Class 12-science RD SHARMA Solutions Maths Chapter 15 – Mean Value Theorems

- Mean Value Theorems Exercise Ex. 15.1
- Solution 1(i)
- Solution 1(ii)
- Solution 1(iii)
- Solution 1(iv)
- Solution 1(v)
- Solution 1(vi)
- Solution 2(i)
- Solution 2(iii)
- Solution 2(iv)
- Solution 2(v)
- Solution 2(vi)
- Solution 2(vii)
- Solution 2(viii)
- Solution 3(i)
- Solution 3(ii)
- Solution 3(iii)
- Solution 3(iv)
- Solution 3(v)
- Solution 3(vi)
- Solution 3(vii)
- Solution 3(viii)
- Solution 3(ix)
- Solution 3(x)
- Solution 3(xi)
- Solution 3(xii)
- Solution 3(xiii)
- Solution 3(xiv)
- Solution 3(xv)
- Solution 3(xvi)
- Solution 3(xvii)
- Solution 3(xviii)
- Solution 7
- Solution 8(i)
- Solution 8(ii)
- Solution 8(iii)
- Solution 9
- Solution 10
- Solution 11
- Solution 2(ii)

- Mean Value Theorems Exercise Ex. 15.2
- Solution 1(i)
- Solution 1(ii)
- Solution 1(iii)
- Solution 1(iv)
- Solution 1(v)
- Solution 1(vi)
- Solution 1(vii)
- Solution 1(viii)
- Solution 1(ix)
- Solution 1(x)
- Solution 1(xi)
- Solution 1(xii)
- Solution 1(xiii)
- Solution 1(xiv)
- Solution 1(xv)
- Solution 1(xvi)
- Solution 2
- Solution 3
- Solution 4
- Solution 5
- Solution 6
- Solution 7
- Solution 8
- Solution 9
- Solution 10
- Solution 11

- Mean Value Theorems Exercise MCQ

## Mean Value Theorems Exercise Ex. 15.1

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 2(i)

### Solution 2(iii)

### Solution 2(iv)

### Solution 2(v)

### Solution 2(vi)

### Solution 2(vii)

### Solution 2(viii)

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 3(iv)

### Solution 3(v)

### Solution 3(vi)

### Solution 3(vii)

Here,

### Solution 3(viii)

### Solution 3(ix)

### Solution 3(x)

### Solution 3(xi)

### Solution 3(xii)

### Solution 3(xiii)

### Solution 3(xiv)

### Solution 3(xv)

### Solution 3(xvi)

### Solution 3(xvii)

### Solution 3(xviii)

### Solution 7

x = 0 then y = 16

Therefore, the point on the curve is (0, 16)

### Solution 8(i)

x = 0, then y = 0

Therefore, the point is (0, 0)

### Solution 8(ii)

### Solution 8(iii)

x = 1/2, then y = – 27

Therefore, the point is (1/2, – 27)

### Solution 9

### Solution 10

### Solution 11

### Solution 2(ii)

Given function is

As the given function is a polynomial, so it is continuous and differentiable everywhere.

Let’s find the extreme values

Therefore, f(2) = f(6).

So, Rolle’s theorem is applicable for f on [2, 6].

Let’s find the derivative of f(x)

Take f'(x) = 0

As 4 ∈ [2, 6] and f'(4) = 0.

Thus, Rolle’s theorem is verified.

## Mean Value Theorems Exercise Ex. 15.2

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 1(vii)

### Solution 1(viii)

### Solution 1(ix)

### Solution 1(x)

### Solution 1(xi)

### Solution 1(xii)

### Solution 1(xiii)

### Solution 1(xiv)

### Solution 1(xv)

### Solution 1(xvi)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

## Mean Value Theorems Exercise MCQ

### Solution 1

Correct option: (c)

### Solution 2

Correct option: (c)

### Solution 3

Correct option: (b)

### Solution 4

Correct option: (c)

Using statement of Lagrange’s mean value theorem function is continuous on [a,b], differentiable on (a,b) then there exists c such that a < x1< b.

### Solution 5

Correct option: (b)

ϕ(x) is continuous and differentiable function then using statement of Rolle’s theorem f(a)=f(b). Hence, here sin 0=0 also sin п=0. The answer is [0, ].

### Solution 6

Correct option: (a)

### Solution 7

Correct option: (a)

### Solution 8

Correct answer: (c)

### Solution 9

Correct option: (d)

### Solution 10

Correct option: (a)

### Solution 11

Correct option: (d)