SSC CGL 202051)\((a+b+c-d)^2-(a-b-c+d)^2=\space?\)
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 202052)यदि \(x-{1\over x}=11\) है, तो \((x^3-{1\over x^3})\) का मान बताइए ।
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 202056)\({x^2(x-4)^2\over(x+4)^2-4x}\div {(x^2-4x)^3\over(x+4)^2}\times {64-x^3\over16-x^2}\) का मान क्या होगा ?
\(x+4\over x(x-4)\)
\({x^2(x-4)^2\over(x+4)^2-4x}\div {(x^2-4x)^3\over(x+4)^2}\times {64-x^3\over16-x^2}={x^2(x-4)^2\over x^2+16+4x}\times {(x+4)^2\over x^3(x-4)^3}\times {(4-x)(x^2+16+4x)\over(4-x)(4+x)}\) = \(x+4\over x(x-4)\)
SSC CGL 202057)यदि \(x^{4} + x^{2} y^{2} + y^{4} = 273\space\)और \( x^{2} - xy + y^{2} = 13\) है, तो xy का मान क्या होगा ?
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 202058)यदि \(20x^{2} — 30x + 1 = 0\) है, तो \(25x^{2}+\frac{1}{16x^{2}}\) का मान क्या होगा ?
\(53\frac{3}{4}\)
\(20x^2-30x+1=0;\)
Dividing by x
\(20x-30+1/x=0;\)
\(20x+1/x=30;\)
\(5x+1/4x=15/2;\)
\(25x^{2}+\frac{1}{16x^{2}}=
(5x+{1\over4x})^2-2\times5x\times{1\over4x}\)\(=({20x^2+1\over4x})^2-{5\over2}\)
\(=({30x\over4x})^2-{5\over2}= (\frac{15}{2})^2 - \frac{5}{2}
= \frac{225}{4} - \frac{5}{2}
= \frac{215}{4} = 53\frac{3}{4}\)
SSC CGL 202059)यदि \(x^2-2\sqrt5x +1=0\) है, तो \((x^5+{1\over x^5})\)का मान क्या है ?
\(610\sqrt5\)
\(x^2-2\sqrt5x +1=0\) ; ⇒ \(x^2+1=2\sqrt5x\) ;
⇒ \({x^2+1\over x}={2\sqrt5x\over x}\) ; ⇒ \(x+{1\over x}=2\sqrt5\) ___(i) ; \(x^2+{1\over x^2}= 18\) ____(ii) ;
Again, \((x+{1\over x})^3= (2\sqrt5)^3\); ⇒ \(x^3+{1\over x^3}=34\sqrt5\) ___(iii) ;
Multiplying equation (ii) by (iii), we have \((x^2+{1\over x^2})(x^3+{1\over x^3}) = 18\times 34\sqrt5\) ;
⇒ \(x^5+{1\over x^5}+x+{1\over x}=612\sqrt5\) ; \(x^5+{1\over x^5} = 610\sqrt5\)
SSC CGL 202060)यदि \(16a^4 + 36a^2b^2\)\( + 81b^4 = 91\) और \(4a^2 + 9b^2\)\( - 6ab = 13\) है, तो 3ab का मान क्या है ?
\(-{3\over2}\)
\(4a^2 + 9b^2 - 6ab = 13\) ;
⇒ \((4a^2 + 9b^2 - 6ab)^2 = (13)^2\);
\((\because(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac))\)
⇒\( (4a^2)^2 + (9b^2)^2 + (6ab)^2 +2(4a^2.9b^2 - 9b^2.6ab - 6ab.4a^2) = 169\) ;
⇒ \(16a^4 + 36a^2b^2 + 81b^4 + 2(36a^2b^2 - 54ab^3 - 24a^3b) = 169\) ;
⇒ \(36a^2b^2 - 54ab^3 - 24a^3b = \frac{169 - 91}{2}\) ;
⇒ \(6ab(6ab - 9b^2 - 4a^2) = 39\) ;
⇒ 6ab = -3 ; ⇒ 3ab = -3/2