Maths M.C.Q (1 Marks)

2. $\sqrt3$

Solution:

We have

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$

Clearly, f(x) is continuous on [1, 3] and derivable on (1, 3).

Thus, both the conditions of Lagrange's theorem is satisfied.

Concequently there exists $\text{c}\in(1,3)$ such that

$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$

Now, $\text{f}(\text{x})=\frac{\text{x}^2+1}{\text{x}}$

$\text{f}'(\text{x})=\frac{\text{x}^2-1}{\text{x}^2},\text{f}(1)=2,\text{f}(3)=\frac{10}{3}$

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$

$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{4}{6}$

$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{2}{3}$

$\Rightarrow3\text{x}^2-3=2\text{x}^2$

$\Rightarrow\text{x}=\pm\sqrt3$

Thus, $\text{c}=\sqrt3\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}.$

3. $\frac{1-\sqrt5}{2}$

Solution:

$\text{f}(\text{x})=\frac{\text{x}(\text{x}+1)}{\text{e}^{\text{x}}}$ defined on [-1, 0]

⇒ f(-1) = 0 also f(0) = 0

Now, f(x) = e^-x(x² + x)

⇒ f'(x) = e^-x(2x + 1) - (x² + x)e^-x

⇒ f'(x) = e^-x(2x + 1 - x² + x)

⇒ f'(x) = e^-x(-x² + x - 1)

⇒ f'(x) = 0

⇒ e^-x(-x² + x - 1) = 0

⇒ -x² + x - 1 = 0

⇒ x² - x + 1 = 0

$\Rightarrow\text{x}=\frac{1\pm\sqrt5}{2}$

As, $\text{x}\in[-1,0]$

$\text{x}=\frac{1-\sqrt5}{2}$

1. $\text{e}^{\frac{1}{1}-\text{e}}$

Solution:

Given:

$\text{y}=\text{f}(\text{x})=\text{x}\log\text{x}$

Differentiating the given function with respect to x, we get

$\text{f}'(\text{x})=1+\log\text{x}$

⇒ Slope of the tangent to the curve $=1+\log\text{x}$

Also,

Slope of the chord joining the points (1, 0) and (e, e), $(\text{m})=\frac{\text{e}}{\text{e}-1}$

The tangent to the curve is parallel to chord joining the points (1, 0) and (e, e).

1. 2

Solution:

f(x) = 2x³ - 5x² - 4x + 3

Differentiating the given function with respect to x, we get

f'(x) = 6x² - 10x - 4

⇒ f'(c) = 6c² - 10c - 4

$\therefore$ f'(c) =0

⇒ 3c² - 5c - 2 = 0

⇒ 3c² - 6c + c - 2 = 0

⇒ 3c(c - 2) + c - 2 = 0

⇒ (3c + 1)(c - 2) = 0

$\Rightarrow\text{c}=2, \frac{-1}{3}$

$\therefore\ \text{c}=2\in\Big(\frac{1}{3},3\Big)$

Thus, $\text{c}=2\in\Big(\frac{1}{3},3\Big)$ for which Rolle's theorem holds.

Hence, the required value of c is 2.

3. $\lambda>\frac{1}{2}$

Solution:

$\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$

$\text{f}'(\text{x})=-\sin\text{x}-2\lambda$

For f(x) to be decreasing, we must have

$\text{f}'(\text{x})<0$

$\Rightarrow-\sin\text{x}-2\lambda<0$

$\Rightarrow\sin\text{x}+2\lambda>0$

$\Rightarrow2\lambda>-\sin\text{x}$

We know that the maximum value of $-\sin\text{x}$ is 1.

$\Rightarrow2\lambda>1$

$\Rightarrow\lambda>\frac{1}{2}$

3. $\text{a}<\text{x}_1<\text{b}$

Solution:

We have

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$

In the Lagrange's mean value theorem, $\text{c}\in(\text{a},\text{b})$ such that $\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$

So, if there is x₁ such that $\text{f}'(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then $\text{x}_1\in(\text{a},\text{b})$

$\Rightarrow\text{a}<\text{x}_1<\text{b}$

3. At least one root.

Solution:

We observe that, $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ is the derivative of the polynomial $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0$Polynomial function is continuous everywhere in R and concequently derivative in R.

Therefore, $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0$ is continuous on $\alpha,\beta$ and derivative on $\alpha,\beta.$

Hence, it is satisfies the both the conditions of Rolle's theorem.

By algebric interpretation of Roll's theorem, we know that between any two roots of a function f(x), there exists atleast one root of its derivative.

Hence, the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ will have atleast one root between $\alpha$ and $\beta.$

4. $\frac{3}{2}$

Solution:

We have

 f(x) = x(x - 2)

It can be rewritten as f(x) = x² - 2x

We know that a polynomial function is everywhere continuous and differentiable.

Since, f(x) is polynomial, it is continuous on [1, 2] and differentiable on [1, 2].

So, there must exist at least one real number $\text{c}\in(1,2)$ such that

$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$

$=\frac{\text{f}(2)-\text{f}(1)}{1}$

Now, f(x) = x² - 2x

⇒ f'(x) = 2x - 2

and f(1) = -1, f(2) = 0

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$

$\Rightarrow\text{f}(\text{x})=\frac{0+1}{1}$

$\Rightarrow2\text{x}-2=1$

$\Rightarrow\text{x}=\frac{3}{2}$

$\therefore\ \text{c}=\frac{3}{2}\in(1,2)$

1. $1$

Solution:

f(x) = x³ - 3x in the interval $\big[0,\sqrt3\big]$

$\text{f}(0)=0$ and $\text{f}\big(\sqrt3\big)=0$

f'(x) = 3x² - 3

f'(c) = 3c² - 3

f'(c) = 0

3c² - 3 = 0

3c² = 3

c² = 1

$\text{c}=\pm1$

$\text{x}\in\big[0,\sqrt3\big]$

Hence, x = 1

4. $\frac{3\pi}{4}$

Solution:

$\text{f}(\text{x})=\text{e}^{\text{x}}\sin\text{x}$

$\text{f}'(\text{x})=\text{e}^{\text{x}}\cos\text{x}+\text{e}^{\text{x}}\sin\text{x}$

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

$\text{e}^\text{c}(\cos\text{c}+\sin\text{c})=0$

$\cos\text{c}+\sin\text{c}=0$

$\cos\text{c}=-\sin\text{c}$

$\tan\text{c}=-1$

$\text{c}=\frac{3\pi}{4}\in(0,\pi)$

2. Any interval $[0,\pi]$

Solution:

$\phi(\text{x})$ is continuous and differentiable function then using statement of Roll's theorem f(a) = f(b). Hence, here $\sin0=0$ also $\sin\pi=0.$