CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

28.For all real values of x, the minimum value of1S(A) 0(B) 1(C) 3D) 3

Open in App
Solution

The given function is,

f(x)= 1x+ x 2 1+x+ x 2 (1)

Differentiate the above equation,

f (x)= ( 1+x+ x 2 ) d dx ( 1x+ x 2 )( 1x+ x 2 ) d dx ( 1+x+ x 2 ) ( 1+x+ x 2 ) 2 f (x)= ( 1+x+ x 2 )( 1+2x )( 1x+ x 2 )( 1+2x ) ( 1+x+ x 2 ) 2 = 1+2xx+2 x 2 x 2 +2 x 3 12x+x+2 x 2 x 2 2 x 3 ( 1+x+ x 2 ) 2 f (x)= 2+2 x 2 ( 1+x+ x 2 ) 2 (2)

For maximum and minimum value,

f ( x )=0 2( 1 x 2 ) ( 1+x+ x 2 ) 2 =0 2( 1 x 2 )=0 x=±1

Differentiate both sides of equation (2).

f ( x )= 2[ ( 1+x+ x 2 ) 2 ( 2x )( x 2 1 )( 2 )( 1+x+ x 2 )( 1+2x ) ] ( 1+x+ x 2 ) 4 = 4( 1+x+ x 2 )[ ( 1+x+ x 2 )( x )( x 2 1 )( 1+2x ) ] ( 1+x+ x 2 ) 4 = 4[ x+ x 2 + x 3 x 2 2 x 3 +1+2x ] ( 1+x+ x 2 ) 3 = 4( 1+3x x 3 ) ( 1+x+ x 2 ) 3

Since x=±1, therefore, at x=1,

f ( 1 )= 4( 1+3( 1 ) ( 1 ) 3 ) ( 1+1+ ( 1 ) 2 ) 3 = 4( 3 ) ( 3 ) 3 = 4 9 >0

At x=1,

f ( 1 )= 4( 1+3( 1 ) ( 1 ) 3 ) ( 11+ ( 1 ) 2 ) 3 = 4( 1 ) ( 1 ) 3 =4 <0

Since the value is minimum at x=1, therefore the minimum value is given by,

f( 1 )= 11+1 1+1+1 = 1 3

Thus, correct option is (D).


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image