SSC CGL 202041)If \(1-64x^3-12x+px^2=(1-4x)^3\), then the value of p is:
48
\(1-64x^3-12x+px^2=(1-4x)^3\) ⇒ \(1-64x^3-12x+px^2=(1)^3-(4x)^3-3\times1\times4x(1-4x)\) ⇒ \(1-64x^3-12x+px^2=1-64x^3-12x+48x^2\); Therefore, p = 48
SSC CGL 202042)If b + c = ax, c + a = by, a + b = cz then the value of \({1\over9}[{1\over x+1}+{1\over y+1}+{1\over z+1}]\) is:
\(1\over9\)
Put, a = b = c 1 . then x = 2 ; y = 2 ; z = 2. \({1\over9}[{1\over x+1}+{1\over y+1}+{1\over z+1}]= {1\over9}[{1\over3}+{1\over3}+{1\over3}]={1\over9}\)
SSC CGL 202043)If \(a^3+{1\over a^3}=52\) then the value of \(2(a+{1\over a})\) is :
8
\(a^3+{1\over a^3}=a^3+{1\over a^3}+3(a+{1\over a})\) ; ⇒\((a+{1\over a})^3-3(a+{1\over a}) =52\); From options, If, \(a+{1\over a}=4\), then \((4)^3-3\times4 = 52\); \(2(a+{1\over a}) = 2\times4 = 8\)
SSC CGL 202044)If \(x^2-4x+4=0\), then the value of \(16({x^4}-{1\over x^4})\) is :
255
\(x^2-4x+4=0\); ⇒ \((x-2)^2=0\); ⇒ x = 2; \(\therefore 16({x^4}-{1\over x^4}) =16({2^4}-{1\over 2^4})= 255\)
SSC CGL 202045)The coefficient of x in \((x-3y)^3\) is :
\(27y^2\)
\((x-3y)^3= x^3-(3y)^3-3x\times 3y(x-3y)=x^3-27y^3-9x^2y+27xy^2\); Co-efficient of x = \(27y^2\)
SSC CGL 202046)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 202047)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 202048)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 202049)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 202050)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