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Differentiation

Section: Mathematics

Question : 31 of 79

Marks: +1, -0

Let

f

g

R

f''(x)=g''(x)+6x

f'(1)=4g'(1)-3=9

f(2)=3g(2)=12

$g(-2)-f(-2)=20$
If $-1 < x < 2$ , then $|f(x)-g(x)| < 8$
$|f'(x)-g'(x)| < 6 \Rightarrow -1 < x < 1 \mid$
There exists $x_0 \in \left(1, \;\frac{3}{2}\right)$ such that $f(x_0) = g(x_0)$

Solution:

\;f''(x)=g''(x)+6x

. . . (1)

\;f'(1)=4g'(1)-3=9

. . . (2)

\;f(2)=3g(2)=12

. . . (3)

By integrating (1)

f'(x)=g'(x)+6\;\frac{x^2}{2}+C

\;\text{At}\;\;x=1

\;f'(1)=g'(1)+3+C

\Rightarrow 9=4+3+C \Rightarrow C=3

\therefore f'(x)=g'(x)+3x^2+3

Again by integrating,

\;f(x)=g(x)+\;\frac{3x^3}{3}+3x+D

\;\;\text{At}\;x=2

\;f(2)=g(2)+8+3(2)+D

\;\Rightarrow 12=4+8+6+D \Rightarrow D=-6

So,

f(x)=g(x)+x^3+3x-6

\;\Rightarrow f(x)-g(x)=x^3+3x-6

\;\;\text{At}\;x=-2

\;\Rightarrow g(-2)-f(-2)=20\;\;\;\text{(Option (1) is true)}\;

Now, for

-1 < x,2

\;h(x)=f(x)-g(x)=x^3+3x-6

\;\Rightarrow h'(x)=3x^2+3

\;\Rightarrow h(x) \uparrow

\;\;\text{So,}\;h(-1) < h(x) < h(2)

\;\Rightarrow -10 < h(x) < 8

\;\Rightarrow |h(x)| < 10\;\;\;\text{(option (2) is NOT true)}\;

Now,

h'(x)=f'(x)-g'(x)=3x^2+3

|h'(x)| < 6 \Rightarrow |3x^2+3| < 6

\Rightarrow 3x^2+3 < 6

\Rightarrow x^2 < 1

\Rightarrow -1 < x < 1\;\;

(option (3) is True)

x \in (-1,1) |f'(x)-g'(x)| < 6

option (3) is true and now to solve

\;f(x)-g(x)=0

\;\Rightarrow x^3+3x-6=0

\;h(x)=x^3+3x-6

here,

h(1)=-\text{ve}

and

h\left(\;\frac{3}{2}\right)=+\text{ve}

So there exists

x_0 \in \left(1, \;\frac{3}{2}\right)

such that

f(x_0) = g(x_0)

(option (4) is true)

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