Jadiel Bowers

2023-02-25

Let $f$ be a twice differentiable function such that $g\text{'}\left(x\right)=-f\left(x\right)$ and $f\text{'}\left(x\right)=g\left(x\right)$, $h\left(x\right)={\left(f\left(x\right)\right)}^{2}+{\left(g\left(x\right)\right)}^{2}$. If $h\left(5\right)=11$, then $h\left(10\right)$ is equal to

A. $22$

B. $11$

C. $0$

D. $20$

A. $22$

B. $11$

C. $0$

D. $20$

Gabriel Valencia

Beginner2023-02-26Added 5 answers

Identify the value of $h\left(10\right)$.

Consider the given equation as: $h\left(x\right)={\left(f\left(x\right)\right)}^{2}+{\left(g\left(x\right)\right)}^{2}$

Then, $h\text{'}\left(x\right)=2f\left(x\right).f\text{'}\left(x\right)+2g\left(x\right).g\text{'}\left(x\right)......\left(1\right)$

From the given data

$g\left(x\right)=f\text{'}\left(x\right)g\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)f\text{'}\text{'}\left(x\right)=-f\left(x\right)$

Replace the aforementioned values in the equation.$\left(1\right)$

$h\text{'}\left(x\right)=2f\left(x\right).f\text{'}\left(x\right)-2g\left(x\right).f\left(x\right)h\text{'}\left(x\right)=2f\left(x\right).g\left(x\right)-2g\left(x\right).f\left(x\right)h\text{'}\left(x\right)=0$

Since, $h\left(x\right)$ has some constant value.

Then, the value of $h\left(10\right)$ is given as:

$h\left(10\right)=h\left(5\right)=11h\left(10\right)=11$

Therefore, The correct answer is Option B

Consider the given equation as: $h\left(x\right)={\left(f\left(x\right)\right)}^{2}+{\left(g\left(x\right)\right)}^{2}$

Then, $h\text{'}\left(x\right)=2f\left(x\right).f\text{'}\left(x\right)+2g\left(x\right).g\text{'}\left(x\right)......\left(1\right)$

From the given data

$g\left(x\right)=f\text{'}\left(x\right)g\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)f\text{'}\text{'}\left(x\right)=-f\left(x\right)$

Replace the aforementioned values in the equation.$\left(1\right)$

$h\text{'}\left(x\right)=2f\left(x\right).f\text{'}\left(x\right)-2g\left(x\right).f\left(x\right)h\text{'}\left(x\right)=2f\left(x\right).g\left(x\right)-2g\left(x\right).f\left(x\right)h\text{'}\left(x\right)=0$

Since, $h\left(x\right)$ has some constant value.

Then, the value of $h\left(10\right)$ is given as:

$h\left(10\right)=h\left(5\right)=11h\left(10\right)=11$

Therefore, The correct answer is Option B