Answer
Verified
468.9k+ views
Hint:For any function \[f\left( x \right)\] if at some point p function is positive and any other point q function is negative that is \[f\left( p \right) > 0\] and \[f\left( q \right) < 0\] then there is at least one root of \[f\left( x \right)\] between p and q.
Complete step-by-step answer:
We are given that for every pair of continuous function \[f,g:\left[ {0,{\text{ }}1} \right] \to R\] such that\[max\{ f\left( x \right):x \in \left[ {0,1} \right]\} = max\{ g\left( x \right):x \in \left[ {0,1} \right]\} \].
Let us take one function at a time. First we take \[f\left( x \right)\].
Let the value of which the function \[f\left( x \right)\] is maximum by ${c_1}$. This is shown as –
\[max\{ f\left( x \right):x \in \left[ {0,1} \right]\} \]=\[f\left( {{c_1}} \right)\]
Similarly, we take the second function \[g\left( x \right)\] and we get,
Let the value of which the function \[g\left( x \right)\] is maximum by ${c_2}$. This is shown as –
\[max\{ g\left( x \right):x \in \left[ {0,1} \right]\} = g({c_2})\]
Now let us take one other function \[h\left( x \right)\] as shown –
\[h\left( x \right)\]=\[f\left( x \right)\]-\[g\left( x \right)\]
Let us find Function \[h\left( x \right)\] for $x = {c_1}$. We get,
\[h\left( {{c_1}} \right)\]=\[f\left( {{c_1}} \right)\]-\[g\left( {{c_1}} \right)\]
Since, \[f\left( {{c_1}} \right)\] is greater than \[g\left( {{c_1}} \right)\] because at $x = {c_1}$ we get maximum value of \[f\left( x \right)\] whereas in case of \[g\left( x \right)\] at $x = {c_1}$ we do not get the maximum value. Therefore,
\[h\left( {{c_1}} \right) > 0\]
Let us find Function \[h\left( x \right)\] for $x = {c_2}$. We get,
\[h\left( {{c_2}} \right)\]=\[f\left( {{c_2}} \right)\]-\[g\left( {{c_2}} \right)\]
Since, \[g\left( {{c_2}} \right)\] is greater than \[f\left( {{c_2}} \right)\] because at $x = {c_2}$ we get maximum value of \[g\left( x \right)\] whereas in case of \[f\left( x \right)\] at \[x = {c_2}\] we do not get the maximum value. Therefore,
\[h\left( {{c_2}} \right) < 0\]
Therefore, for any arbitrary \[c\] which lie in between ${c_1}$ and ${c_2}$ \[h\left( x \right)\] has a root that is –
\[h\left( c \right) = 0\] For \[c \in \left[ {0,{\text{ }}1} \right]\].
Which implies,
\[f(c) - g\left( c \right) = 0\] For \[c \in \left[ {0,{\text{ }}1} \right]\]
\[ \Rightarrow f(c) = g\left( c \right)\]…………. (1)
Now, squaring both sides we get –
\[ \Rightarrow {[f(c)]^2} = {[g\left( c \right)]^2}\]…….. (2)
Multiplying equation (1) by 3 we get,
\[ \Rightarrow 3f(c) = 3g\left( c \right)\]………… (3)
Adding both the equations (2) and (3) we get,
\[{\left[ {f\left( c \right)} \right]^2} + {\text{3}}f\left( c \right) = {\left[ {g\left( c \right)} \right]^2} + 3g\left( c \right)\]…….. (4)
Therefore, from equations (2) and (4) we get that options (A) and (D) are correct options.
So, the correct answer is “Option A and D”.
Note:When we use the property that for any function \[f\left( x \right)\] if at some point p function is positive and any other point q function is negative that is \[f\left( p \right) > 0\] and \[f\left( q \right) < 0\] then there is at least one root of \[f\left( x \right)\] between p and q. It should be noted that the number of roots lying between p and q are atleast one and not equal to one.
Complete step-by-step answer:
We are given that for every pair of continuous function \[f,g:\left[ {0,{\text{ }}1} \right] \to R\] such that\[max\{ f\left( x \right):x \in \left[ {0,1} \right]\} = max\{ g\left( x \right):x \in \left[ {0,1} \right]\} \].
Let us take one function at a time. First we take \[f\left( x \right)\].
Let the value of which the function \[f\left( x \right)\] is maximum by ${c_1}$. This is shown as –
\[max\{ f\left( x \right):x \in \left[ {0,1} \right]\} \]=\[f\left( {{c_1}} \right)\]
Similarly, we take the second function \[g\left( x \right)\] and we get,
Let the value of which the function \[g\left( x \right)\] is maximum by ${c_2}$. This is shown as –
\[max\{ g\left( x \right):x \in \left[ {0,1} \right]\} = g({c_2})\]
Now let us take one other function \[h\left( x \right)\] as shown –
\[h\left( x \right)\]=\[f\left( x \right)\]-\[g\left( x \right)\]
Let us find Function \[h\left( x \right)\] for $x = {c_1}$. We get,
\[h\left( {{c_1}} \right)\]=\[f\left( {{c_1}} \right)\]-\[g\left( {{c_1}} \right)\]
Since, \[f\left( {{c_1}} \right)\] is greater than \[g\left( {{c_1}} \right)\] because at $x = {c_1}$ we get maximum value of \[f\left( x \right)\] whereas in case of \[g\left( x \right)\] at $x = {c_1}$ we do not get the maximum value. Therefore,
\[h\left( {{c_1}} \right) > 0\]
Let us find Function \[h\left( x \right)\] for $x = {c_2}$. We get,
\[h\left( {{c_2}} \right)\]=\[f\left( {{c_2}} \right)\]-\[g\left( {{c_2}} \right)\]
Since, \[g\left( {{c_2}} \right)\] is greater than \[f\left( {{c_2}} \right)\] because at $x = {c_2}$ we get maximum value of \[g\left( x \right)\] whereas in case of \[f\left( x \right)\] at \[x = {c_2}\] we do not get the maximum value. Therefore,
\[h\left( {{c_2}} \right) < 0\]
Therefore, for any arbitrary \[c\] which lie in between ${c_1}$ and ${c_2}$ \[h\left( x \right)\] has a root that is –
\[h\left( c \right) = 0\] For \[c \in \left[ {0,{\text{ }}1} \right]\].
Which implies,
\[f(c) - g\left( c \right) = 0\] For \[c \in \left[ {0,{\text{ }}1} \right]\]
\[ \Rightarrow f(c) = g\left( c \right)\]…………. (1)
Now, squaring both sides we get –
\[ \Rightarrow {[f(c)]^2} = {[g\left( c \right)]^2}\]…….. (2)
Multiplying equation (1) by 3 we get,
\[ \Rightarrow 3f(c) = 3g\left( c \right)\]………… (3)
Adding both the equations (2) and (3) we get,
\[{\left[ {f\left( c \right)} \right]^2} + {\text{3}}f\left( c \right) = {\left[ {g\left( c \right)} \right]^2} + 3g\left( c \right)\]…….. (4)
Therefore, from equations (2) and (4) we get that options (A) and (D) are correct options.
So, the correct answer is “Option A and D”.
Note:When we use the property that for any function \[f\left( x \right)\] if at some point p function is positive and any other point q function is negative that is \[f\left( p \right) > 0\] and \[f\left( q \right) < 0\] then there is at least one root of \[f\left( x \right)\] between p and q. It should be noted that the number of roots lying between p and q are atleast one and not equal to one.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
If the mean of the set of numbers x1x2xn is bar x then class 10 maths JEE_Main
What is the meaning of celestial class 10 social science CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE