Maths 4 Marks

f(x) = (x - 1)(x - 2)²

i.e. f(x) = x³ + 4x - 4x² - x² - 4 + 4x

i.e. f(x) = x³ - 5x² + 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) = x³ + - 5x² - 8x - 4

⇒ f'(x) = 3x² + 8 - 10x

$\therefore$ f'(x) = 0 ⇒ 3x² - 10x + 8 = 0

⇒ 3x² - 6x - 4x + 8 = 0

⇒ 3x(x - 2) - 4(x - 2) = 0

⇒ (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.

f(x) = x(x - 1)²

⇒ f(x) = x(x² - 2x + 1)

$\therefore$ f(x) = (x³ - 2x² + 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) = x³ - 2x² + x

⇒ f'(x) = 3x² - 4x + 1

$\therefore$ f'(x) = 0 ⇒ 3x² - 4x + 1 = 0

⇒ 3x² - 3x - x + 1 = 0

⇒ 3x(x - 1) - 1(x - 1) = 0

⇒ (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.

We 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.

$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}\text{ on }[0,\pi]$

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.

$\text{f}(\text{x})=\sqrt{25-\text{x}^2}$

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.

f(x) = x² - 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) = x² - 2x + 4

⇒ f'(x) = 2x - 2

⇒ f(5) = 25 - 10 + 4 = 19

⇒ 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.

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$

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.

$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$

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.

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.

f(x) = 2x - x² 

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 - x² 

⇒ f'(x) = 2 - 2x,

⇒ f(1) = 2 - 1

⇒ f(1) = 1,

⇒ 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.

$\text{f}(\text{x})=\cos2{\text{x}}\text{ on }[0,\pi]$

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.

f(x) = x(x + 4)²

⇒ f(x) = x(x² + 16 + 8x)

⇒ f(x) = x³ + 8x² + 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) = x³ + 8x² + 16x

⇒ f(x) = 3x² + 16x + 16,

⇒ f(4) = 64+ 128 + 64 = 256,

⇒ 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.

f(x) = x³ - 5x² - 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) = x³ - 5x² - 3x

f'(x) = 3x² - 10x - 3

⇒ f(3) = -27

⇒ 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.

f(x) = x³ - 2x² - 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

$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$

$\Rightarrow3\text{c}^2-4\text{c}-1=\frac{[(1)^3-2(1)^2-(1)+3]-3}{1}$

⇒ 3c² - 4c - 1 = 1^-2

⇒ 3c² - 4c + 1 = 0

⇒ 3c² - 3c - c + 1 = 0

⇒ 3c(c - 1) - 1(c - 1) = 0

⇒ (3c - 1)(c - 1) = 0

$\Rightarrow\text{c}=\frac{1}{3}\in(0,1)$

Hence, Lagrange's mean value theorem is verified.

y = x³ − 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.

f(x) = (x² - 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

⇒ 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) = (x² - 1)(x - 2)

f'(x) = 2x(x - 2) + (x² - 1)

= 2x² - 4 + x² - 1

f'(x) = 3x² - 5

Now,

f'(c) = 0

⇒ 3x² - 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.

1. f is continuous on [-5, 5].
2. 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.

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).

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. $\frac{7}{2},\frac{1}{4},$ is a point on the given curve where the tangent is parallel to the chord joining the points (3, 0) and (4, 1).

Also, $\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$ exists for all $\text{x}\in[1,4],$

$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$

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) = x² - 8x + 12

⇒ f'(x) = 2x - 8

$\therefore$ f'(x) = 0

⇒ 2x - 8 = 0

⇒ x = 4

Thus, $\text{c}=4\in(2,6)$ such that f'(c) = 0

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,

We have

We have

$\Rightarrow\text{c}=0\in\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$

Thus, Rolle's theorem verified.

$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$

$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}$