Question

1.Given \\f(x)=3x(x-1)^{5}. Compute \\(f\'\'\'(x)\\)
a.\\(f\'\'\'(x)=80(2x-1)^{2}(x-1)\\)
b.\\(f\'\'\'(x)=100(x-1)^{2}(4x-1)\\)
c.\\(f\'\'\'(x)=180(x-1)^{2}(2x-1)\\)
d.\\(2i-j\\)

2.Let \\(f(x)=x^{4}-2x^{2}\\). Find the all \\(c\\) (where \\(c\\) is the interception on the x-axis ) in the interval (-2, 2) such that \\(f\'(x)=0\\). ( Hint use Rolle\'s theorem )
a.(-1, 0, 2)
b.(-1, 0, 1)
c.(-1, 1, 1)
d.(-1, 2, 1)

Answer 1

1) $f(x)=3x(x-1)^{5}$

$f'(x)=3(x-1)^5+15x(x-1)^4$

$f''(x)=15(x-1)^4+15(x-1)^4+60x(x-1)^3$ $=30(x-1)^4+60x(x-1)^3$

$f'''(x)=120(x-1)^3+60(x-1)^3+180x(x-1)^2$

$=60(x-1)^2[2(x-1)+x-1+3x]$

$=60(x-1)^2[6x-3]=180(2x-1)(x-1)^2$

Hence option "C" is correct.

2) $f(x)=x^{4}-2x^{2}$

This is a polynomial function which is differentiable as well as continuous over $R$

so,

a) $f(x)$ is differentiable over the interval $(-2, 2)$

b) $f(x)$ is continuous over the interval $(-2, 2)$

c) $f(a)=f(2)=2^4-2\times 2^2=8$

$f(b)=f(-2)=(-2)^4-2\times (-2)^2=8$

$f(a)=f(b)$

So,Rolle's theorem holds good.

hence, function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval (a, b) such that$f(a) = f(b)$ , then$f'(x) = 0$ for some c with $a ≤ c ≤ b.$

$f'(x)=4x^3-4x=0$

$4x(x^2-1)=4x(x-1)(x+1)=0$

$x=(-1,0,1)$

Hence option "B" is correct.