Differentiation Question 4

Question 4 - 29 January - Shift 2

Let $f$ and $g$ be twice differentiable functions on $R$ such that

$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$

$f^{\prime}(1)=4 g^{\prime}(1)-3=9$

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

Then which of the following is NOT true ?

(1) $g(-2)-f(-2)=20$

(2) If $-1<x<2$, then $|f(x)-g(x)|<8$

(3) $|f^{\prime}(x)-g^{\prime}(x)|<6 \Rightarrow-1<x<1 \mid$

(4) There exists $x_0 \in(1, \frac{3}{2})$ such that $f(x_0)=g(x_0)$

Show Answer

Answer: (2)

Solution:

Formula: If $f $ and $ g $ are functions of $x$ such that $g^{\prime}(x) = f(x) $ then, Successive differentiation

$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x $ …..$(1)$

$f^{\prime}(1)=4 g^{\prime}(1)-3=9 $ …..$(2)$

$f(2)=3 g(2)=12 $ …..$(3)$

By integrating (1)

$ f^{\prime}(x)=g^{\prime}(x)+6 \frac{x^{2}}{2}+C $

At $x=1$,

$ f^{\prime}(1)=g^{\prime}(1)+3+C $

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

$\therefore f^{\prime}(x)=g^{\prime}(x)+3 x^{2}+3$

Again by integrating,

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

At $x=2$,

$f(2)=g(2)+8+3(2)+D$

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

So, $f(x)=g(x)+x^{3}+3 x-6$

$\Rightarrow f(x)-g(x)=x^{3}+3 x-6$

At $x=-2$,

$\Rightarrow g(-2)-f(-2)=20 \quad$ (Option (1) is true)

Now, for $-1<x<2$

$h(x)=f(x)-g(x)=x^{3}+3 x-6$

$\Rightarrow h^{\prime}(x)=3 x^{2}+3$

$\Rightarrow h(x) \uparrow$

So, $h(-1)<h(x)<h(2)$

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

$\Rightarrow|h(x)|<10 \quad$ (option (2) is NOT true)

Now, $h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=3 x^{2}+ 3$

If $|h^{\prime} (x) | < 6 \Rightarrow | 3x + 3|<6$

$\Rightarrow 3 x^2 + 3 <6$

$\Rightarrow x^2 < 1$

$-1 < x < 1$ (option (3) is True)