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

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Chapter 15 - Mean Value Theorems Exercise Ex. 15.1

Question 1(i)
Solution 1(i)
Question 1(ii)
Solution 1(ii)
Question 1(iii)
Solution 1(iii)
Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)
Question 2(i)

Solution 2(i)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

Solution 3(vii)

Here,

            f open parentheses x close parentheses equals fraction numerator sin x over denominator e to the power of x end fraction space o n space x space element of open square brackets 0 comma space straight pi close square brackets
W e space k n o w space t h a t comma space e x p o n e n t i a l space a n d space sin e space b o t h space f u n c t i o n s space a r e space c o n t i n u o u s space a n d space d i f f e r e n t i a b l e
e v e r y space w h e r e comma space s o space f open parentheses x close parentheses space i s space c o n t i n u o u s space i s space open square brackets 0 comma space straight pi close square brackets space a n d space d i f f e r e n t i a b l e space i s space open square brackets 0 comma space straight pi close square brackets

Now comma space

space space space space space space space space space space space space space straight f open parentheses 0 close parentheses equals fraction numerator sin space 0 over denominator straight e to the power of 0 end fraction equals 0
space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space straight f open parentheses straight pi close parentheses equals fraction numerator sin space straight pi over denominator straight e to the power of straight pi end fraction equals 0

rightwards double arrow straight f open parentheses 0 close parentheses equals straight f open parentheses straight pi close parentheses
Since space Rolle apostrophe straight s space theorem space applicable comma space therefore space there space must space exist space straight a space point space straight c element of open square brackets 0 comma space straight pi close square brackets
such space that space straight f apostrophe open parentheses straight c close parentheses equals 0

Now comma
space space space space space space space space space space straight f open parentheses straight x close parentheses equals sinx over straight e to the power of straight x

space space space space space space space space space space rightwards double arrow straight f apostrophe open parentheses straight x close parentheses equals fraction numerator straight e to the power of straight x open parentheses cosx close parentheses minus straight e to the power of straight x open parentheses sinx close parentheses over denominator open parentheses straight e to the power of straight x close parentheses squared end fraction
Now comma
space space space space space space space space space space space space space space space space straight f apostrophe open parentheses straight c close parentheses equals 0
space space space space space space space space space space space rightwards double arrow straight e to the power of straight c open parentheses cosc minus sinc close parentheses equals 0
space space space space space space space space space space space rightwards double arrow space straight e to the power of straight c not equal to 0 space and space cosc minus sinc equals 0
space space space space space space space space space space space rightwards double arrow space tanc equals 1
space space space space space space space space space space space space space space space straight c equals straight pi over 4 element of open square brackets 0 comma straight pi close square brackets
Hence comma space Rolle apostrophe straight s space theorem space is space verified.

Question 3(viii)

Solution 3(viii)

Question 3(ix)

Solution 3(ix)

Question 3(x)

Solution 3(x)

Question 3(xi)

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 3(xiii)

Solution 3(xiii)

Question 3(xiv)

Solution 3(xiv)

Question 3(xv)

Solution 3(xv)

Question 3(xvi)

Solution 3(xvi)

Question 3(xvii)

Solution 3(xvii)

Question 3(xviii)

Verify Rolle's theorem for function f(x) = sin x - sin 2x on [0, pi] on the indicated intervals.

Solution 3(xviii)

Question 7

Solution 7

x = 0 then y = 16

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

Question 8(i)

Solution 8(i)

x = 0, then y = 0

Therefore, the point is (0, 0)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

x = 1/2, then y = - 27

Therefore, the point is (1/2, - 27)

Question 9
Solution 9
Question 10

Solution 10

Question 11

Solution 11

Question 2(ii)

Verify Rolle's theorem for each of the following functions on the indicated intervals:

  

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.

Chapter 15 - Mean Value Theorems Exercise Ex. 15.2

Question 1(i)
Solution 1(i)
Question 1(ii)

Solution 1(ii)

Question 1(iii)
Solution 1(iii)
Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)

Question 1(vii)
Solution 1(vii)
Question 1(viii)

Solution 1(viii)

Question 1(ix)
Solution 1(ix)
Question 1(x)

Solution 1(x)

Question 1(xi)
Solution 1(xi)
Question 1(xii)
Solution 1(xii)

Question 1(xiii)

Solution 1(xiii)

Question 1(xiv)
Solution 1(xiv)
Question 1(xv)
Solution 1(xv)
Question 1(xvi)

Solution 1(xvi)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5
Solution 5
Question 6
Solution 6

Question 7
Solution 7
Question 8

Solution 8

Question 9
Solution 9
Question 10

Solution 10

Question 11

Solution 11

Chapter 15 - Mean Value Theorems Exercise MCQ

Question 1

If the polynomial equation a0xn + an-1xn - 1 + an - 2xn - 2 + ……+ a2x2 + a1x + a0 = 0 n being a positive integer, has two different real roots α and β, then between α and β the equation n anxn-1 + (n - 1)an - 1xn - 2 + …. + a1 = 0 has

 

  1. exactly one root
  2. almost one root
  3. at least one root
  4. no root

 

Solution 1

Correct option: (c)

 

  

Question 2

If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval

 

  1. (0, 1)
  2. (1, 2)
  3. (0, 2)
  4. none of these

 

Solution 2

Correct option: (c)

 

  

Question 3

  1. 1
  2. 2
  3. none of these

 

Solution 3

Correct option: (b)

Question 4

  1. a < x1 b
  2. a x1< b
  3. a < x1< b
  4. a x1 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.

Question 5

Rolle's theorem is applicable in case of ϕ(x) = asin x, a > 0 in

 

  1. any interval
  2. the interval [0, π]
  3. the interval (0, π/2)
  4. none of these

 

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,  ].

Question 6

The value of c in Rolle's theorem when f(x) = 2x3 - 5x2 - 4x + 3, is x [1/3, 3]

 

  1. 2
  2. -1/3
  3. -2
  4. 2/3

 

Solution 6

Correct option: (a)

 

  

Question 7

When the tangent to the curve y = x log x is parallel to the chord joining the points (1,0) and (e, e), the value of x is

 

  1. e1/1 - e
  2. e(e - 1)(2e - 1)
Solution 7

Correct option: (a)

  

Question 8

Solution 8

Correct answer: (c)

 

  

Question 9

The value of c in Lagrange's mean value theorem for the function f(x) = x(x - 2) where x [1,2] is

 

  1. 1
  2. 1/2
  3. 2/3
  4. 3/2

 

Solution 9

Correct option: (d)

 

  

Question 10

The value of c in Rolle's theorem for the function f(x) = x3 - 3x in the interval   is

 

  1. 1
  2. -1
  3. 3/2
  4. 1/3

 

Solution 10

Correct option: (a)

 

  

Question 11

If f(x) = ex sin x in [0, π], then c in Rolle's theorem is

 

  1. π/6 
  2. π/4
  3. π/2
  4. 3π/4

 

Solution 11

Correct option: (d)

 

  

Chapter 15 - Mean Value Theorems Exercise Ex. 15VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5