Solve the Following Question.(3 Marks) - Maths STD 12 Questions - Vidyadip

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Solve the Following Question.(3 Marks)

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 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.
$f(x) = 2x^2 - 3x + 1$ on $[1, 3]$

Answer

Here,$f(x) = 2x^2 - 3x + 1$ on $[1, 3]$
We know that a polynomial function is continuous and differentiable.
So, f(x) is continuous in $[1, 3]$ and f(x) differentiable in $(1, 3)$.
So, Lagrange's mean value theorem is applicable.
So, there must exist at least one real number $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(-1)}{3-1}$
$\Rightarrow4\text{c}-3=\frac{(2(3)^2-3(3)+1)-(2-3+1)}{3-1}$
$\Rightarrow4\text{c}-3=\frac{10}{2}$
$\Rightarrow4\text{c}=5+3$
$\Rightarrow4\text{c}=8$
$\Rightarrow\text{c}=2\in(1,3)$
Hence, Lagrange's mean value theorem is verified.

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Question 23 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.
$f(x) = x^2 - 3x + 2$ on $[-1, 2]$

Answer

We have $f(x) = x^2 - 3x + 2$ 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 are satisfied.
So, there must exist at least one real number $\text{c}\in-1,2$ such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{2+1}$
$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{3}$
Now, $f(x) = x^2 - 3x + 2 $
$\Rightarrow f'(x) = 2x - 3 $
$\Rightarrow f(2) = 0 $
$\Rightarrow f(-1) = (-1)^2 - 3(-1) + 2$
$ \Rightarrow f(-1) = 6$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(-1)}{3}$ $\Rightarrow2\text{x}-3=-2$
$\Rightarrow2\text{x}-1=0$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,2)$ such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{2-(-1)}$
Hence, Lagrange's mean value theorem is verified.

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Question 33 Marks

Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=3+(\text{x}-2)^{\frac{2}{3}}\text{ on }[1,3]$

Answer

The given function is $\text{f}(\text{x})=3+(\text{x}-2)^{\frac{2}{3}}$
Defferentiating with respect to x, we get
$\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{2}{3}-1}$
$\Rightarrow\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{-1}{3}}$
$\Rightarrow\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{1}{3}}$
Clearly, we observe that for $\text{x}=2\in[1,3],\text{f}'(\text{x})$ does not exist.
Therefore, f(x) is not derivable on [1, 3].
Hence, Rolle's theorem is not applicable for the given function.

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Question 43 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.
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}\text{ on }[2,4]$

Answer

We have,$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$
Here, f(x) will exist,
if
$\text{x}^2-4\geq0$
$\Rightarrow\text{x}\leq-2\text{ or }\text{x}\geq2$
Since, for each $\text{x}\in2,4,$ the function f(x) attains a unique definite value.
So, f(x) is continuous on 2, 4
Also,
$\text{f}'(\text{x})=\frac{1}{2\sqrt{\text{x}^2-4}}(2\text{x})=\frac{\text{x}}{\sqrt{\text{x}^2-4}}$
Exists for all $\text{x}\in2,4$
So, f(x) is differentiable on 2, 4.
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists some $\text{c}\in2,4$ such that
$\text{f}'(\text{x})=\frac{1}{2\sqrt{\text{x}^2-4}}(2\text{x})=\frac{\text{x}}{\sqrt{\text{x}^2-4}}$
Now,
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$
$\text{f}'(\text{x})=\frac{1}{\sqrt{\text{x}^2-4}},\text{f}(4)=2\sqrt3,\text{f}(2)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(2)}{4-2}$
$\Rightarrow\frac{\text{x}}{\sqrt{\text{x}^2-4}}=\frac{2\sqrt3}{2}$
$\Rightarrow\frac{\text{x}}{\sqrt{\text{x}^2-4}}=\sqrt3$
$\Rightarrow\frac{\text{x}^2}{\text{x}^2-4}=3$
$\Rightarrow\text{x}^2=3\text{x}^2-12$
$\Rightarrow\text{x}^2=6$
$\Rightarrow\text{x}=\pm\sqrt6$
Thus, $\text{c}=\sqrt6\in(2,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(2)}{4-2}$
Hence, Lagrange's theorem is verified.

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Question 53 Marks

Show that the Lagrange's mean value theorem is not applicable to the function
$\text{f}(\text{x})=\frac{1}{\text{x}}\text{ on }[-1,1]$

Answer

Given,
$\text{f}(\text{x})=\frac{1}{\text{x}}$
Clearly, f(x) is does not exist for x = 0
Thus, the given function is discontinuous on [-1, 1]
Hence, Lagrange's mean value theorem is not applicable for the given function on [-1, 1].

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Question 63 Marks

State Lagrange's mean value theorem.

Answer

Lagrange's Mean Value Theorem:
Let f(x) be a function defined on [a, b] such that

It is continuous on [a, b] and
It is differentiable on (a, b).

Then, there exists a real number $\text{c}\in(\text{a},\text{b})$ such that $\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}.$

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Question 73 Marks

State Rolle's theorem.

Answer

Rolle's theorem: Let f(x) be a real value function defined on the closed interval [a, b] such that

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

Then, there exists a real number $\text{c}\in(\text{a},\text{b})$ such that f'(c) = 0.

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Question 83 Marks

Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$

Answer

Here, $\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$
$\text{f}'(\text{x})=\frac{2}{3\text{x}^{\frac{1}{3}}}$
$\text{f}'(0)=\frac{2}{3(0)^{\frac{1}{3}}}$
$\text{f}'(0)=\infty$
So, f'(x) does not exist at $\text{x}=0\in(-1,1)$
⇒ f(x) is not differentialble in $\text{x}\in(-1,1)$
So, Rolle's theorem is not applicable on f(x) in [-1, 1].

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Question 93 Marks

At what points on the following curves, is the tangent parallel to x-axis?
$\text{y}=\text{x}^2\text{ on }[-2,2]$

Answer

Let $f(x) = x^2$
Since f(x) is a polynomial function, it is continuous on [-2, 2] and differentiable on (-2, 2)
Also, $f(2) = f(-2) = 4$
Thus, all the conditions of Rolle's theorem are satisfied.
Concequently, there exists at least one point $\text{c}\in(-2,2)$ for which f'(c) = 0.
But $\text{f}'(\text{c})=0$
$\Rightarrow2\text{c}=0$
$\Rightarrow\text{c}=0$
$\therefore\text{f}_\text{c}=\text{f}_0=0$
By the geometrical interpretetion of Rolle's theorem, (0, 0) is the point on $y = x^2$, where the tangent is parallel to the x-axis.

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Question 103 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.
$f(x) = x^2 + x - 1$ on $[0, 4]$

Answer

Here, $f(x) = x^2 + x - 1$ on $[0, 4]$
f(x) is polynomial, so it is continuous is $[0, 4]$ and differentiable in $(0, 4)$
as every polynomial is continuous and differentiable everywhere. So, Lagrange's mean value theorem is applicable, so there exist a point $\text{c}\in[0,4]$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow2\text{c}+1=\frac{\big((4)^2+4-1\big)-(0-1)}{4}$
$\Rightarrow2\text{c}+1=\frac{19+1}{4}$
$\Rightarrow2\text{c}+1=5$
$\Rightarrow\text{c}=2\in(0,4)$
Hence, Lagrange's mean value theorem is verified.

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Question 113 Marks

Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=[\text{x}]\text{ for }-1\leq\text{x}\leq1,$ where [x] denotes the greatest integer not exceeding x.

Answer

Here, $\text{f}(\text{x})=[\text{x}]$ and $\text{x}\in[-1,1],$ at n = 1 $\text{LHL}=\lim\limits_{\text{x}\rightarrow(1-\text{h})}[\text{x}]$ $=\lim\limits_{\text{h}\rightarrow0}[1-\text{h}]$ $=0$ $\text{RHL}=\lim\limits_{\text{x}\rightarrow(1+\text{h})}[\text{x}]$ $=\lim\limits_{\text{h}\rightarrow0}[1+\text{h}]$ $=1$ $\text{LHL}\neq\text{RHL}$So, f(x) is not continuos at $1\in[-1,1]$
Hence, Rolle's theorem is not applicable on f(x) in [-1, 1].

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Question 123 Marks

Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\begin{cases}-4\text{x}+5,&0\leq\text{x}\leq1\\2\text{x}-3,&1<\text{x}\leq2\end{cases}$

Answer

The given function is
$\text{f}(\text{x})=\begin{cases}-4\text{x}+5,&0\leq\text{x}\leq1\\2\text{x}-3,&1<\text{x}\leq2\end{cases}$
At x = 0, we have
$\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim\limits_{\text{h}\rightarrow0}[-4(1-\text{h}+5)]=1$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})=\lim\limits_{\text{h}\rightarrow0}[2(1+\text{h}-3)]=-1$
$\therefore\ \lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})$
Thus, f(x) is discontinuous at x = 1.
Hence, Rolle's theorem is not applicable for the given function.

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Question 133 Marks

Using Rolle's theorem, find points on the curve $\text{y}=16-\text{x}^2,\text{x}\in[-1,1],$ where tagent is parallel to x-axis.

Answer

The equation of the curve is,
$\text{y}=16-\text{x}^2\ ....(1)$
Let $P(x_1,y_1)$ be a point on it where the tangent is parallel to x-axis.
Then,
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=0\ ....(2)$
Differentiating (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=-2\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=-2\text{x}_1$
$\Rightarrow-2\text{x}_1=0$ (from(2))
$\Rightarrow\text{x}_1=0$
$P(x_1, y_1)$ lies on the curve $y = 16 - x^2$
$\therefore\text{y}_1=16-\text{x}_1^2$
When $x_1 = 0,$
$y_1 = 16$
Hence, (0, 16) is the required point.

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Question 143 Marks

Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{x}^2-5\text{x}+4\text{ on }[1,4]$

Answer

According to Rolle’s theorem, if f(x) is a real valued function defined on [a, b] such that it is continuous on [a, b], it is differentiable on (a, b) and f(a) = f(b), then there exists a real number $\text{c}\in(\text{a},\text{b})$ such that f(c)= 0.
Now, f(x) is defined for all $\text{x}\in[1,4].$
At each point of [1, 4], the limit of f(x) is equal to the value of the function.
Therefore, f(x) is continuous on [1, 4].
Also, f'(x) = 2x - 5 exists for all $\text{x}\in(1,4).$
So, f(x) is differentiable on (1, 4).
Also,
f(1) = f(4) = 0
Thus, all the three conditions of Rolle’s theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(1,4)$ such that f'(c) = 0.
We have
f'(x) = 2x - 5
$\therefore$ f'(x) = 0
⇒ 2x - 5 = 0
$\Rightarrow\text{x}=\frac{5}{2}$
$\Big[\text{Since }\text{c}=\frac{5}{2}\in(1,4)\text{ such that }\text{f}'(\text{c})=0\Big]$
Hence, Rolle's theorem is verified.

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Question 153 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.$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}\text{ on }[0,\pi]$

Answer

We have,$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}$
Since, $\sin\text{x},\sin2\text{x}\ \&\ \text{x}$ are everywhere continuous and differentiable.
Therefore, f(x) is continuous on $[0,\pi]$ and differentiable on $(0,\pi)$
Concequently, there exist some $\text{c}\in(0,\pi)$such that
$\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}=\frac{\text{f}(\pi)-\text{f}(0)}{\pi}$
Now, $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}$
$\text{f}'(\text{x})=\cos\text{x}-2\cos2\text{x}-1,\text{f}(\pi)=-\pi,\text{f}(0)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}$
$\Rightarrow\cos\text{x}-2\cos2\text{x}-1=-1$
$\Rightarrow\cos\text{x}-2\cos2\text{x}=0$
$\Rightarrow\cos\text{x}-4\cos^2\text{x}=-2$
$\Rightarrow4\cos^2\text{x}-\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1}{8}\big(1\pm\sqrt{33}\big)$
$\Rightarrow\text{x}=\cos^{-1}\Big[\frac{1}{8}\big(1\pm\sqrt{33}\big)\Big]$
Thus, $\text{c}=\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)\in(0,\pi)$ such that $\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}.$
Hence, Lagrange's mean value theorem is verified.

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