
The length of the longest interval in which Rolle's theorem can be applied for the function $f(x) = \left| {{x^2} - {a^2}} \right|$ where $(a > 0)$ is
A) ${\text{2a}}$
B) ${\text{4}}{{\text{a}}^2}$
C) ${\text{a}}\sqrt 2 $
D) ${\text{a}}$
Answer
590.1k+ views
Hint: Rolle’s Theorem is a special case of the mean-value theorem, which states that any real-valued differential equation whose values are equal at two different points must have a stationary point between those points. In general, it states that if a function is continuous on closed interval \[\left[ {a,b} \right]\]and differential in open interval \[\left( {a,b} \right)\]such that \[f\left( a \right) = f\left( b \right),\]then\[f'\left( x \right) = 0\]where\[a \leqslant x \leqslant b\].
To check whether the given function is continuous in a range, we have to check whether the function is polynomial as we know polynomial functions are continuous everywhere in the range\[\left( { - \infty ,\infty } \right)\].
In this question, we need to find the length of the longest interval in which $f(x) = \left| {{x^2} - {a^2}} \right|$ where $(a > 0)$ which can be easily done by using Rolle ’s Theorem.
Complete step-by-step solution
$f(x) = \left| {{x^2} - {a^2}} \right|$ where $(a > 0)$.
Using Rolle's Theorem for the above function we get,
$
f(x) = \left| {{x^2} - {a^2}} \right| \\
f(x) = \left| {(x - a)(x + a)} \right| \\
$
On plotting $f(x)$, we can see that $f$is continuous on $\left[ { - a,a} \right]$, differentiable on $( - a,a)$ and also $f( - a) = f(a)$, and they also exists a point $c = 0$ in $( - a,a)$.
Therefore, the length of the longest interval becomes $2a$.
Hence option A is the correct answer.
Note:In such type of questions which involves Rolle's Theorem having knowledge about the statement, closed intervals and open intervals for which the given function may be defined is needed. Plotting the function gives the required information to answer the question. Here, we have plotted the graph for the function $f(x) = \left| {{x^2} - {a^2}} \right|$ by taking different values of x and corresponding values of the function. It is always advised to the candidates to follow the table before plotting the graph.
To check whether the given function is continuous in a range, we have to check whether the function is polynomial as we know polynomial functions are continuous everywhere in the range\[\left( { - \infty ,\infty } \right)\].
In this question, we need to find the length of the longest interval in which $f(x) = \left| {{x^2} - {a^2}} \right|$ where $(a > 0)$ which can be easily done by using Rolle ’s Theorem.
Complete step-by-step solution
$f(x) = \left| {{x^2} - {a^2}} \right|$ where $(a > 0)$.
Using Rolle's Theorem for the above function we get,
$
f(x) = \left| {{x^2} - {a^2}} \right| \\
f(x) = \left| {(x - a)(x + a)} \right| \\
$
On plotting $f(x)$, we can see that $f$is continuous on $\left[ { - a,a} \right]$, differentiable on $( - a,a)$ and also $f( - a) = f(a)$, and they also exists a point $c = 0$ in $( - a,a)$.
Therefore, the length of the longest interval becomes $2a$.
Hence option A is the correct answer.
Note:In such type of questions which involves Rolle's Theorem having knowledge about the statement, closed intervals and open intervals for which the given function may be defined is needed. Plotting the function gives the required information to answer the question. Here, we have plotted the graph for the function $f(x) = \left| {{x^2} - {a^2}} \right|$ by taking different values of x and corresponding values of the function. It is always advised to the candidates to follow the table before plotting the graph.
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