If the function f(x)=2x^3-9ax^2+12a^2x+1, a0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x)=2x^3-9ax^2+12a^2x+1$.Find the derivative: $f'(x)=6x^2-18ax+12a^2$.For local extrema,set $f'(x)=0$,so $6(x^2-3ax+2a^2)=0$,which factors

Vedclass · Accepted Answer

$x^2-6x+8=0$

Vedclass · Answer

(A) Given $f(x)=2x^3-9ax^2+12a^2x+1$.Find the derivative: $f'(x)=6x^2-18ax+12a^2$.For local extrema,set $f'(x)=0$,so $6(x^2-3ax+2a^2)=0$,which factors to $6(x-a)(x-2a)=0$.The roots are $x=a$ and $x=2a$.Given the local maximum at $x=\alpha$ and local minimum at $x=\alpha^2$,we have two cases:Case $1$: $\alpha=a$ and $\alpha^2=2a$. Then $a^2=2a \Rightarrow a(a-2)=0$. Since $a>0$,$a=2$.If $a=2$,the roots are $\alpha=2$ and $\alpha^2=4$. The equation is $(x-2)(x-4)=x^2-6x+8=0$.Case $2$: $\alpha=2a$ and $\alpha^2=a$. Then $(2a)^2=a \Rightarrow 4a^2-a=0 \Rightarrow a(4a-1)=0$. Since $a>0$,$a=1/4$.If $a=1/4$,the roots are $\alpha=1/2$ and $\alpha^2=1/4$. The equation is $(x-1/2)(x-1/4)=x^2-(3/4)x+1/8=0$,or $8x^2-6x+1=0$.Checking the second derivative $f''(x)=12x-18a$:For $a=2$,$f''(2)=24-36=-12  0$ (min). This matches.For $a=1/4$,$f''(1/2)=6-18(1/4)=1.5 > 0$ (min) and $f''(1/4)=3-18(1/4)=-1.5 Thus,the correct equation is $x^2-6x+8=0$.

If the function $f(x)=2x^3-9ax^2+12a^2x+1, a>0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$,then $\alpha$ and $\alpha^2$ are the roots of the equation:

Similar Questions

The number of distinct real roots of the equation $3x^{4} + 4x^{3} - 12x^{2} + 4 = 0$ is ..... .

$A$ rectangle with its sides parallel to the $X$-axis and $Y$-axis is inscribed in the region bounded by the curves $y=x^2-4$ and $y=\frac{4-x^2}{2}$. The maximum possible area of such a rectangle is closest to the integer.

For all real $x$,the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

The minimum value of $4e^{2x} + 9e^{-2x}$ is:

$A$ wire of length $2$ units is cut into two parts,which are bent respectively to form a square of side $x$ units and a circle of radius $r$ units. If the sum of the areas of the square and the circle so formed is minimum,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module