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$If the function f (x) = x^{4} + b x^{2} + 8 x + 1 has a horizontal tangent and a point x for f^{''} (x) = 0$ $then the value of b is equal to$

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Correct option is D

f′(x)=0 and f′′(x)=0 for same x=x1 (Say) ⇒f′(x)=4x3+2bx+8⇒f′′(x)=12x2+2b⇒f′x1=22x13+bx1+4=0→(1)⇒f′′x1=26x12+b=0→(2)⇒b=−6x12 Substitute in eqn −>12x13+−6x13+4=0⇒4x13=4⇒x13=1⇒x1=1⇒b=−6

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If the function f(x)=x4+bx2+8x+1 has a horizontal tangent and a point x for f′′(x)=0 then the value of b is equal to

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$If the function f (x) = x^{4} + b x^{2} + 8 x + 1 has a horizontal tangent and a point x for f^{''} (x) = 0$ $then the value of b is equal to$