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Question

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

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Solution

## The given function is, f(x)= 1−x+ x 2 1+x+ x 2 (1) Differentiate the above equation, f ′ (x)= ( 1+x+ x 2 ) d dx ( 1−x+ x 2 )−( 1−x+ x 2 ) d dx ( 1+x+ x 2 ) ( 1+x+ x 2 ) 2 f ′ (x)= ( 1+x+ x 2 )( −1+2x )−( 1−x+ x 2 )( 1+2x ) ( 1+x+ x 2 ) 2 = −1+2x−x+2 x 2 − x 2 +2 x 3 −1−2x+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 ) ( 1−1+ ( −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 )= 1−1+1 1+1+1 = 1 3 Thus, correct option is (D).

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