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If $f(x)$ and $g(x)$ are twice differentiable. functions on (0,3)
satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x), f^{\prime}(1)=4, g^{\prime}(1)=6, \quad f(2)=3$
$g(2)=9,$ then $f(1)-g(1)$ is
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satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x), f^{\prime}(1)=4, g^{\prime}(1)=6, \quad f(2)=3$
$g(2)=9,$ then $f(1)-g(1)$ is
Solution:
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Verified Answer
The correct answer is:
-4
According to question, $f^{\prime \prime}(x)=g^{\prime \prime}(x)$
Integrating w.r.t. $x$, we get $\begin{array}{ll} & f^{\prime}(x)=g^{\prime}(x)+c_{1} \\ \text { Put } & x=1 \Rightarrow f^{\prime}(1)=g^{\prime}(1)+C_{1} \\ \Rightarrow & 4=6+C_{1} \\ \therefore & C_{1}=-2 \\ \therefore \quad & f^{\prime}(x)=g^{\prime}(x)-2\end{array}$
Again, integrating w.r.t. $x$, we get $f(x)=g(x)-2 x+C_{2}$
$\Rightarrow \quad 3=9-4+C_{2} \Rightarrow C_{2}=-2$
$\therefore \quad f(x)=g(x)-2 x-2$
Put $x=1,$ we get $f(1)-g(1)=-2(1)-2=-4$
Integrating w.r.t. $x$, we get $\begin{array}{ll} & f^{\prime}(x)=g^{\prime}(x)+c_{1} \\ \text { Put } & x=1 \Rightarrow f^{\prime}(1)=g^{\prime}(1)+C_{1} \\ \Rightarrow & 4=6+C_{1} \\ \therefore & C_{1}=-2 \\ \therefore \quad & f^{\prime}(x)=g^{\prime}(x)-2\end{array}$
Again, integrating w.r.t. $x$, we get $f(x)=g(x)-2 x+C_{2}$
$\Rightarrow \quad 3=9-4+C_{2} \Rightarrow C_{2}=-2$
$\therefore \quad f(x)=g(x)-2 x-2$
Put $x=1,$ we get $f(1)-g(1)=-2(1)-2=-4$
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