5 Marks Questions

Question 515 Marks

It is given that the Rolle's theorem holds for the function $f(x) = x^3 + bx^2 + cx,$ $\text{x}\in[1,2]$ at the point $\text{x}=\frac{4}{3},$ the values of b and c.

Answer

So, $f(1) = f(2)$
$\Rightarrow (1)^3 + b(1)^2 + c(1) = (2)^3 + b(2)^2 + c(2)$
$\Rightarrow 1 + b + c = 8 + 4b + 2c$
$\Rightarrow 3b + c + 7 = 0 ....(i)$
And $\text{f}'\Big(\frac{4}{3}\Big)=0$
$\Rightarrow3\Big(\frac{4}{3}\Big)^2+2\text{b}\Big(\frac{4}{3}\Big)+\text{c}=0 [$As, $f'(x) = 3x^2 + 2bx + c]$
$\Rightarrow\frac{16}{3}+\frac{8\text{b}}{3}+\text{c}=0$
$\Rightarrow8\text{b}+3\text{c}+16=0\ ....(\text{ii})$
$(ii) - (i) × 3, $we get
$8b - 9b + 16 - 21 = 0$
$⇒ -b - 5 = 0$
$⇒ b = -5$
Substituting b = -5 in (i), we get
$3(-5) + c + 7 = 0$
$⇒ -15 + c + 7 = 0$
$⇒ c = 8$

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Question 525 Marks

Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}\text{ on }0\leq\text{x}\leq\pi$

Answer

The given function is $\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}.$
Since $\cos\text{x}$ and $\text{e}^\text{x}$ are everywhere continuous and differentiable, being a quotient of these two, f(x) is continuous on $[0,\pi]$ and differentiable on $(0,\pi).$
Also,
$\text{f}(\pi)=\text{f}(0)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in(0,\pi)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}$
$\Rightarrow \text{f}'(\text{x})=\frac{\cos\text{x}-\sin\text{x}}{\text{e}^{\text{x}}}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\frac{\cos\text{x}-\sin\text{x}}{\text{e}^{\text{x}}}=0$
$\Rightarrow\cos\text{x}-\sin\text{x}=0$
$\Rightarrow\tan\text{x}=1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Thus, $\text{c}=\frac{\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Hence, Rolle's theorem is verified.

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Question 535 Marks

Explain if Rolle's theorem is applicable to any one of the following functions.

$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$

Can you say something about the converse of Rolle's Theorem from these functions?

Answer

By Rolle’s theorem, for a function $\text{f}:[\text{a},\text{b}]\rightarrow\text{R},$ if

f is continuous on [a, b],
f is differentiable on (a, b) and
f(a) = f(b)

Then there exists some $\text{c}\in(\text{a},\text{b})$ such that f'(c) = 0
Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$

It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9.
Thus, f(x) is not continuous on [5, 9].
Also, f(5) = [5] = 5 and f(9) = [9] = 9
$\therefore\ \text{f}(5)\neq\text{f}(9)$
The differentiability of f on (5, 9) is checked in the following way.
Let n be an integer such that $\text{n}\in(5,9).$
The left hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}0=0$
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, f is not differentiable on (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.
Hence, Rolle’s theorem is not applicable on $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$

$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$

It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = −2 and x = 2.
Thus, f (x) is not continuous on [−2, 2].
Also, f(-2) = [-2] = -2 and f(2) = [2] = 2
$\therefore\ \text{f}(-2)\neq\text{f}(2)$
The differentiability of f on (-2, 2) is checked in the following way.
Let n be an integer such that $\text{n}\in(-2,2).$
The left hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}0=0$
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, f is not differentiable on (-2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.
Hence, Rolle’s theorem is not applicable on $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$

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MATHS 5 Marks Questions