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)
