Question 15 Marks
Find a point on the curve $y = x^3 + 1$ where the tangent is parallel to the chord joining (1, 2) and (3, 28).
AnswerLet, $f(x) = x^2+ 1$ The tangent to the curve is parallel to the chord joining the points (1, 2) and (3, 28).
Assume that the chord joins the points (a, f(a)) and (b, f(b)).
$\therefore a = 1, b = 3$ The polynomial function is everywhere continuous and differentiable.
So, $f(x) = x^2+ 1$ is continuous on [1, 3] and differentiable on (1, 3).
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exists $\text{c}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Now, $f(x) = x^2+ 1 \Rightarrow f'(x) = 3x^2, $
$f(1) = 2, f(3) = 28$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
$\Rightarrow3\text{x}^2=\frac{26}{2}$ $\Rightarrow3\text{x}^2=13$
$\Rightarrow\text{x}=\pm\sqrt{\frac{13}{3}}$
Thus, $\text{c}=\sqrt{\frac{13}{3}}$
such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Clearly, $\text{f}(\text{c})=\bigg[\Big(\frac{13}{3}\Big)^{\frac{3}{2}}+1\bigg]$
Thus, c, f(c), i.e. $\bigg(\sqrt{\frac{13}{3}},1+\Big(\frac{13}{3}\Big)^{\frac{3}{2}}\bigg),$
is a point on the given curve where the tangent is parallel to the chord joining the points (1, 2) and (3, 28).
View full question & answer→Question 25 Marks
Find a point on the curve $y = x^2 + x,$ where the tangent is parallel to the chord joining (0, 0) and (1, 2).
AnswerLet, $f(x) = x^2 + x $ The tangent to the curve is parallel to the chord joining the point (0, 0) and (1, 2).
Assume that the chord joins the points (a, f(a)) and (b, f(b)).
$\therefore$ a = 0, b = 1 The polynomial function is everywhere continuous and differentiable.
So, $f(x) = x^2 + x$ is continuous on [0, 1] and differentiable on (0, 1).
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exists $\text{c}\in(0,1)$
such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Now, $f(x) = x2 + x$
$ ⇒ f'(x) = 2x + 1, f(1) = 2, f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow2\text{x}+1=\frac{2-0}{1-0}$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(0,1)$
such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}.$
Clearly,
$\text{f}(\text{c})=\Big(\frac{1}{2}\Big)^2+\frac{1}{2}=\frac{3}{4}.$
Thus, (c, f(c)), i.e. $\Big(\frac{1}{2},\frac{3}{4}\Big),$ is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).
View full question & answer→Question 35 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = (x - 1) (x - 2)^2$ on $[1, 2]$
AnswerGiven:$f(x) = (x - 1)(x - 2)^2$
i.e. $f(x) = x^3 + 4x - 4x^2 - x^2 - 4 + 4x$
i.e. $f(x) = x^3- 5x^2+ 8x - 4$
We know that a polynominal function is everywhere derivable and hence continuous.
So, being a polynomial function, $f(x)$ is continuous and derivable on $[1, 2]$
Also,
$f(1) = (1)^3 - 5(1)^2 + 8(1) - 4 = 0$
$f(2) = (2)^3 - 5(2)^2 + 8(2) - 4 = 0$
$\therefore f(1) = f(2) = 0$
Thus, all the continuous of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(1,2)$ such that $f'(c) = 0.$
We have,
$f(x) = x^3+ - 5x^2-^8x - 4$
$\Rightarrow f'(x) = 3x^2 + 8 - 10x$
$\therefore f'(x) = 0 \Rightarrow 3x^2- 10x + 8 = 0$
$\Rightarrow 3x^2- 6x - 4x + 8 = 0$
$\Rightarrow 3x(x - 2) - 4(x - 2) = 0$
$\Rightarrow (x - 2)(3x - 4)$
$\Rightarrow\text{x}=2,\frac{4}{3}$
Thus, $\text{c}\in(1,2)$ such that $f'(c) = 0.$
Hence, Rolle's theorem is verified.
View full question & answer→Question 45 Marks
Find a point on the parabola $y = (x − 3)^2,$ where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1).$
AnswerLet,$ f(x) = (x − 3)^2 = x^2 -6x + 9$
The tangent to the curve is parallel to the chord joining the points $(3, 0)$ and $(4, 1).$
Assume that the chord joins the points $(a, f(a))$ and $(b, f(b)).$
$\therefore a = 3, b = 4$ The polynomial function is everywhere continuous and differentiable.
So, $f(x) = x^2 -6x + 9$ is continuous on $(3, 4)$ and differentiable on $(3, 4).$
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exists $\text{c}\in(3,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(3)}{4-3}$
Now, $f(x) = x^2 -6x + 9 $
$\Rightarrow f'(x) = 2x - 6,$
$ f(3) = 0, f(4) = 1$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(3)}{4-3}$
$\Rightarrow2\text{x}-6=\frac{1-0}{4-3}$
$\Rightarrow2\text{x}=7$
$\Rightarrow\text{x}=\frac{7}{2}$
Thus, $\text{c}=\frac{7}{2}\in(3,4)$
such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(3)}{4-3}$
Clearly, $\text{f}(\text{c})=\Big(\frac{7}{2}-3\Big)^2=\frac{1}{4}$
Thus, $c, f(c),$ i.e. $\frac{7}{2},\frac{1}{4},$ is a point on the given curve where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1).$
View full question & answer→Question 55 Marks
Verify the hypothesis and conclusion of Lagrange's mean value theorem for the function
$\text{f}(\text{x})=\frac{1}{4\text{x}-1},1\leq\text{x}\leq4.$
AnswerThe given function $\text{f}(\text{x})=\frac{1}{4\text{x}-1}.$ Clearly, f(x) is does not exist for x = 0 Since for each $\text{x}\in[1,4],$ the function attains a unique definite value, f(x) is continuous on [1, 4].Also, $\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$ exists for all $\text{x}\in[1,4],$
Thus, both the conditions of Lagrange's mean value theorem are verified. Concequently, there exists some $\text{c}\in[1,4]$ such that$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$
$=\frac{\text{f}(4)-\text{f}(1)}{3}$ Now, $\text{f}(\text{x})=\frac{1}{4\text{x}-1}\Rightarrow\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$ $\text{f}(4)=\frac{1}{15},\text{f}(1)=\frac{1}{3}$ $\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$ $\Rightarrow\text{f}'(\text{x})=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}==\frac{-4}{45}$ $\Rightarrow\frac{-4}{(4\text{x}-1)^2}=\frac{-4}{45}$ $\Rightarrow(4\text{x}-1)^2=45$ $\Rightarrow16\text{x}^2-8\text{x}-44=0$ $\Rightarrow4\text{x}^2-2\text{x}-11=0$ $\Rightarrow\text{x}=\frac{1}{4}\big(1+3\sqrt{5}\big)$ Thus, $\text{c}=\frac{1}{4}\big(1+3\sqrt{5}\big)\in(1,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}.$ Hence, Lagrange's theorem is verified.
View full question & answer→Question 65 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
AnswerHere, $f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
$f(x)$ is continuous is $[-1, 2]$ and differentiable in $(-1, 2)$ as it is a polynomial functions.
Now,
$f(-1) = (1-1)(-1-2) = 0$
$f(2) = (4-1)(2-2) = 0$
$\Rightarrow f(-1) = f(2)$
So, Rolle's theorem is applicable on $f(x)$ is $[-1, 2]$ therefore, we have to show that there exist a $\text{c}\in(-1,2)$ such that $f'(c) = 0$
Now,
$f(x) = (x^2- 1)(x - 2)$
$f'(x) = 2x(x - 2) + (x^2 - 1)$
$= 2x^2 - 4 + x^2 - 1$
$f'(x) = 3x^2 - 5$
Now,
$f'(c) = 0$
$\Rightarrow 3x^2 - 5 = 0$
$\Rightarrow\text{x}=-\sqrt{\frac{5}{3}}$ or $\text{x}=\sqrt{\frac{5}{3}}\in(-1,2)$
Thus, Rolle's theorem is verified.
View full question & answer→Question 75 Marks
Let C be a curve defined parametrically as $\text{x}=\text{a}\cos^3\theta,\text{y}=\text{a}\sin^3\theta,0\leq\theta\leq\frac{\pi}{2}.$ Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a).
AnswerAs, $\text{x}=\text{a}\cos^3\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta$
And, $\text{y}=\text{a}\sin^3\theta$
$\Rightarrow\frac{\text{dy}}{\text{d}\theta}=3\text{a}\sin^2\theta\cos\theta$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{3\text{a}\sin^2\theta\cos\theta}{-3\text{a}\cos^2\theta\sin\theta}=-\tan\theta$
For the tangent to be parallel to the chord joining the points (a, 0) and (0, a).
$\frac{\text{dy}}{\text{dx}}=\frac{0-\text{a}}{\text{a}-0}$
$\Rightarrow-\tan\theta=-1$
$\Rightarrow\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
Now,
$\text{x}=\text{a}\cos^3\frac{\pi}{4}=\text{a}\Big(\frac{1}{\sqrt{2}}\Big)^3=\frac{\text{a}}{2\sqrt2}$ and
$\text{y}=\text{a}\sin^3\frac{\pi}{4}=\text{a}\Big(\frac{1}{\sqrt{2}}\Big)^3=\frac{\text{a}}{2\sqrt2}$
So, the point P on the curve C is $\Big(\frac{\text{a}}{2\sqrt2},\frac{\text{a}}{2\sqrt2}\Big).$
View full question & answer→Question 85 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x(x - 2)^2$ on the interval $[0, 2]$
AnswerGiven function is $f(x) = x(x^- 2)^2.$ Which can be rewritten as $f(x) = x^3 - 4x^2 + 4x.$
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function $f(x) $ is continuous and derivable on $[0, 2].$
Also,
$f(0) = f(2) = 0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,2]$ such that $ f'(c) = 0.$
We have
$f(x) = x^3- 4x^2 + 4x$
$\Rightarrow f'(x) = 3x^2- 8x + 4$
When, $f'(x) = 0$
$3x^2- 8x + 4 = 0$
$\Rightarrow 3x^2- 6x - 2x + 4 = 0$
$\Rightarrow 3x(x - 2) - 2(x - 2) = 0$
$\Rightarrow (x - 2)(3x - 2)$
$\Rightarrow\text{x}=2,\frac{2}{3}$
Thus, $\text{c}=\frac{2}{3}\in(0,2)$ such that $f'(c) = 0.$$$
Hence, Rolle's theorem is verified.
View full question & answer→Question 95 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\tan^{-1}\text{x}\text{ on }[0,1]$
AnswerWe have,$\text{f}(\text{x})=\tan^{-1}\text{x}$
Clearly, f(x) is continuous on 0, 1 and derivable on 0, 1
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exist some $\text{c}\in-3,4$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$\text{f}(\text{x})=\tan^{-1}\text{x}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}^2},\text{f}(1)=\frac{\pi}{4},\text{f}(0)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\pi}{4}-0$
$\Rightarrow49\Big(\frac{\pi}{4}-1\Big)=\text{x}^2$
$\Rightarrow\text{x}=\pm\sqrt{\frac{4-\pi}{\pi}}$
Thus, $\text{c}=\sqrt{\frac{4-\pi}{\pi}}\in(0,1)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 105 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x^2+ 5 x + 6$ on the interval $[-3, -2]$
AnswerHere, $ f(x) = x^2+ 5 x + 6$ on $[-3, -2]$
$f(x)$ is continuous is $[-3, -2]$ and $f(x)$ is differentiable is $(-3, -2)$ since it is a polynomial function.
Now,
$f(x) = x^2+ 5x + 6$
$f(-3) = (-3)^2+5(-3) + 6$
$= 9 - 15 + 6$
$f(-3) = 0 ....(i)$
$f(-2) = (-2)^2+ 5(-2) + 6$
$= 4 - 10 + 6$
$f(-2) = 20 ....(ii)$
From equation $(i)$ and $(ii),$
$f(-3) = f(-2)$
So, Rolle's theorem is applicable is $[-3, -2],$ we have to show that
$f'(c) = 0$ as $\text{c}\in (-3,-2)$
Now,
$f(x) = x^2 + 5x + 6$
$f'(x) = 2x + 5$
$\Rightarrow f'(c) = 0$
$2c + 5 = 0$
$\text{c}=\frac{-5}{2}\in(-3,-2)$
So, Rolle's theorem is verified.
View full question & answer→Question 115 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}\text{ on }[0,\pi]$
AnswerHere,$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}\text{ on }[0,\pi]$
We know that sine function is continuous and differentiable every where, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$
Now,
$\text{f}(0)=2\sin0+\sin0=0$
$\text{f}(\pi)=2\sin\pi+\sin2\pi=0$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0.
Now,
$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}$
$\text{f}'(\text{x})=2\cos\text{x}+2\cos2\text{x}$
Now,
$\text{f}'(\text{c})=0$
$2\cos\text{c}+2\cos2\text{c}=0$
$\Rightarrow2(\cos\text{c}+2\cos^2\text{c}-1)=0$
$\Rightarrow(2\cos^2+2\cos\text{c}-\cos\text{c}-1)=0$
$\Rightarrow(2\cos\text{c}-1)(\cos\text{c}+1)=0$
$\Rightarrow\cos\text{c}=\frac{1}{2},\cos\text{c}=-1$
$\Rightarrow\tan\text{c}=1$
$\text{c}=\frac{\pi}{3}\in(0,\pi),\text{c}=\pi$
Hence, Rolle's theorem is verified.
View full question & answer→Question 125 Marks
If the value of c prescribed in Roll's theorem for the function$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$ on the interval $\big[0,2\sqrt3\big]$ is $\frac{3}{4},$ write the value of n (a positive integers).
AnswerWe have,
$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$
Differentiating the given function with respect to x, we get
$\text{f}'(\text{x})=2\big[\text{xn}(\text{x}-3)^{\text{n}-1}+(\text{x}-3)^{\text{n}}\big]$
$\Rightarrow\text{f}'(\text{x})=2(\text{x}-3)^{\text{n}}\Big[\frac{\text{xn}}{(\text{x}-3)}+1\Big]$
$\Rightarrow\text{f}'(\text{c})=2(\text{c}-3)^{\text{n}}\Big[\frac{\text{cn}}{(\text{c}-3)}+1\Big]$
Given:
$\text{f}'\Big(\frac{3}{4}\Big)=0$
$\therefore\ 2-\Big(\frac{9}{4}\Big)^{\text{n}}\Bigg[\frac{\frac{3}{4}\text{n}}{\big(\frac{-9}{4}\big)}+1\Bigg]=0$
$\Rightarrow2-\Big(\frac{9}{4}\Big)^{\text{n}}\Big[\frac{-\text{n}}{3}+1\Big]=0$
$\Rightarrow\Big[\frac{-\text{n}}{3}+1\Big]=0$
$\Rightarrow-\text{n}+3=0$
$\Rightarrow\text{n}=3$
View full question & answer→Question 135 Marks
At what points on the following curves, is the tangent parallel to x-axis?
$y = 12(x + 1)(x - 2)$ on $[-1, 2]$
AnswerLet $f(x) = 12(x + 1)(x - 2) ...(1)$
$\Rightarrow f(x) = 12(x^2 - x - 2)$
$\Rightarrow f(x) = 12x^2 - 12x - 24$
Since f(x) is a polynomial function, f(x) is continuous on [-1, 2] and differentiable on (1, 2).
Also,
f(2) = f(-1) = 0
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point $\text{c}\in(-1,2)$ for which f'(c) = 0.
But $\text{f}'(\text{c})=0$
$\Rightarrow24\text{c}-12=0$
$\Rightarrow\text{c}=\frac{1}{2}$
$\therefore\ \text{f}(\text{c})=\text{f}\Big(\frac{1}{2}\Big)=-12\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)=-27$
By the geometrical interepretetion of Rolle's theorem, $\Big(\frac{1}{2}\Big),-27$ is the point on y = 12(x + 1)(x - 2) where the tangent is parallel to the x-axis.
View full question & answer→Question 145 Marks
Verify Rolle's theorem for the following function on the indicated intervals
$f(x) = x^2 - 8x + 12$ on $[2, 6]$
AnswerGiven: $f(x) = x^2 - 8x + 12$ We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on [2, 6].
$f(2) = (2)^2 - 8(2) + 12 = 4 - 16 + 12 = 0$
$f(6) = (6)^2 - 8(6) + 12 = 36 - 48 + 12 = 0$
$\therefore f(2) = f(6) = 0$
Thus, all the conditions of rolle's theorem are satisfied.
Now, we have to show that there exist $\text{c}\in(2, 6)$ such that f'(c) = 0
We have
$f(x) = x^2 - 8x + 12$
$\Rightarrow f'(x) = 2x - 8$
$\therefore f'(x) = 0$
$\Rightarrow 2x - 8 = 0$
$\Rightarrow x = 4$
Thus, $\text{c}=4\in(2,6)$ such that f'(c) = 0
View full question & answer→Question 155 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{\text{x}}\sin{\text{x}}\text{ on }[0,\pi]$
AnswerThe given function is $\text{f}(\text{x})=\text{e}^{\text{x}}\sin{\text{x}}$
Since $\sin\text{x}\ \&\ \text{e}^{\text{x}}$ are everywhere continuous and differentiable.
Therefore, being a product 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})=\text{e}^{\text{x}}\sin{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})=0$
$\Rightarrow\sin\text{x}+\cos\text{x}=0$
$\Rightarrow\tan\text{x}=-1$
$\Rightarrow\text{x}=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
Since $\text{c}=\frac{3\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Thus, Rolle's theorem verified.
View full question & answer→Question 165 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\sqrt{25-\text{x}^2}\text{ on }[-3,4]$
AnswerWe have,$\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
Here, f(x) will exist, if
$25-\text{x}^2\geq0$
$\Rightarrow\text{x}^2\leq25$
$\Rightarrow-5\leq\text{x}\leq5$
Since for each $\text{x}\in[-3,4],$ the function f(x) attains a unique definite value.
So, f(x) is continuous on [-3, 4]
Also, $\text{f}'(\text{x})=\frac{1}{2\sqrt{25-\text{x}^2}}(-2\text{x})=\frac{-\text{x}}{\sqrt{25-\text{x}^2}}$ exists for all $\text{x}\in(-3,4)$
So, f(x) is differentiable on (-3, 4).
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exist some $\text{c}\in(-3,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(-3)}{4+3}=\frac{\text{f}(4)-\text{f}(-3)}{7}$
Now, $\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
$\text{f}'(\text{x})=\frac{-\text{x}}{\sqrt{25-\text{x}^2}},\text{f}(-3)=4,\text{f}(4)=3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(-3)}{4+3}$
$\Rightarrow\frac{-\text{x}}{\sqrt{25-\text{x}^2}}=\frac{3-4}{7}$
$\Rightarrow49\text{x}^2=25-\text{x}^2$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt2}$
Thus, $\text{c}=\pm\frac{1}{\sqrt2}\in(-3,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(-3)}{4-(-3)}.$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 175 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^2 - 2x + 4$ on $[1, 5]$
AnswerWe have,$f(x) = x^2 - 2x + 4$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [1, 5] and differentiable on (1, 5).
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in(1,5)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-1)}{5-1}=\frac{\text{f}(5)-\text{f}(-1)}{4}$
Now, $f(x) = x^2 - 2x + 4$
$\Rightarrow f'(x) = 2x - 2$
$\Rightarrow f(5) = 25 - 10 + 4 = 19$
$\Rightarrow f(1) = 1 - 2 + 4 = 3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(5)-\text{f}(-1)}{4}$
$\Rightarrow2\text{x}-2=\frac{19-3}{4}$
$\Rightarrow2\text{x}-2-4=0$
$\Rightarrow\text{x}=\frac{6}{2}=3$
Thus, $\text{c}=3\in(1,5)$ such that $\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-1)}{5-1}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 185 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$
AnswerHere,$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$
f(x) attains a unique value for each $\text{x}\in[1,3],$ so it is continuous
$\text{f}'(\text{x})=1-\frac{1}{\text{x}^2}$ is defined for each $\text{x}\in(1,3)$
⇒ f(x) is differentiable in (1,3), so Lagrange's mean value theorem is applicable, so there exist a point $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{\Big(3+\frac{1}{3}-(1+1)\Big)}{2}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{\frac{10}{3}-2}{2}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{4}{3\times2}$
$\Rightarrow1-\frac{2}{3}=\frac{1}{\text{c}^2}$
$\Rightarrow\text{c}^2=3$
$\text{c}=\sqrt{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
So, Lagrange's mean value theorem is verified.
View full question & answer→Question 195 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\log(\text{x}^2+2)-\log3\text{ on }[-1,1]$
AnswerThe given function is $\text{f}(\text{x})=\log(\text{x}^2+2)-\log3,$ which can be rewritten as$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$
We know that, logarithmic function is differentiable and so continuous in its domain $\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$ is continuous is $[-1,1]$ and differentiable is $(-1,1).$
Also,
f(1) = f(-1) = 0
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have show that there must exists $\text{c}\in(-1,1)$ such that f'(c) = 0.
We have,
$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$
$\text{f}'(\text{x})=\frac{3(2\text{x})}{\text{x}^2+2}=\frac{6\text{x}}{\text{x}^2+2}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\frac{6\text{x}}{\text{x}^2+2}=0$
$\Rightarrow\text{x}=0$
Thus, $\text{c}=0\in(-1,1)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
View full question & answer→Question 205 Marks
Find the points on the curve $y = x^3 - 3x,$ where the tangent to the curve is parallel to the chord joining (1, -2) and (2, 2).
AnswerHere,$y = x^3 − 3x$
y is polynomial function, so it is continuous and differentiable, so Lagrange's mean value theorem is applicable thus there exists a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow3\text{c}^2-3=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow3\text{c}^2-3=\frac{2+2}{1}$
$\Rightarrow3\text{c}^2=7$
$\Rightarrow\text{c}=\pm\sqrt{\frac{7}{3}}$
$\Rightarrow\text{y}=\pm\frac{2}{3}\sqrt{\frac{7}{3}}$
So, $(\text{c},\text{y})=\Big(\pm\sqrt{\frac{7}{3}},\pm\frac{2}{3}\sqrt{\frac{7}{3}}\Big)$ is the required point.
View full question & answer→Question 215 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
Answerf(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
We know that, polynomial function is continuous and differentiable everywhere. So, f(x) is continuous in [0, 4] and differentiable in (0, 4). So Lagrange's mean value theorem is applicable. Thus, there exist a point $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow(\text{c}-1)(\text{c}-2)(\text{c}-3)+(\text{c}-1)(\text{c}-3)\\=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4-1}$
$\Rightarrow\text{c}^2-3\text{c}+2+\text{c}^2+5\text{c}+6+\text{c}^2-4\text{c}+3=\frac{6+6}{4}$
$\Rightarrow3\text{c}^2-12\text{c}+11=3$
$\Rightarrow3\text{c}^2-12\text{c}+8=0$
$\Rightarrow\text{c}=\frac{-(-12)\pm\sqrt{144-4\times3\times8}}{6}$
$\Rightarrow\text{c}=\frac{12\pm\sqrt{48}}{6}$
$\Rightarrow\text{c}=2\pm\frac{2\sqrt3}{3}\in(0,4)$
$\Rightarrow\text{c}=2\pm\frac{2}{\sqrt3}\in(0,4)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 225 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
Since $\sin\text{x}$ and $\cos\text{x}$ are everywhere continuous and differentiable $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big]$ and differentiable on $\Big(0,\frac{\pi}{2}\Big).$
Also,
$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=1$
Thus, f(x) satisfies all the conditionss of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
$\Rightarrow\text{f}'(\text{x})=4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\sin^3\text{x}\cos\text{x}-\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\tan^3\text{x}-\tan\text{x}=0$
$\Rightarrow\tan\text{x}(\tan^2\text{x}-1)=0$
$\Rightarrow\tan\text{x}=0,\tan^2\text{x}=1$
$\Rightarrow\tan\text{x}=0,\tan\text{x}=\pm1$
$\Rightarrow\text{x}=0,\text{x}=\frac{\pi}{4},\frac{3\pi}{4}$
Since $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c)=0
Hence, Rolle's theorem is verified.
View full question & answer→Question 235 Marks
Verify Rolle's theorem for the following function on the indicated intervals
$f(x) = x(x - 1)^2$^ on $[0, 1]$
AnswerGiven $f(x) = x(x - 1)^2$
$\Rightarrow f(x) = x(x^2 - 2x + 1)$
$\therefore f(x) = (x^3- 2x^2+ x)$
We know that a polynominal function is everywhere derivable and hence continuous.
So, being a polynomial function, f(x) is continuous and derivable on [0, 1]
Also,
f(0) = f(1) = 0
Thus, all the continuous of Solids theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(0,1)$ such that f'(c) = 0.
We have
$f(x) = x^3 -^2x^2 +^x$
$\Rightarrow f'(x) = 3x^2 - 4x + 1$
$\therefore f'(x) = 0 \Rightarrow 3x^2- 4x + 1 = 0$
$\Rightarrow 3x^2- 3x - x + 1 = 0$
$\Rightarrow 3x(x - 1) - 1(x - 1) = 0$
$\Rightarrow (x - 1)(3x - 1) = 0$
$\Rightarrow\text{x}=1,\frac{1}{3}$
Thus, $\text{c}=\frac{1}{3}\in(0,1)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
View full question & answer→Question 245 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$Since, $\sin\text{x}$ and $\cos\text{x}$ are everywhere differentiable and continuous,
$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big]$ and differentiable on $\Big(0,\frac{\pi}{2}\Big)$
Also,
$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=1$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have show that there must exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sin\text{x}$
$\therefore\ \text{f}'(\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}=0\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
View full question & answer→Question 255 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. Find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^2- 1$ on $[2, 3]$
AnswerWe have
$f(x) = x^2- 1$
Since a polynomial function is everywhere continuous and differentiable, f(x) is continuous on 2, 3 and differentiable on 2, 3.
Thus, both conditions of Lagrange's mean value theorem is satisfied.
So, there must exist at least one real number $\text{c}\in2,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
Now,
$f(x) = x^2 - 1$
$\Rightarrow f'(x) = 2x,$
$\Rightarrow f(3) = (3)^2 - 1 = 8$
$\Rightarrow f(2) = (2)^2 - 1 = 3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
$\Rightarrow2\text{x}=\frac{8-3}{1}$
$\Rightarrow\text{x}=\frac{5}{2}$
Thus,
$\text{c}=\frac{5}{2}\in(2,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
View full question & answer→Question 265 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = 2x - x^2$ on $[0, 1]$
AnswerWe have,$f(x) = 2x - x^2$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on 0, 1 and differentiable on 0,1
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in0,1$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$f(x) = 2x - x^2$
$\Rightarrow f'(x) = 2 - 2x,$
$\Rightarrow f(1) = 2 - 1$
$\Rightarrow f(1) = 1,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow2-2\text{x}=\frac{1-0}{1}$
$\Rightarrow-2\text{x}=1-2$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,0)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 275 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\cos2{\text{x}}\text{ on }[0,\pi]$
AnswerHere$\text{f}(\text{x})=\cos2{\text{x}}\text{ on }[0,\pi]$
We know that, cosine function is continuous and differentiable every where, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$
Now,
$\text{f}(0)=\cos0=1$
$\text{f}(\pi)=\cos(2\pi)=1$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0.
Now,
$\text{f}(\text{x})=\cos2\text{x}$
$\text{f}'(\text{x})=-2\sin2\text{x}$
So, $\text{f}'(\text{c})=0$
$\Rightarrow-2\sin2\text{c}=0$
$\Rightarrow\sin2\text{c}=0$
$\Rightarrow2\text{c}=0$ or $2\text{c}=\pi$
$\Rightarrow\text{c}=0$ or $\text{c}=\frac{\pi}{2}\in(0,\pi)$
Hence, Rolle's theorem is verified.
View full question & answer→Question 285 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x(x + 4)^2$ on $[0,4]$
AnswerHere,$f(x) = x(x + 4)^2$
$\Rightarrow f(x) = x(x^2+ 16 + 8x)$
$\Rightarrow f(x) = x^3 + 8x^2+ 16x$
Since f(x) is a polynomial function which is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [0, 4] and derivable on (0, 4)
Thus, both the conditions of Lagrange's theorem is satisfied.
Consequently, there exists some $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}=\frac{\text{f}(4)-\text{f}(0)}{4}$
Now,$ f(x) = x^3 + 8x^2+ 16x$
$\Rightarrow f(x) = 3x^2+ 16x + 16,$
$\Rightarrow f(4) = 64+ 128 + 64 = 256,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow3\text{x}^2+16\text{x}+16=\frac{256}{4}$
$\Rightarrow3\text{x}^2+16\text{x}-48=0$
$\Rightarrow\text{x}=-\frac{4}{3}\big(2+\sqrt{13}\big),\frac{4}{3}\big(\sqrt{13}-2\big)$
Thus, $\text{c}=\frac{-8+4\sqrt{13}}{3}\in(0,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 295 Marks
Verify Rolle's theorem for the following function on the indicated intervals
$f(x) = x^2 -4x + 3$ on $[1, 3]$
AnswerThe given function is $f(x) = x^2 -4x + 3$
f, being a pollynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x - 4.
$f(1) = 1^2 - 4 \times 1 + 3 = 0$
$f(4) = 4^2 - 4 \times 4 + 3 = 3$
$\therefore\ \frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}=\frac{\text{f}(4)-\text{f}(1)}{4-1}=\frac{3-(0)}{3}=\frac{3}{3}=1$
Mean Value Theorem states that there is a point $\text{c}\in(1,4)$ such that f'(c) = 1
$f'(c) = 1$
$⇒ 2c - 4 = 1$
$\Rightarrow\text{c}=\frac{5}{2},$ where $\text{c}=\frac{5}{2}\in(1,4)$
Hence, Mean Value Theorem is verified for the given function.
View full question & answer→Question 305 Marks
If the value of c prescribed bye Lagrange's mean value theorem for the function
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$ defined on [2, 3].
AnswerHere,
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$ defined on [2, 3].
We have to find c prescribed by Lagrange's mean value theorem, so
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow\frac{2\text{c}}{2\sqrt{\text{c}^2-4}}=\frac{(\sqrt{9-4})-(\sqrt{4-4})}{3-2}$
$\Rightarrow\frac{\text{c}}{\sqrt{\text{c}^2-4}}=\frac{\sqrt5-0}{1}$
$\Rightarrow\frac{\text{c}}{\sqrt{\text{c}^2-4}}=\sqrt5$
Squaring both sides,
$\Rightarrow c^2 = (c^2- 4)5$
$\Rightarrow 5c^2 - c^2 = 20$
$\Rightarrow 4c^2 = 20$
$\Rightarrow c^2 = 5$
$\Rightarrow\text{c}=\pm\sqrt5$
but $\text{c}=\sqrt5\text{ as }\sqrt5\in(2,3).$
View full question & answer→Question 315 Marks
Verify Rolle's theorem for the following function on the indicated intervals$\text{f}(\text{x})=\sin2\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin2\text{x}$ Since, $\sin2\text{x}$ is everywhere continuous and differentiable. Therefore $\sin2\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big],$ and differentiable on $\Big(0,\frac{\pi}{2}\Big)$ Also, $\text{f}\Big(\frac{\pi}{2}\Big)=\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\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\sin2\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$ $\Rightarrow\text{f}'(\text{x})=0$ $\Rightarrow2\cos2\text{x}=0$ $\Rightarrow\cos2\text{x}=0$ $\Rightarrow\text{x}=\frac{\pi}{4}$ Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$ Hence, Rolle's theorem is verified .
View full question & answer→Question 325 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}\text{ on }[0,\pi]$
AnswerSince trignometric functions are differentiable and continuous, the given function, $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}$ is also continuous and differentiable.
Now $\text{f}(0)=\sin0-2\times0=0$
and
$\text{f}(\pi)=\sin\pi-\sin2\times\pi=0$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
Thus, f(x) satisfies conditions of the Rolle's Theorem on $[0,\pi].$
Therefore there exist $\text{c}\in[0,\pi]$ such that f'(c)=0
Now $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}$
$\Rightarrow\text{f}'(\text{x})\cos\text{x}-2\cos2\text{x}=0$
$\Rightarrow\cos\text{x}=2\cos2\text{x}$
$\Rightarrow\cos\text{x}=2(2\cos^2\text{x}-1)$
$\Rightarrow\cos\text{x}=4\cos^2\text{x}-2$
$\Rightarrow4\cos^2\text{x}-\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1\pm\sqrt{33}}{8}=0.8431\text{ or }-0.5931$
$\Rightarrow\text{x}=\cos^{-1}(0.8431)\text{ or }\cos^{-1}(-0.5931)$
$\Rightarrow\text{x}=\cos^{-1}(0.8431)\text{ or }180^{\circ}-\cos^{-1}(0.5931)$ $\big[\because\ \cos^{-1}(-\text{x})=\pi-\cos^{-1}(\text{x})\big]$
$\Rightarrow\text{x}=32^\circ32'\text{ or }\text{x}=126^\circ23'$
Both $32^\circ32'$ and $126^\circ23'\in[0,\pi]$ such that f'(c) = 0.
Hence Rolle's theorem is verified.
View full question & answer→Question 335 Marks
Using Lagrange's mean value theorem, prove that
$(\text{b}-\text{a})\sec^2\text{a}<\tan\text{b}-\tan\text{a}<(\text{b}-\text{a})\sec^2\text{b}$
where $0<\text{a}<\text{b}<\frac{\pi}{2}.$
AnswerConsider the function as
$\text{f}(\text{x})=\tan\text{x},$ $\Big\{\text{x}\in[\text{a},\text{b}]\text{ such that }0<\text{a}<\text{b}<\frac{\pi}{2}\Big\}$
We know that $\tan\text{x}$ is continuous and differentiable in $\Big(0,\frac{\pi}{2}\Big),$ so, Lagrange's mean value theorem is applicable on (a, b), so there exist a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow\sec^2\text{c}=\frac{\tan\text{b}-\tan\text{a}}{\text{b}-\text{a}}\ ....(\text{i})$
Now,
$\text{c}\in(\text{a},\text{b})$
$\Rightarrow\text{a}<\text{c}<\text{b}$
$\Rightarrow\sec^2\text{a}<\sec^2\text{c}<\sec^2\text{b}$
$\Rightarrow\sec^2\text{a}<\Big(\frac{\tan\text{b}-\tan\text{a}}{\text{b}-\text{a}}\Big)<\sec^2\text{b}$
Using equation (i),
$(\text{b}-\text{a})\sec^2\text{a}<(\tan\text{b}-\tan\text{a})<(\text{b}-\text{a})\sec^2\text{b}$
View full question & answer→Question 345 Marks
If $\text{f}:[-5,5]\rightarrow\text{R}$ is differentiable and if f'(x) does not vanish anywhere, then prove that $\text{f}(-5)\pm\text{f}(5).$
AnswerIt is given that $\text{f}:[-5,5]\rightarrow\text{R}$ is a differentiable function.
Since every differentiable function is continuous function, we obtain
- f is continuous on [-5, 5].
- f is differentiable on (-5, 5).
Therefore, by the Mean Value Theorem, there exists $\text{c}\in(-5,5)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-5)}{5-(-5)}$
$\Rightarrow10\text{f}'(\text{c})=\text{f}(5)-f(-5)$
It is also given that f'(x) does not vanish anywhere.
$\therefore\ \text{f}'(\text{c})\neq0$
$\Rightarrow10\text{f}'(\text{c})\neq0$
$\Rightarrow\text{f}(5)-\text{f}(-5)\neq0$
$\Rightarrow\text{f}(5)\neq\text{f}(-5)$
Hence, proved. View full question & answer→Question 355 Marks
Discuss the applicability of Lagrange's mean value theorem for the function:
f(x) = |x| on [−1, 1]
AnswerHere,
f(x) = |x| on [−1, 1]
$\text{f}(\text{x})=\begin{cases}-\text{x},&\text{x}<0\\\text{x},&\text{x}\geq0\end{cases}$
For differentiability at x = 0
$\text{LHD}=\lim_\limits{\text{x}\rightarrow0^-}\frac{\text{f}(0-\text{h})-\text{f}(0)}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-(0-\text{h})-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{-\text{h}}$
$\text{LHD}=-1$
$\text{RHD}=\lim_\limits{\text{x}\rightarrow0^+}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0+\text{h})-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}$
$=1$
$\text{LHD}\neq\text{RHD}$
⇒ f(x) is not differentiable at $\text{x}=0\in(-1,1)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 365 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=4^{\sin\text{x}}\text{ on }[0,\pi]$
AnswerHere, $\text{f}(\text{x})=4^{\sin\text{x}}\text{ on }[0,\pi]$We know that exponential and $\sin\text{x}$ both are continuous and differentiable, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$
Now,
$\text{f}(0)=4^{\sin0}=4^0=1$ $\text{f}(\pi)=4^{\sin\pi}=4^0=1$ $\Rightarrow\text{f}(0)=\text{f}(\pi)$ So, Rolle's theorem is applicable, there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0. Now, $\text{f}(\text{x})=4^{\sin\text{x}}$ $\text{f}'(\text{x})=4^{\sin\text{x}}\log4\times\cos\text{x}$ Now, $\text{f}'(\text{c})=0$ $4^{\sin\text{c}}\times\cos\times\text{c}\log4=0$ $\Rightarrow\cos\text{c}=0$ $\Rightarrow\text{c}=\frac{\pi}{2}\in(0,\pi)$ Hence, Rolle's theorem is verified.
View full question & answer→Question 375 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^3- 5x^2 - 3x$ on $[1, 3]$
AnswerWe have$,f(x) = x^3- 5x^2 - 3x$
Since, polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on 1, 3 and differentiable on 1, 3
Thus, both the conditions of Lagrange's theorem is satisfied.
Concequently, there exist some $\text{c}\in1,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$
Now, $f(x) = x^3- 5x^2 - 3x$
$f'(x) = 3x^2 - 10x - 3$
$\Rightarrow f(3) = -27$
$\Rightarrow f(1) = -7$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$
$\Rightarrow3\text{x}^2-10\text{x}-3=\frac{-20}{2}$
$\Rightarrow3\text{x}^2-10\text{x}+7=0$
$\Rightarrow\text{x}=1,\frac{7}{3}$
Thus, $\text{c}=\frac{7}{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Hence, Lagrange's theorem is verified.
View full question & answer→Question 385 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}\text{ on }[-1,0]$
AnswerThe given function is $\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}$ Since $\sin\text{x}\ \&\ \frac{\text{x}}{2}$ are everywhere continuous and differentiable, f(x) is continuous on [-1, 0] and differentiable on (-1, 0). Also, f(-1) - f(0) = 0 Thus, f(x) satisfies all the conditions of Rolle's theorem. Now, we have to show that there must exist a point $\text{c}\in(-1,0)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}$ $\Rightarrow\text{f}'(\text{x})=\frac{1}{2}-\frac{\pi}{6}\cos\frac{\pi\text{x}}{6}$ $\therefore\ \text{f}'(\text{x})=0$ $\Rightarrow\frac{1}{2}-\frac{\text{x}}{6}\cos\frac{\pi\text{x}}{6}=0$ $\Rightarrow\cos\frac{\pi\text{x}}{6}=\frac{3}{\pi}$ $\Rightarrow\text{x}=\frac{-6}{\pi}\cos^{-1}\Big(\frac{3}{\pi}\Big)$ Thus, $\text{c}=\frac{-6}{\pi}\cos^{-1}\Big(\frac{3}{\pi}\Big)\in(-1,0)$ such that f'(c) = 0. Hence, Rolle's theorem is verified.
View full question & answer→Question 395 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}\text{ on }\Big[0,\frac{\pi}{6}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}$ Since $\sin\text{x}\ \&\ \text{x}$ are everywhere continuous and differentiable, f(x) is continuous on $\Big[0,\frac{\pi}{6}\Big]$ and differentiable on $\Big(0,\frac{\pi}{6}\Big)$ Also, $\text{f}\Big(\frac{\pi}{6}\Big)=\text{f}(0)=0$ Thus, f(x) satisfies all the conditions of Rolle's theorem. Now, we have to show that there must exist a point $\text{c}\in\Big(0,\frac{\pi}{6}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}$ $\Rightarrow\text{f}'(\text{x})=\frac{6}{\pi}-8\sin\text{x}\cos\text{x}$ $\therefore\ \text{f}'(\text{x})=0$ $\Rightarrow\frac{6}{\pi}-8\sin\text{x}\cos\text{x}=0$ $\Rightarrow\sin2\text{x}=\frac{3}{2\pi}$ $\Rightarrow\text{x}=\frac{1}{2}\sin^{-1}\Big(\frac{3}{2\pi}\Big)$ Thus, $\text{c}=\frac{1}{2}\sin^{-1}\Big(\frac{3}{2\pi}\Big)\in\Big(0,\frac{\pi}{6}\Big)$ such that f'(c) = 0. Hence, Rolle's theorem is verified.
View full question & answer→Question 405 Marks
If $f(x) = Ax^2 + Bx + C$ is such that f(a) = f(b), then write the value of c in Rolle's theorem.
AnswerWe have,
$f(x) = Ax^2 + Bx + C$
Differentiating the given function with respect to x, we get
$f'(x) = 2Ax + B$
$⇒ f'(c) = 2Ac + B$
$\therefore f'(c) = 0$
$⇒ 2Ac + B = 0$
$\Rightarrow\text{c}=\frac{-\text{b}}{2\text{A}}\ ...1$
$\because$ f(a) = f(b)
$\therefore Aa^2+ Ba + C = Ab^2+ bB + C$
$\Rightarrow Aa^2+ Ba = Ab^2+ bB$
$\Rightarrow A(a^2- b^2) + B(a - b) = 0$
$\Rightarrow A(a^- b)(a + b) + B(a - b) = 0$
$\Rightarrow (a^- b)[A(a + b) + B] = 0$
$\Rightarrow\text{a}=\text{b},\text{A}=\frac{-\text{B}}{\text{a}+\text{b}}$
$\Rightarrow(\text{a}+\text{b})=\frac{-\text{B}}{\text{A}}$ $(\because\ \text{a}\neq\text{b})$
From (1) we have,
$\text{c}=\frac{\text{a}+\text{b}}{2}$
View full question & answer→Question 415 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin3\text{x}\text{ on }[0,\pi]$
AnswerThe given function is $\text{f}(\text{x})=\sin3\text{x}$
Since $\sin3\text{x}$ everywhere continuous and differentiable,
$\sin3\text{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})=\sin3\text{x}$
$\Rightarrow \text{f}'(\text{x})=3\cos3\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow3\cos3\text{x}=0$
$\Rightarrow\cos3\text{x}=0$
$\Rightarrow3\text{x}=\frac{\pi}{2},\frac{3\pi}{2},....$
$\Rightarrow\text{x}=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$
Since, $\text{c}=\frac{\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Hence, Rolle's theorem is verified.
View full question & answer→Question 425 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x(x^- 4)^2$ on the interval $[0, 4]$
AnswerGiven function is f(x) = x(x^- 4)^2. Which can be rewritten as $f(x) = x^3 - 8x^2 + 16x$.
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on [0, 4].
Also,
$f(0) = f(4) = 0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,4]$ such that f'(c) = 0.
We have
$f(x) = x^3- 8x^2 + 16x$
$\Rightarrow f'(x) = 3x^2- 16x + 16$
$\therefore f'(x) = 0$
$\Rightarrow 3x^2- 16x + 16 = 0$
$\Rightarrow 3x^2- 12x - 4x + 16 = 0$
$\Rightarrow 3x(x - 4) - 4(x - 4) = 0$
$\Rightarrow (x - 4)(3x - 4)$
$\Rightarrow\text{x}=4,\frac{4}{3}$
Thus, $\text{c}=\frac{4}{3}\in(0,4)$ such that $f'(c) = 0.$
Hence, Rolle's theorem is verified.
View full question & answer→Question 435 Marks
At what points on the following curves, is the tangent parallel to x-axis?
$\text{y}=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
AnswerLet $\text{f}(\text{x})=\text{e}^{1-\text{x}^2}$
Since f(x) is a exponential function, which is continuous and differentiable on its domain.
Also,
f(1) = f(-1) = 1
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point $\text{c}\in(-1,1)$ for which f'(c) = 0.
But $\text{f}'(\text{c})=0\Rightarrow2\text{c}\text{e}^ {1-\text{c}^2}=0\Rightarrow\text{c}=0$ $\Big(\because\ \text{e}^{1-\text{c}^2}\neq0\Big)$
$\therefore$ f(c) = f(0) = e
By the geometrical interpretetion of Rolle's theorem, (0, e) is the point on $\text{y}=\text{e}^{1-\text{x}^2}$where the tangent is parallel to the x-axis.
View full question & answer→Question 445 Marks
Find a point on the parabola $y = (x - 4)^2$, where the tangent is parallel to the chord joining (4, 0) and (5, 1).
AnswerHere,
curve is $y = (x - 4)^2$
Since, it is a polynomial function so it is differentiable and continuous. So, it Lagrange's mean value theorem is applicable, so, there exist a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow2(\text{c}-4)=\frac{\text{f}(5)-\text{f}(4)}{5-4}$
$\Rightarrow2\text{c}-8=\frac{1-0}{1}$
$\Rightarrow2\text{c}=9$
$\Rightarrow\text{c}=\frac{9}{2}$
$\Rightarrow\text{y}=\Big(\frac{9}{2}-4\Big)^2$
$\Rightarrow\text{y}=\frac{1}{4}$
Thus, $(\text{c},\text{y})=\Big(\frac{9}{2},\frac{1}{4}\Big)$ is a required point.
View full question & answer→Question 455 Marks
Verify Rolle's theorem for the following function on the indicated intervals$\text{f}(\text{x})=\cos2\Big(\text{x}-\frac{\pi}{4}\Big)\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\cos2\Big(\text{x}-\frac{\pi}{4}\Big)$ $=\cos\Big(2\text{x}-\frac{\pi}{2}\Big)=\sin2\text{x}.$ Thus, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\sin2\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$ $\Rightarrow\text{f}'(\text{x})=0$ $\Rightarrow2\cos2\text{x}=0$ $\Rightarrow\cos2\text{x}=0$ $\Rightarrow\text{x}=\frac{\pi}{4}$ Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$ Hence, Rolle's theorem is verified .
View full question & answer→Question 465 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\cos2\text{x}\text{ on }\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$
AnswerHere $\text{f}(\text{x})=\cos2\text{x}\text{ on }\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$ We know that $\cos\text{x}$ is continuous and differentiable everywhere. So, f(x) is continuous in $\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$ and differentiable is $\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$. Now, $\text{f}\Big(-\frac{\pi}{4}\Big)=\cos2\Big(-\frac{\pi}{4}\Big)=\cos\Big(-\frac{\pi}{2}\Big)=0$ $\text{f}\Big(\frac{\pi}{4}\Big)=\cos2\Big(\frac{\pi}{4}\Big)=\cos\Big(\frac{\pi}{2}\Big)=0$ $\Rightarrow\text{f}\Big(-\frac{\pi}{4}\Big)=\text{f}\Big(\frac{\pi}{4}\Big)$ So, Rolle's theorem is applicable, so, there must exist a $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0 Now, $\text{f}'(\text{x})=2\sin2\text{x}$ $\text{f}'(\text{c})=2\sin2\text{c}=0$ $\Rightarrow\sin2\text{c}=0$ $\Rightarrow2\text{c}=0$$\Rightarrow\text{c}=0\in\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$
Thus, Rolle's theorem verified.
View full question & answer→Question 475 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x(x - 1)$ on $[1, 2]$
AnswerWe have
$f(x) = x(x - 1) $
which can be rewritten as $f(x) = x^2 - x$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [1, 2] and differentiable on (1, 2).
Thus, both conditions of Lagrange's mean value theorem is satisfied.
So, there must exist at least one real number $\text{c}\in(1,2)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
Now,$ f(x) = x^2 - x$
$⇒ f'(x) = 2x - 1,$
$⇒ f(2) = 2, f(1) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow2\text{x}-1=\frac{2-0}{2-1}$
$\Rightarrow2\text{x}-1-2=0$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
Thus, $\text{c}=\frac{3}{2}\in(1,2)$ such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 485 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
AnswerHere,$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
We know that, exponantial function is continuous and differentiable everywhere. So, f(x) is continuous IS $[-1,1]$ and differentiable is $(-1,1).$ Now, $\text{f}(-1)=\text{e}^{1-1}=1$ $\text{f}(1)=\text{e}^{1-1}=1$ $\Rightarrow\text{f}(-1)=1$ So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(-1,1)$ such that f'(c) = 0. Now,$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}$
$\text{f}'(\text{x})=\text{e}^{1-\text{x}^2}(-2\text{x})$ Now, $\text{f}'(\text{c})=0$ $-2\text{ce}^{1-\text{c}^2}=0$ $\Rightarrow\text{c}=0$ or $\text{e}^{1-\text{c}^2}=0$ $\Rightarrow\text{c}=0\in(-1,1)$ Hence, Rolle's theorem is verified.
View full question & answer→Question 495 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}\text{ on }\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}$
Since $\cos\text{x}\ \&\ \text{e}^{\text{x}}$ are everywhere continuous and differentiable.
Therefore, f(x) being a product of these two, is continuous on $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$ and differentiable on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
Also,
$\text{f}\Big(\frac{-\pi}{2}\Big)=\text{f}\Big(\frac{\pi}{2}\Big)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})=0$
$\Rightarrow\sin\text{x}+\cos\text{x}=0$
$\Rightarrow\tan\text{x}=-1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Since, $\text{c}=\frac{\pi}{4}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$ such that f'(c) = 0
Hence, Rolle's theorem verified.
View full question & answer→Question 505 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^3 - 2x^2 - x + 3$ on $[0, 1]$
AnswerHere,$f(x) = x^3 - 2x^2 - x + 3$
Since f(x) is polynomial function. So, f(x) is continuous in [0, 1] and differentiable in (0, 1).
Thus, both conditions of Lagrange's mean value theorem is appplicable.
Thus, there exist a point $\text{c}\in(0,1)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow3\text{c}^2-4\text{c}-1=\frac{[(1)^3-2(1)^2-(1)+3]-3}{1}$
$\Rightarrow 3c^2 - 4c - 1 = 1^{-2}$
$\Rightarrow 3c^2 - 4c + 1 = 0$
$\Rightarrow 3c^2- 3c - c + 1 = 0$
$\Rightarrow 3c(c - 1) - 1(c - 1) = 0$
$\Rightarrow (3c - 1)(c - 1) = 0$
$\Rightarrow\text{c}=\frac{1}{3}\in(0,1)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→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.
AnswerSo, $f(1) = f(2)$
$\Rightarrow (1)^3 + b(1)^2 + c(1) = (2)^3 + b(2)^2 + c(2)$
$\Rightarrow 1 + b + c = 8 + 4b + 2c$
$\Rightarrow 3b + c + 7 = 0 ....(i)$
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
$3(-5) + c + 7 = 0$
$⇒ -15 + c + 7 = 0$
$⇒ c = 8$
View full question & answer→Question 525 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}\text{ on }0\leq\text{x}\leq\pi$
AnswerThe 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.
View full question & answer→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? AnswerBy 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 $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$ View full question & answer→