Maths 3 Marks

$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$

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.

f(x) = x² + 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.

$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}$

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.

f(x) = 2x² - 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.

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

1. It is continuous on [a, b]
2. It is differentiable on (a, b)
3. f(a) = f(b)

So, f(x) is not continuos at $1\in[-1,1]$

Therefore, f(x) is continuous on -1, 2 and differentiable on -1, 2.