Question

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Answer 1

11.

Solution : It is given that the numeric value of x + 1 / x is 5.

On squaring both sides :

= > ( x + 1 / x )^2 = 5^2

= > x^2 + 1 / x^2 + 2( x × 1 / x ) = 25 { We know, ( a + b )^2 = a^2 + b^2 + 2ab }

= > x^2 + 1 / x^2 + 2( 1 ) = 25

= > x^2 + 1 / x^2 + 2 = 25

= > x^2 + 1 / x^2 = 25 - 2

= > x^2 + 1 / x^2 = 23 ...( 1 )

On cubing both sides :

= > ( x + 1 / x )^3 = 5^3

= > x^3 + 1 / x^3 + 3( x × 1 / x )( x + 1 / x ) = 125 { ( a + b )^3 = a^3 + b^3 + 3ab( a + b ) }

= > x^3 + 1 / x^3 + 3( 1 )( 5 ) = 125

= > x^3 + 1 / x^3 + 15 = 125

= > x^3 + 1 / x^3 = 125 - 15

= > x^3 + 1 / x^3 = 110 ...( 2 )

Then,

= > x^3 - 5x^2 + x + 1 / x^3 - 5 / x^2 + 1 / x

= > x^3 + 1 / x^3 - 5x^2 - 5 / x^2 + x + 1 / x

= > ( x^3 + 1 / x^3 ) - 5( x^2 + 1 / x^2 ) + ( x + 1 / x )

Substituting values from above ( 1 ) & ( 2 ) :

= > 110 - 5( 23 ) + 5

= > 115 - 115

= > 0

Hence option ( B ) is correct.

12.

Solution : Given, a + b + c = 0, so b + c = - a

On squaring both sides :

= > ( a + b + c )^2 = 0^2

= > a^2 + b^2 + c^2 + 2( ab + bc + ca ) = 0

= > a^2 + b^2 + c^2 + 2{ ab + ac + bc } = 0

= > a^2 + b^2 + c^2 + 2{ a( b + c ) + bc } = 0 { from above, b + c = - a }

= > a^2 + b^2 + c^2 + 2{ a( - a ) + bc } = 0

= > a^2 + b^2 + c^2 = - 2{ - a^2 + bc }

= > a^2 + b^2 + c^2 = 2( a^2 - bc )

On comparing this with a^2 + b^2 + c^2 = k( a^2 - bc ).

= > 2( a^2 - bc ) = k( a^2 - bc )

= > 2 = k

Hence the required correct option is ( C ).$\:$