CGL 202027)If \(x-{1\over x}=11\), then the value of \((x^3-{1\over x^3})\) is :
1364
\(x-{1\over x}=11\);
Cubing both sides,
\((x-{1\over x})^3= 11^3\) ; ⇒ \(x^3-{1\over x^3}-3\times x\times{1\over x}(x-{1\over x})=1331\) ;
⇒ \(x^3-{1\over x^3}=1331+33\); ⇒ \(x^3-{1\over x^3}=1364\)
SSC CGL 202028)What is the value of \({x^2(x-4)^2\over(x+4)^2-4x}\div {(x^2-4x)^3\over(x+4)^2}\times {64-x^3\over16-x^2}\) ?
\(x+4\over x(x-4)\)
\({x^2(x-4)^2\over(x+4)^2-4x}\div {(x^2-4x)^3\over(x+4)^2}\times {64-x^3\over16-x^2}={x^2(x-4)^2\over x^2+16+4x}\times {(x+4)^2\over x^3(x-4)^3}\times {(4-x)(x^2+16+4x)\over(4-x)(4+x)}\) = \(x+4\over x(x-4)\)
SSC CGL 202029)If \(x^{4} + x^{2} y^{2} + y^{4} = 273\space and \space x^{2} - xy + y^{2} = 13\), then the value of xy is :
4
\(x^{4} + x^{2} y^{2} + y^{4} =( x^{2} +xy + y^{2}) ( x^{2} - xy + y^{2}) \); ⇒ \(273=13( x^{2} +xy + y^{2}) \) ; ⇒
\(( x^{2} + xy + y^{2}) =21\);
\(\therefore ( x^{2} + xy + y^{2}) -( x^{2} - xy + y^{2}) =21-13\); ⇒ 2xy = 8 ; ⇒ xy = 4
SSC CGL 202030)If \(20x^{2} — 30x + 1 = 0\), then what is the value of \(25x^{2}+\frac{1}{16x^{2}}\) :
\(53\frac{3}{4}\)
\(20x^2-30x+1=0;\)
Dividing by x
\(20x-30+1/x=0;\)
\(20x+1/x=30;\)
\(5x+1/4x=15/2;\)
\(25x^{2}+\frac{1}{16x^{2}}=
(5x+{1\over4x})^2-2\times5x\times{1\over4x}\)\(=({20x^2+1\over4x})^2-{5\over2}\)
\(=({30x\over4x})^2-{5\over2}= (\frac{15}{2})^2 - \frac{5}{2}
= \frac{225}{4} - \frac{5}{2}
= \frac{215}{4} = 53\frac{3}{4}\)