Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Let f be a twice differentiable such that ${f}''\left( x \right)=-f\left( x \right)$ and ${f}'\left( x \right)=g\left( x \right)$. If $h\left( x \right)={{\left\{ f\left( x \right) \right\}}^{2}}+{{\left\{ g\left( x \right) \right\}}^{2}}$, where h (5) = 11. Find the value of h (10).(a) 1(b) 10(c) 11(d) 100

Last updated date: 21st Apr 2024
Total views: 35.7k
Views today: 1.35k
Verified
35.7k+ views
Hint: First, differentiate the equation ${f}'\left( x \right)=g\left( x \right)$ and compare it with ${f}''\left( x \right)=-f\left( x \right)$ to get ${g}'\left( x \right)=-f\left( x \right)$. Next, differentiate the equation $h\left( x \right)={{\left\{ f\left( x \right) \right\}}^{2}}+{{\left\{ g\left( x \right) \right\}}^{2}}$ and substitute ${f}'\left( x \right)=g\left( x \right)$ and ${g}'\left( x \right)=-f\left( x \right)$ to get ${h}'\left( x \right)=0$. Integrate this to get $h\left( x \right)=c$ where c is a constant. Compare this with $h\left( 5 \right)=11$ to get the final answer.

Complete step-by-step solution -
In this question, we are given that f is a twice differentiable such that ${f}''\left( x \right)=-f\left( x \right)$ and ${f}'\left( x \right)=g\left( x \right)$. $h\left( x \right)={{\left\{ f\left( x \right) \right\}}^{2}}+{{\left\{ g\left( x \right) \right\}}^{2}}$, where h (5) = 11.
We need to find the value of h (10).
First, we will start with ${f}'\left( x \right)=g\left( x \right)$.
On differentiating both sides of this expression, we will get the following:
${f}''\left( x \right)={g}'\left( x \right)$
But we are given that ${f}''\left( x \right)=-f\left( x \right)$. On comparing these two equations, we will get the following:
${g}'\left( x \right)=-f\left( x \right)$ …(1)
Now, we will look into $h\left( x \right)={{\left\{ f\left( x \right) \right\}}^{2}}+{{\left\{ g\left( x \right) \right\}}^{2}}$.
On differentiating both sides of this expression, we will get the following:
${h}'\left( x \right)=2f\left( x \right)\times {f}'\left( x \right)+2g\left( x \right)\times {g}'\left( x \right)$
Her, we are given that ${f}'\left( x \right)=g\left( x \right)$ and we deduced in equation (1) that ${g}'\left( x \right)=-f\left( x \right)$.
Substituting these in the above equation, we will get the following:
${h}'\left( x \right)=2f\left( x \right)\times g\left( x \right)+2g\left( x \right)\times \left( -f\left( x \right) \right)$
$h\left( 5 \right)=11$
${h}'\left( x \right)=0$
Now, we know that a function whose derivative is zero is a constant function.
Using the fact above, we will get the following:
$h\left( x \right)=c$, where c is a constant.
We need to find the value of this c.
We are given that $h\left( 5 \right)=11$. On comparing this with the above equation, we will get the following:
c = 11
So, $h\left( 10 \right)=11$
Hence, option (c) is correct.

Note:- In this question, it is very important to identify the functions which will be beneficial to differentiate. You have to differentiate the right functions in order to get an expression which can be used further in the solution.