Answer
Verified
37.2k+ 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.
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.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
Other Pages
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
Identify which of the above shown graphs represent class 12 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main