Rd Sharma XII Vol 1 2020 Solutions for Class 12 Commerce Math Chapter 14 Mean Value Theorems are provided here with simple step-by-step explanations. These solutions for Mean Value Theorems are extremely popular among Class 12 Commerce students for Math Mean Value Theorems Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Commerce Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 14.17:

Question 1:

Verify Lagrange's mean value theorem for the following functions on the indicated intervals. In each case find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem
(i) f(x) = x2 − 1 on [2, 3]
(ii) f(x) = x3 − 2x2x + 3 on [0, 1]
(iii) f(x) = x(x −1) on [1, 2]
(iv) f(x) = x2 − 3x + 2 on [−1, 2]
(v) f(x) = 2x2 − 3x + 1 on [1, 3]
(vi) f(x) = x2 − 2x + 4 on [1, 5]
(vii) f(x) = 2xx2 on [0, 1]
(viii) f(x) = (x − 1)(x − 2)(x − 3) on [0, 4]
(ix) fx=25-x2 on [−3, 4]
(x) f(x) = tan1 x on [0, 1]
(xi) fx=x+1x on[1, 3]
(xii) f(x) = x(x + 4)2 on [0, 4]
(xiii) fx=x2-4 on[2, 4]
(xiv) f(x) = x2 + x − 1 on [0, 4]
(xv) f(x) = sin x − sin 2xx on [0, π]
(xvi) f(x) = x3 − 5x2 − 3x on [1, 3]

Answer:

(i) We have

fx=x2-1

Since a polynomial function is everywhere continuous and differentiable, fx is continuous on 2, 3 and differentiable on 2, 3.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​c2, 3 such that
f'c=f3-f23-2

Now, fx=x2-1
f'x=2x ,  f3=32-1=8 ,  f2=22-1=3

∴  f'x=f3-f23-2

2x=8-31x=52
Thus, c=522, 3 such that f'c=f3-f23-2.

Hence, Lagrange's theorem is verified.

 

(ii) We have,

fx=x3-2x2-x+3=0

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx 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 ​c0, 1 such that
f'c=f1-f01-0

 

Now, fx=x3-2x2-x+3=0
f'x=3x2-4x-1 ,  f1= 1 ,  f0=3

∴  f'x=f1-f01-0

3x2-4x-1=1-313x2-4x-1+2=03x2-4x+1=03x2-3x-x+1=03x-1x-1=0x=13, 1
Thus, c=130, 1 such that f'c=f1-f01-0.

Hence, Lagrange's theorem is verified.
 

(iii) We have,

fx=xx-1 which can be rewritten as fx=x2-x

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx is continuous on 1, 2 and differentiable on 1, 2.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​c1, 2 such that
f'c=f2-f12-1

Now, fx=x2-x

f'x=2x-1 ,  f2= 2 ,  f1=0

∴  f'x=f2-f12-1

2x-1=2-02-12x-1-2=02x=3x=32
Thus, c=321, 2 such that f'c=f2-f12-1.

Hence, Lagrange's theorem is verified.

 

(iv) We have,

fx=x2-3x+2

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx is continuous on -1, 2 and differentiable on -1, 2.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​c-1, 2 such that
f'c=f2-f-12+1=f2-f-13
 

Now, fx=x2-3x+2
f'x=2x-3 ,  f2=0 ,  f-1=-12-3-1+2=6

∴  f'x=f2-f-13

2x-3=-22x-1=0x=12
Thus, c=12-1, 2 such that f'c=f2-f-12--1.

Hence, Lagrange's theorem is verified.

 

(v) We have,

fx=2x2-3x+1

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx is continuous on 1, 3 and differentiable on 1, 3.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​c1, 3 such that
f'c=f3-f13-1=f3-f12
 

Now, fx=2x2-3x+1
f'x=4x-3 ,  f3=10 ,  f1=212-31+1=0

∴  f'x=f3-f12

4x-3=10-024x-3-5=0x=2
Thus, c=21, 3 such that f'c=f3-f13-1.

Hence, Lagrange's theorem is verified.

 

(vi) We have,

fx=x2-2x+4

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx 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 ​c1, 5 such that
f'c=f5-f15-1=f5-f14
 

Now, fx=x2-2x+4
f'x=2x-2 ,  f5=25-10+4=19 ,  f1=1-2+4=3

∴  f'x=f5-f14

2x-2=19-342x-2-4=0x=62=3
Thus, c=31, 5 such that f'c=f5-f15-1.

Hence, Lagrange's theorem is verified.


 

(vii) We have,

fx=2x-x2

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx 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 ​c0, 1 such that
f'c=f1-f01-0=f1-f01
 

Now, fx=2x-x2
f'x=2-2x ,  f1=2-1=1 ,  f0=0

∴  f'x=f1-f01

2-2x=1-01-2x=1-2x=12
Thus, c=120, 1 such that f'c=f1-f01-0.

Hence, Lagrange's theorem is verified.

 

(viii) We have,

fx=x-1x-2x-3 which can be rewritten as fx=x3-6x2+11x-6 

Since a polynomial function is everywhere continuous and differentiable.

Therefore, fx is continuous on 0, 4 and differentiable on 0, 4.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​c0, 4 such that
f'c=f4-f04-0=f4-f04
 

Now, fx=x3-6x2+11x-6
f'x=3x2-12x+11 ,  f0=-6 ,  f4=64-96+44-6=6

∴  f'x=f4-f04-0

3x2-12x+11=6+643x2-12x+8=0x=2-23, 2+23 
Thus, c=2±230, 4 such that f'c=f4-f04-0.

Hence, Lagrange's theorem is verified.


 

(ix) We have,

fx=25-x2 

Here, fx will exist,
if
 25-x20x225-5x5

Since for each x-3, 4, the function fx attains a unique definite value.
So, fx is continuous on -3, 4

Also, f'x=1225-x2-2x=-x25-x2 exists for all x-3, 4

So, fx is differentiable on -3, 4.

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c-3, 4 such that

f'c=f4-f-34+3=f4-f-37
 

Now, fx=25-x2

f'x=-x25-x2 ,  f-3=4 ,  f4=3

∴  f'x=f4-f-34+3

-x25-x2=3-4749x2=25-x2x=±12 
Thus, c=±12-3, 4 such that f'c=f4-f-34--3.

Hence, Lagrange's theorem is verified.

 

(x) We have,

fx=tan-1x 

Clearly,  fx is continuous on 0, 1 and derivable on 0,1

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c-3, 4 such that

f'c=f1-f01-0=f1-f01
 

Now, fx=tan-1x

f'x=11+x2 ,  f1=π4 ,  f0=0

∴  f'x=f1-f01-0

11+x2=π4-04π-1=x2x=±4-ππ 
Thus, c=4-ππ0, 1 such that f'c=f1-f01-0.

Hence, Lagrange's theorem is verified.
 

(xi) We have,

fx=x+1x=x2+1x 


Clearly,  fx is continuous on 1, 3 and derivable on 1, 3

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c1, 3 such that

f'c=f3-f13-1=f3-f12
 

Now, fx=x2+1x

f'x=x2-1x2 ,  f1=2 ,  f3=103

∴  f'x=f3-f12

x2-1x2=46x2-1x2=233x2-3=2x2x=±3 
Thus, c=31, 3 such that f'c=f3-f13-1.

Hence, Lagrange's theorem is verified.


 

(xii) We have,

fx=xx+42=xx2+16+8x=x3+8x2+16x 

Since fx is a polynomial function which is everywhere continuous and differentiable.

Therefore,  fx is continuous on 0, 4 and derivable on 0,4

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c0, 4 such that

f'c=f4-f04-0=f4-f04
 

Now, fx=x3+8x2+16x

f'x=3x2+16x+16 ,  f4=64+128+64=256 ,  f0=0

∴  f'x=f4-f04-0

3x2+16x+16=25643x2+16x-48=0x=-432+13, 4313-2
Thus, c=-8+41330, 4 such that f'c=f4-f04-0.

Hence, Lagrange's theorem is verified.

 

(xiii) We have,

fx=x2-4 

Here, fx will exist,
if
 x2-40x-2 or x2

Since for each x2, 4, the function fx attains a unique definite value.
So, fx is continuous on 2, 4

Also, f'x=12x2-42x=xx2-4 exists for all x2, 4

So, fx is differentiable on 2, 4.

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c2, 4 such that

f'c=f4-f24-2=f4-f22
 

Now, fx=x2-4

f'x=xx2-4 ,  f4=23 ,  f2=0

∴  f'x=f4-f24-2

xx2-4=232xx2-4=3x2x2-4=3 x2=3x2-12x2=6x=±6
Thus, c=62, 4 such that f'c=f4-f24-2.

Hence, Lagrange's theorem is verified.

 

(xiv) We have,

fx=x2+x-1 

Since polynomial function is everywhere continuous and differentiable.

Therefore, fx is continuous on 0, 4 and differentiable on 0, 4

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c0, 4 such that

f'c=f4-f04-0=f4-f04
 

Now, fx=x2+x-1

f'x=2x+1 ,  f4=19 ,  f0=-1

∴  f'x=f4-f04-0

2x+1=2042x+1=52x=4 x=2
Thus, c=20, 4 such that f'c=f4-f04-0.

Hence, Lagrange's theorem is verified.

 

(xv) We have,

fx=sinx-sin2x-x 

Since sinx, sin2x & x  are everywhere continuous and differentiable

Therefore, fx is continuous on 0,π and differentiable on 0, π

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c0, π such that

f'c=fπ-f0π-0=fπ-f0π
 

Now, fx=sinx-sin2x-x

f'x=cosx-2cos2x-1 ,  fπ=-π ,  f0=0

∴  f'x=fπ-f0π-0

cosx-2cos2x-1=-1cosx-2cos2x=0cosx-4cos2x=-2 4cos2x-cosx-2=0cosx=181±33x=cos-1181±33
Thus, c=cos-11±3380, π such that f'c=fπ-f0π-0.

Hence, Lagrange's theorem is verified.


 

(xvi) We have,

fx=x3-5x2-3x 

Since polynomial function is everywhere continuous and differentiable

Therefore, fx is continuous on 1, 3 and differentiable on 1, 3

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some c1, 3 such that

f'c=f3-f13-1=f3-f12
 

Now, fx=x3-5x2-3x

f'x=3x2-10x-3 ,  f3=-27 ,  f1=-7

∴  f'x=f3-f12
3x2-10x-3=-2023x2-10x+7=0x=1, 73
  
Thus, c=731, 3 such that f'c=f3-f13-1.

Hence, Lagrange's theorem is verified.



Page No 14.18:

Question 2:

Discuss the applicability of Lagrange's mean value theorem for the function
f(x) = | x | on [−1, 1]

Answer:

Given:
fx=x

If Lagrange's theorem is applicable for the given function, then fx is continuous on -1, 1 and differentiable on -1, 1.
 
But it is known that fx=x is not differentiable at x=0-1, 1.

Thus, our supposition is wrong.
Therefore, Lagrange's theorem is not applicable for the given function.

Page No 14.18:

Question 3:

Show that the lagrange's mean value theorem is not applicable to the function
f(x) = 1x on [−1, 1]

Answer:

Given:
fx=1x

Clearly, fx does not exist for x = 0

Thus, the given function is discontinuous on -1, 1.

Hence, Lagrange's mean value theorem is not applicable for the given function on -1, 1.1x

Page No 14.18:

Question 4:

Verify the  hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = 14x-1, 1≤ x ≤ 4.

Answer:

The given function is fx=14x-1.

Since for each x1, 4, the function attains a unique definite value, fx is continuous on 1, 4.

Also, f'x=-44x-12 exists for all x1, 4

Thus, both the conditions of Lagrange's mean value theorem are satisfied.

Consequently, there exists some c1, 4 such that 
f'c=f4-f14-1=f4-f13

Now, 
fx=14x-1f'x=-44x-12f4=115, f1=13

 f'x=f4-f14-1
f'x=115-134-1=-445-44x-12=-4454x-12=4516x2-8x-44=04x2-2x-11=0x=141±35

Thus, c=141+351, 4 such that f'c=f4-f14-1.

Hence, Lagrange's theorem is verified.

Page No 14.18:

Question 5:

Find a point on the parabola y = (x − 4)2, where the tangent is parallel to the chord joining (4, 0) and (5, 1).

Answer:

​Let:
fx=x-42=x2-8x+16

The tangent to the curve is parallel to the chord joining the points 4, 0 and 5, 1.

Assume that the chord joins the points a, fa and b, fb.

 a=4, b=5

The polynomial function is everywhere continuous and differentiable.

So, x2-8x+16 is continuous on 4, 5 and differentiable on 4, 5.

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

Consequently, there exists c4, 5 such that f'c=f5-f45-4.

Now, 
fx=x2-8x+16f'x=2x-8f5=1, f4=0

f'x=f5-f45-42x-8=112x=9x=92

Thus, c=924, 5 such that ​f'c=f5-f45-4.

Clearly,
 fc=92-42=14

Thus, c, fc, i.e.​  92,14, is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

Page No 14.18:

Question 6:

Find a point on the curve y = x2 + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2).

Answer:

​Let:
fx=x2+x

The tangent to the curve is parallel to the chord joining the points 0, 0 and 1, 2.

Assume that the chord joins the points a, fa and b, fb.

 a=0, b=1

The polynomial function is everywhere continuous and differentiable.

So, fx=x2+x is continuous on 0, 1 and differentiable on 0, 1.

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

Consequently, there exists c0, 1 such that f'c=f1-f01-0.

Now, 
fx=x2+xf'x=2x+1f1=2, f0=0

f'x=f1-f01-02x+1=2-01-02x=1x=12

Thus, c=120,1 such that ​f'c=f1-f01-0.

Clearly,
 fc=122+12=34.

Thus, c, fc, i.e.​  12, 34, is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

Page No 14.18:

Question 7:

Find a point on the parabola y = (x − 3)2, where the tangent is parallel to the chord joining (3, 0) and (4, 1).

Answer:

​Let:
fx=x-32=x2-6x+9

The tangent to the curve is parallel to the chord joining the points 3, 0 and 4, 1.

Assume that the chord joins the points a, fa and b, fb.

 a=3, b=4

The polynomial function is everywhere continuous and differentiable.

So, fx=x2-6x+9 is continuous on 3, 4 and differentiable on 3, 4.

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

Consequently, there exists c3, 4 such that f'c=f4-f34-3.

Now, 
fx=x2-6x+9f'x=2x-6f3=0, f4=1

f'x=f4-f34-32x-6=1-04-32x=7x=72

Thus, c=723, 4 such that ​f'c=f4-f34-3.

Clearly,
 fc=72-32=14

Thus, c, fc, i.e.  72, 14, is a point on the given curve where the tangent is parallel to the chord joining the points 3, 0 and 4, 1.

Page No 14.18:

Question 8:

Find the points on the curve y = x3 − 3x, where the tangent to the curve is parallel to the chord joining (1, −2) and (2, 2).

Answer:

​Let:
fx=x3-3x

The tangent to the curve is parallel to the chord joining the points 1, -2 and 2, 2.

Assume that the chord joins the points a, fa and b, fb.

 a=1, b=2

The polynomial function is everywhere continuous and differentiable.

So, fx=x3-3x is continuous on 1, 2 and differentiable on 1, 2.

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

Consequently, there exists c1, 2 such that f'c=f2-f12-1.

Now, 
fx=x3-3xf'x=3x2-3f1=-2, f2=2

f'x=f2-f12-13x2-3=2+22-13x2=7x=±73

Thus, c=±73 such that ​f'c=f2-f12-1.

Clearly,
 f73=7332-373=7373-3=73-23=-2373 and f-73=2373

∴ fc=2373

Thus, c, fc, i.e.​  ±73, 2373, are points on the given curve where the tangent is parallel to the chord joining the points 1, -2 and 2, 2.

Page No 14.18:

Question 9:

Find a point on the curve y = x3 + 1 where the tangent is parallel to the chord joining (1, 2) and (3, 28).

Answer:

​Let:
fx=x3+1

The tangent to the curve is parallel to the chord joining the points 1, 2 and 3, 28.

Assume that the chord joins the points a, fa and b, fb.

 a=1, b=3

The polynomial function is everywhere continuous and differentiable.

So, fx=x3+1 is continuous on 1, 3 and differentiable on 1, 3.

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

Consequently, there exists c1, 3 such that f'c=f3-f13-1.

Now, 
fx=x3+1f'x=3x2f1=2, f3=28

f'x=f3-f13-13x2=2623x2=13x=±133

Thus, c=133 such that ​f'c=f3-f13-1.

Clearly, 
fc=13332+1 

Thus, c, fc, i.e.​  133, 1+13332, is a point on the given curve where the tangent is parallel to the chord joining the points 1, 2 and 3, 28.

Page No 14.18:

Question 10:

Let C be a curve defined parametrically as x=acos3θ, y=asin3θ, 0θπ2. Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a).                                                                                                                                 [CBSE 2014]

Answer:

As, x=acos3θdxdθ=-3acos2θsinθAnd, y=asin3θdydθ=3asin2θcosθSo, dydx=dydθdxdθ=3asin2θcosθ-3acos2θsinθ=-tanθFor the tangent to be parallel to the chord joining the points a,0 and 0,a,dydx=0-aa-0-tanθ=-1tanθ=1θ=π4Now,x=acos3π4=a123=a22 andy=asin3π4=a123=a22So, the point P on the curve C is a22,a22.

Page No 14.18:

Question 11:

Using Lagrange's mean value theorem, prove that

(ba) sec2 a < tan b − tan a < (ba) sec2 b

where 0 < a < b < π2.

Answer:

​Consider, the function
fx=tanx, xa, b, 0<a<b<π2

Clearly, fx is continuous on a, b and derivable on a, b.

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

Consequently, ca, b such that f'c=fb-fab-a.

Now, 
fx=tanx f'x=sec2xfa=tana, fb=tanb

f'c=fb-fab-a sec2c=tanb-tanab-a     ...1


Now, 
ca, ba<c<bsec2a<sec2c<sec2b   sec2x is increasing in 0, π2sec2a<tanb-tanab-a<sec2b   from 1b-asec2a<tanb-tana<b-asec2b

Hence proved.



Page No 14.19:

Question 1:

If the polynomial equation
a0 xn+an-1 xn-1+an-2 xn-2+...+a2x2+a1 x+a0=0
n positive integer, has two different real roots α and β, then between α and β, the equation
n anxn-1+n-1 an-1 xn-2+...+a1=0 has
(a) exactly one root
(b) almost one root
(c) at least one root
(d) no root

Answer:

(c) at least one root

We observe that, nanxn-1+n-1an-1xn-2+...+a1=0 is the derivative of the
polynomial anxn+an-1xn-1+an-2xn-2+...+a2x2+a1x+a0=0

Polynomial function is continuous every where in R and consequently derivative in R
Therefore, anxn+an-1xn-1+an-2xn-2+...+a2x2+a1x+a0 is continuous on α, β and derivative on α, β.
Hence, it satisfies the both the conditions of Rolle's theorem.

By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function fx, there exists at least one root of its derivative.

Hence, the equation nanxn-1+n-1an-1xn-2+...+a1=0 will have at least one root between α and β.

Page No 14.19:

Question 2:

If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these

Answer:

(c) (0, 2)

Letfx=ax3+bx2+cx+d            .....1f0=df2=8a+4b+2c+d      =24a+2b+c+d      =d                         4a+2b+c=0

f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

By Rolle's Theorem,
f'α=0    for 0<α<2Now, f'x=3ax2+2bx+cf'α=3aα2+2bα+c=0Equation 1 has atleast one root in the interval 0, 2.Thus, f'x must have root in the interval 0, 2.

Page No 14.19:

Question 3:

For the function f (x) = x + 1x, x ∈ [1, 3], the value of c for the Lagrange's mean value theorem
is
(a) 1
(b) 3
(c) 2
(d) none of these

Answer:

(b)3

We have
fx=x+1x=x2+1x 

Clearly,  fx is continuous on 1, 3 and derivable on 1, 3.

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

Consequently, there exists c1, 3 such that

f'c=f3-f13-1=f3-f12
 

Now, fx=x2+1x

f'x=x2-1x2 ,  f1=2 ,  f3=103

∴  f'x=f3-f12

x2-1x2=46x2-1x2=233x2-3=2x2x=±3 
Thus, c=31, 3 such that f'c=f3-f13-1.

Page No 14.19:

Question 4:

If from Lagrange's mean value theorem, we have

f' x1=f' b-f ab-a, then
(a) a < x1b
(b) ax1 < b
(c) a < x1 < b
(d) ax1b

Answer:

(c) a < x1 < b

In the Lagrange's mean value theorem, ca, b such that f'c=fb-fab-a.

So, if there is x1 such that f'x1=fb-fab-a, then x1a, b.
a<x1<b

Page No 14.19:

Question 5:

Rolle's theorem is applicable in case of ϕ (x) = asin x, a > a in
(a) any interval
(b) the interval [0, π]
(c) the interval (0, π/2)
(d) none of these

Answer:

(b) the interval [0, π]152

The given function is ϕx=asinx, where > 0.

Differentiating the given function with respect to x, we get

f'x=loga cosx asinx   

 f'c=loga cosc asinc 

Let  f'c=0 loga cosc asinc=0 cosc asinc=0 cosc=0c=π2
∴  c0, π

Also, the given function is derivable and hence continuous on the interval 0, π.

Hence, the Rolle's theorem is applicable on the given function in the interval ​0, π.

Page No 14.19:

Question 6:

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

(a) 2

(b) -13

(c) −2

(d) 23

Answer:

(a) 2

Given:
fx=2x3-5x2-4x+3

Differentiating the given function with respect to x, we get 

f'x=6x2-10x-4f'c=6c2-10c-4f'c=0  3c2-5c-2=0 3c2-6c+c-2=0 3cc-2+c-2=0 3c+1c-2=0 c=2, -13 c=213, 3

Thus, c=213,3 for which Rolle's theorem holds.

Hence, the required value of c is 2.

Page No 14.19:

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

(a) e1/1−e

(b) e(e−1)(2e−1)

(c) e2e-1e-1

(d) e-1e

Answer:

(a) e1/1−e

Given:
y=fx=xlogx

Differentiating the given function with respect to x,  we get

f'x=1+logx   

 Slope of the tangent to the curve = 1+logx

Also,
Slope of the chord joining the points 1, 0 and e, e, (m) = ee-1

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

  m=1+logx  

 ee-1=1+logx

 ee-1-1=logx e-e+1e-1=logx 1e-1=logx x=e1e-1

Page No 14.19:

Question 8:

The value of c in Rolle's theorem for the function fx=xx+1ex defined on [−1, 0] is
(a) 0.5

(b) 1+52

(c) 1-52

(d) −0.5

Answer:

(c) 1-52152

Given:
fx=xx+1ex

Differentiating the given function with respect to x, we get 

f'x=ex2x+1-xx+1exex2f'x=2x+1-xx+1exf'x=2x+1-x2-xex f'x=-x2+x+1exf'c=-c2+c+1ec f'c=0  -c2+c+1ec=0 c2-c-1=0  c=1-52, 1+52 c=1-52-1, 0


Hence, the required value of c is 1-52.



Page No 14.20:

Question 9:

The value of c in Lagrange's mean value theorem for the function f (x) = x (x − 2) when x ∈ [1, 2] is
(a) 1
(b) 1/2
(c) 2/3
(d) 3/2

Answer:

(d)32

We have
 f (x) = x (x − 2)

It can be rewritten as fx=x2-2x.

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

Since fx is a polynomial , it is continuous on 1, 2 and differentiable on 1, 2.

Thus, fx satisfies both the conditions of Lagrange's theorem on 1, 2.

So, there must exist at least one real number 1, 2 such that 

f'c=f2-f12-1=f2-f11
 
Now, fx=x2-2x 
    f'x=2x-2,
and f1=-1, f2=0

 f'x=f2-f12-1
f'x=0+112x-2=1x=32

∴ c=321, 2

Page No 14.20:

Question 10:

The value of c in Rolle's theorem for the function f (x) = x3 − 3x in the interval [0, 3] is
(a) 1
(b) −1
(c) 3/2
(d) 1/3

Answer:

(a) 1

The given function is fx=x3-3x.
This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0, 3] and derivable on (0, 3 ).

Differentiating the given function with respect to x, we get 

f'x=3x2-3f'c=3c2-3 f'c=0  3c2-3=0c2=1c=±1

Thus, c=10, 3 for which Rolle's theorem holds.

Hence, the required value of c is 1.

Page No 14.20:

Question 11:

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

(a) π/6

(b) π/4

(c) π/2

(d) 3π/4

Answer:

(d) 3π/4

The given function is fx=exsinx.

Differentiating the given function with respect to x, we get 

f'x=excosx+sinxexf'c=eccosc+sincecNow , excosx is continuous and derivable in R.Therefore, it is continuous on 0, π and derivable on 0, π. f'c=0  eccosc+sinc=0 cosc+sinc=0     ec0 tanc=-1 c=3π4, 7π4, ... c=3π40, π

Thus, c=3π40,π for which Rolle's theorem holds.

Hence, the required value of c is 3π/4.

Page No 14.20:

Question 1:

A function (x) = 1 + 1x is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is ______________.

Answer:


The function fx=1+1x is defined on the interval [1, 3].

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

So, by mean value theorem there must exist at least one real number c ∈ (1, 3) such that

f'c=f3-f13-1

-1c2=43-23-1             fx=1+1xf'x=-1x2

-1c2=-232

1c2=13

c2=3

c=±3

Thus, c=31,3 such that f'c=f3-f13-1.

Hence, a point in the interval where the given function satisfies the mean value theorem is 3.


A function (x) = 1 + 1x is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is       3      .

Page No 14.20:

Question 2:

For the function f(x) = 8x2 - 7x + 5, x ∈ [-6, 6], the value of c for the lagrange's mean value theorem is __________________.

Answer:


The given function is fx=8x2-7x+5.

f(x) is a polynomial function.

We know that a polynomial function is everywhere continuous and differentiable. So, f(x) is continuous on [−6, 6] and differentiable on (−6, 6). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (−6, 6) such that

f'c=f6-f-66--6

Now, fx=8x2-7x+5

f'x=16x-7

f'c=f6-f-66--6

16c-7=8×62-7×6+5-8×-62-7×-6+512

16c-7=-8412=-7

16c=0

c=0

Thus, c = 0 ∈ (−6, 6) such that f'c=f6-f-66--6.

Hence, the value of c is 0.


For the function f(x) = 8x2 − 7x + 5, x ∈ [−6, 6], the value of c for the Lagrange's mean value theorem is ___0___.

Page No 14.20:

Question 3:

If the function  f(x) = x3 – 6x2 + ax + b defined on [1, 3] satisfies Roll's theorem for c2+13, then a = ___________, b = __________.

Answer:


The given function is f(x) = x3 − 6x2 + ax + b.

It is given that f(x) defined on [1, 3] satisfies Rolle's theorem for c=2+13.

f1=f3 and f'c=0

Now,

f1=f3

1-6+a+b=27-54+3a+b

-5+a=-27+3a

2a=22

a=11

Also,

f(x) = x3 − 6x2 + ax + b

f'x=3x2-12x+a


f'c=0

3c2-12c+a=0

3c2-12c+11=0      (a = 11)

Since both equations f(1) = f(3) and f'c=3c2-12c+11=0 are independent of b, so b can taken any real value.

a = 11 and b ∈ R

Thus, if the function  f(x) = x3 – 6x2 + ax + b defined on [1, 3] satisfies Roll's theorem for c=2+13, then a = 11 and b ∈ R.


If the function  f(x) = x3 – 6x2 + ax + b defined on [1, 3] satisfies Roll's theorem for c2+13, then a = ___11___, b = ___R___.

Page No 14.20:

Question 4:

It is given that for the function  f(x) = x3 - 6x2 + ax + b on [1, 3], Rolle's theorem holds with c2+13. If f(1) = f(3) = 0, then a =_______, b =________.

Answer:


The given function is f(x) = x3 − 6x2 + ax + b.

It is given that Rolle's theorem holds for f(x) defined on [1, 3] with c=2+13.

f1=f3=0     (Given)

f1=0

1-6+a+b=0

a+b=5        .....(1)

Also,

f3=0

27-54+3a+b=0

3a+b=27     .....(2)

Solving (1) and (2), we get

a = 11 and b = −6

It can be verified that for a = 11 and b = −6, f'2+13=0.

Thus, the values of a and b are 11 and −6, respectively.


It is given that for the function  f(x) = x3 − 6x2 + ax + b on [1, 3], Rolle's theorem holds with c2+13. If f(1) = f(3) = 0, then a = ___11___, b =___−6___.

Page No 14.20:

Question 5:

For the function f(x) = logex, ∈ [1, 2], the value of c for the lagrange's mean value theorem is _______________.

Answer:


The given function is f(x) = logex.

Now, f(x) = logex is differentiable and so continuous for all x > 0. So, f(x) is continuous on [1, 2] and differentiable on (1, 2). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (1, 2) such that

f'c=f2-f12-1

f(x) = logex

f'x=1x


f'c=f2-f12-1

1c=loge2-loge12-1

1c=loge2-0             loga1=0, a>0

c=1loge2=log2e       logba=1logab

c=log2e1,2        2<e<4log22<log2e<log241<log2e<2

Thus, c=log2e1,2 such that f'c=f2-f12-1.

Hence, the value of c is log2e.


For the function f(x) = logex, ∈ [1, 2], the value of c for the Lagrange's mean value theorem is       log2e      .

Page No 14.20:

Question 6:

The value of c in Rolle's theorem for the function  f(x) = x3 - 3x in the interval [0, 3] is _______________.

Answer:


The given function is f(x) = x3 − 3x.

f(x) is a polynomial function. We know that a polynomial function is everywhere continuous and differentiable.

So, f(x) is continuous on 0,3 and differentiable on 0,3.

Also, f(0) = 0 and f3=33-33=33-33=0

f0=f3

Thus, all the conditions of Rolle's theorem are satisfied.

So, there exist a real number c0,3 such that f'c=0.

f(x) = x3 − 3x

f'x=3x2-3


f'c=0

3c2-3=0

c2=1

c=±1

Thus, c = 1 ∈ 0,3 such that f'c=0.

Hence, the value of c is 1.


The value of c in Rolle's theorem for the function  f(x) = x3 − 3x in the interval 0,3 is ___1___.

Page No 14.20:

Question 1:

If f (x) = Ax2 + Bx + C is such that f (a) = f (b), then write the value of c in Rolle's theorem.

Answer:

We have
fx=Ax2+Bx+C

Differentiating the given function with respect to x, we get
f'x=2Ax+B
f'c=2Ac+B
  f'c=0 2Ac+B=0c=-B2A    ...1

fa=fb Aa2+Ba+C=Ab2+bB+C Aa2+Ba=Ab2+bB Aa2-b2+Ba-b=0 Aa-ba+b+Ba-b=0 a-bAa+b+B=0 a=b, A=-Ba+b a+b=-BA     ab

From (1), we have

c=a+b2

Page No 14.20:

Question 2:

State Rolle's theorem.

Answer:

Rolle's Theorem:

Let be a real valued function  defined on the closed interval a,b such that
(i) it is continuous on the closed interval ​a,b,
(ii) it is differentiable on the open interval a, b, and
(iii) fa=fb
Then, there exists a real number ca,b such that f'c=0.

Page No 14.20:

Question 3:

State Lagrange's mean value theorem.

Answer:

Lagrange's Mean Value Theorem:

Let fx be a function defined on a,b such that 
(i) it is continuous on ​a,b and
(ii) it is differentiable on a,b.

Then, there exists a real number ca,b such that f'c=fb-fab-a.

Page No 14.20:

Question 4:

If the value of c prescribed in Rolle's theorem for the function
f (x) = 2x (x − 3)n on the interval [0, 23] is 34, write the value of n (a positive integer).

Answer:

We have
fx=2xx-3n

Differentiating the given function with respect to x, we get

f'x=2xnx-3n-1+x-3nf'x=2x-3nxnx-3+1f'c=2c-3ncnc-3+1

Given:
f'34=0

 2-94n34n-94+1=0 2-94n-n3+1=0-n3+1=0-n+3=0n=3

Page No 14.20:

Question 5:

Find the value of c prescribed by Lagrange's mean value theorem for the function
fx=x2-4 defined on [2, 3].

Answer:

We have

fx=x2-4 

Here, fx will exist, if 
x2-40x-2 or x2

Since for each x2, 3, the function fx attains a unique definite value, fx is continuous on 2, 3.

Also, f'x=12x2-42x=xx2-4 exists for all x2, 3.

So, fx is differentiable on 2, 3.

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists c2, 3 such that

f'c=f3-f23-2=f3-f21
 

Now,
 fx=x2-4

f'x=xx2-4 ,  f3=5 ,  f2=0

∴  f'x=f3-f23-2

xx2-4=5x2x2-4=5 x2=5x2-204x2=20x=±5
Thus, c=52, 3 such that f'c=f3-f23-2.

Hence, Lagrange's theorem is verified.
 



Page No 14.8:

Question 1:

Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
(i) f(x) = 3 + (x − 2)2/3 on [1, 3]

(ii) f(x) = [x] for −1 ≤ x ≤ 1, where [x] denotes the greatest integer not exceeding x

(iii) f(x) = sin1xfor −1 ≤ x ≤ 1

(iv) f(x) = 2x2 − 5x + 3 on [1, 3]

(v) f(x) = x2/3 on [−1, 1]

(vi) fx=-4x+5,0x12x-3,1<x2

Answer:

(i) The given function is fx=3+x-223.

Differentiating with respect to x, we get

f'x=23x-223-1f'x=23x-2-13f'x=23x-213

Clearly, we observe that for x=21, 3f'x does not exist.

Therefore, fx is not derivable on 1, 3.

Hence, Rolle's theorem is not applicable for the given function.

(ii) The given function is fx=x.
The domain of is given to be -1, 1.

Let c-1, 1 such that is not an integer.
Then,
limxcfx=fc

Thus, fx is continuous at x=c.

Now, let c=0.

Then,

limx0-fx=-10=f0

Thus, is discontinuous at = 0.

Therefore, fx is not continuous in -1, 1.

Rolle's theorem is not applicable for the given function.

(iii) The given function is fx=sin1x.
The domain of is given to be -1, 1.

It is known that limx0sin1x does not exist.

Thus, fx is discontinuous at x = 0 on -1, 1.

Hence, Rolle's theorem is not applicable for the given function.


(iv) The given function is fx=2x2-5x+3 on 1, 3.
The domain of is given to be 1, 3.
It is a polynomial function.
Thus, it is everywhere derivable and hence continuous.

But
f1=0 and f3=6f3f1

Hence, Rolle's theorem is not applicable for the given function.

(v) The given function is fx=x23 on -1, 1.
The domain of is given to be -1, 1.

Differentiating fx with respect to x, we get

f'x=23x-13
We observe that at x=0f'x is not defined.
Hence, Rolle's theorem is not applicable for the given function.

(vi) The given function is 
fx=-4x+5, 0x12x-3, 1<x2

At x = 0, we have

limx1-fx=limh0f1-h=limh0-41-h+5=1
And
limx1+fx=limh0f1+h=limh021+h-3=-1

 limx1-fxlimx1+fx

Thus, fx is discontinuous at x=1.
Hence, Rolle's theorem is not applicable for the given function.



Page No 14.9:

Question 2:

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = x2 − 8x + 12 on [2, 6]

(ii) f(x) = x2 − 4x + 3 on [1, 3]

(iii) f(x) = (x − 1) (x − 2)2 on [1, 2]

(iv) f(x) = x(x − 1)2 on [0, 1]

(v) f(x) = (x2 − 1) (x − 2) on [−1, 2]

(vi) f(x) = x(x − 4)2 on the interval [0, 4]

(vii)
f(x) = x(x −2)2 on the interval [0, 2]

(viii) 
f(x) = x2 + 5x + 6 on the interval [−3, −2]

Answer:

(i) Given:
fx=x2-8x+12

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on 2, 6

Also,
 f2=22-82+12=4-16+12=0f6=62-86+12=36-48+12=0 f2=f6=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c2, 6 such that f'c=0.

We have
fx=x2-8x+12f'x=2x-8 f'x=0 2x-8=0x=4

Thus, c=42, 6 such that f'c=0.

Hence, Rolle's theorem is verified.

(ii) Given:
fx=x2-4x+ 3

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on 1, 3

Also,
 f1=12-41+3=1-4 + 3=0f3=32-43+3=9-12+3=0 f1=f3=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c1, 3 such that f'c=0.

We have
fx=x2-4x+3f'x=2x-4 f'x=0 2x-4=0x=2

Thus, c=21, 3 such that f'c=0.

Hence, Rolle's theorem is verified.

(iii) Given:
fx=x-1x-22 
i.e. fx=x3+4x-4x2-x2-4+4x
i.e. fx=x3-5x2+8x-4

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on 1, 2

Also,
 f1=13-512+81-4=0f2=23-522+82-4=0 f1=f2=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c1, 2 such that f'c=0.

We have
fx=x3+8x-5x2-4f'x=3x2+8-10x f'x=0 3x2-10x+8=0                  3x2-6x-4x+8=0                  3xx-2-4x-2=0                  x-23x-4                  x=2, 43

Thus, c=431, 2 such that f'c=0.

Hence, Rolle's theorem is verified.

(iv) Given:
fx=xx-12 
 fx=xx2-2x+1
  fx=x3-2x2+x

We know that a polynomial function is everywhere derivable and hence continuous.

So, fx being a polynomial function is continuous and derivable on 0, 1

Also,
 f0=f1=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c0, 1 such that f'c=0.

We have
fx=x3-2x2+xf'x=3x2-4x+1 f'x=0 3x2-4x+1=0                 3x2-3x-x+1=0                 3xx-1-1x-1=0                 x-1 3x-1=0                 x=1, 13

Thus, c=130, 1 such that f'c=0.

Hence, Rolle's theorem is verified.

(v) Given:
fx=x2-1x-2 
i.e. fx=x3-2x2-x+2

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on -1, 2

Also,
 f-1=f2=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c-1, 2 such that f'c=0.

We have
fx=x3-x-2x2+2f'x=3x2-4x-1 f'x=0 3x2-4x-1=0                 x=--4±-42-4×3×-12×3                 x=4±16+126                 x=4±286                 x=4±276                 x=2±73                 x=132-7, 132+7

Thus, c=132-7, 132+7-1, 2 such that f'c=0.

Hence, Rolle's theorem is verified.

(vi) Given function is 
fx=xx-42, which can be rewritten as fx=x3-8x2+16x.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on 0, 4

Also,
 f0=f4=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c0, 4 such that f'c=0.

We have
fx=x3-8x2+16xf'x=3x2-16x+16f'x=0  3x2-16x+16=03x2-12x-4x+16=03xx-4-4x-4=0x-43x-4x=4, 43

Thus, c=430, 4 such that f'c=0.

Hence, Rolle's theorem is verified.

(vii) The given function is 
fx=xx-22, which can be rewritten as fx=x3-4x2+4x.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on 0, 2

Also,
 f0=f2=0

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists c0, 2 such that f'c=0.

We have
fx=x3-4x2+4xf'x=3x2-8x+4When f'x=0  3x2-8x+4=03x2-6x-2x+4=03xx-2-2x-2=0x-23x-2x=2, 23

Thus, c=230, 2 such that f'c=0.

Hence, Rolle's theorem is verified.

(viii) Given function is 
fx=x2+5x+6.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, fx is continuous and derivable on -3, -2
Also,
 f-3=-32+5-3+6=9-15+6=0f-2=-22+5-2+6=4-10+6=0 f-3=f-2=0

Thus, all the conditions of the Rolle's theorem are satisfied.

Now, we have to show that there exists c-3, -2 such that f'c=0.

We have
fx=x2+5x+6f'x=2x+5 f'x=0 2x+5=0                x=-52

Thus, c=-52-3, -2 such that f'c=0.

Hence, Rolle's theorem is verified.

Page No 14.9:

Question 3:

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]

(ii) f(x) = sin 2x on [0, π/2]

(iii) f(x) = cos 2x on [−π/4, π/4]

(iv) f(x) = ex sin x on [0, π]

(v) f(x) = ex cos x on [−π/2, π/2]

(vi) f(x) = cos 2x on [0, π]

(vii) f(x) = sin xex on 0 ≤ x ≤ π

(viii) f(x) = sin 3x on [0, π]

(ix) f(x) = e1-x2 on [−1, 1]

(x) f(x) = log (x2 + 2) − log 3 on [−1, 1]

(xi) f(x) = sin x + cos x on [0, π/2]

(xii) f(x) = 2 sin x + sin 2x on [0, π]

(xiii) fx=x2-sinπx6 on[-1, 0]

(xiv) fx=6xπ-4 sin2 x on [0, π/6]

(xv) f(x) = 4sin x on [0, π]

(xvi) f(x) = x2 − 5x + 4 on [1, 4]

(xvii) f(x) = sin4 x + cos4 x on 0, π2

(xviii) f(x) = sin x − sin 2x on [0, π]

Answer:

(i) The given function is fx=cos2x-π4=cos2x-π2=sin2x.

Since sin2x is everywhere continuous and differentiable.

Therefore, sin2x is continuous on 0, π2 and differentiable on 0, π2.

Also,
 fπ2=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π2 such that f'c=0.

We have
fx=sin2xf'x=2cos2x

f'x=02cos2x=0cos2x=0x=π4

Thus, c=π40, π2 such that f'c=0.  

Hence, Rolle's theorem is verified.

(ii) The given function is fx=sin2x.

Since sin2x is everywhere continuous and differentiable.

Therefore, sin2x is continuous on 0, π2 and differentiable on 0, π2.

Also,
 fπ2=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π2 such that f'c=0.

We have
fx=sin2xf'x=2cos2x

f'x=02cos2x=0cos2x=0x=π4

Thus, c=π40, π2 such that f'c=0.

​Hence, Rolle's theorem is verified.


(iii)

 The given function is fx=cos2x.

Since cos2x is everywhere continuous and differentiable, cos2x is continuous on -π4, π4 and differentiable on -π4, π4.

Also,
 fπ4=f-π4=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c-π4, π4 such that f'c=0.

We have
fx=cos2xf'x=-2sin2x

 f'x=0-2sin2x=0sin2x=0sin2x=0x=0

Since c=0-π4, π4 such that f'c=0.

​Hence, Rolle's theorem is verified.


(iv)

 The given function is fx=exsinx.

Since sinx & ex are everywhere continuous and differentiable.

Therefore, being a product of these two, fx is continuous on 0, π and differentiable on 0, π.

Also,
 fπ=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=exsinxf'x=exsinx+cosx

 f'x=0exsinx+cosx=0sinx+cosx=0tanx=-1x=π-π4=3π4

Since c=3π40, π such that f'c=0.

​Hence, Rolle's theorem is verified.


(v)

 The given function is fx=excosx.

Since cosx & ex are everywhere continuous and differentiable, fx being a product of these two is continuous on -π2, π2 and differentiable on -π2, π2.

Also, 
f-π2=fπ2=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c-π2, π2 such that f'c=0.

We have
fx=excosxf'x=excosx-sinx

 f'x=0excosx-sinx=0sinx-cosx=0tanx=1x=π4

Since c=π4-π2, π2 such that f'c=0.

​Hence, Rolle's theorem is verified.


(vi)

 The given function isfx=cos2x.

Since cos2x is everywhere continuous and differentiable.

Therefore, fx  is continuous on 0, π and differentiable on 0, π.

Also,
 fπ=f0=1

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=cos2xf'x=-2sin2x

 f'x=0-2sin2x=0sin2x=02x=πx=π2

Thus, c=π20, π such that f'c=0.

​Hence, Rolle's theorem is verified.


(vii)

 The given function is fx=sinxex.

Since cosx and ex are everywhere continuous and differentiable, being the quotient of these two, fx is continuous on 0, π and differentiable on 0, π.

Also,
 fπ=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=sinxexf'x=cosx-sinxex

 f'x=0cosx-sinxex=0cosx-sinx=0tanx=1x=π4

Thus, c=π40, π such that f'c=0.

​Hence, Rolle's theorem is verified.



(viii)

 The given function isfx=sin3x.

Since sin3x is everywhere continuous and differentiable, sin3x is continuous on 0, π and differentiable on 0,π.

Also,
 fπ=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=sin3xf'x=3cos3x

f'x=03cos3x=0cos3x=03x=π2, 3π2,....x=π6, π2, 5π6

Thus, c=π6, π2, 5π60, π such that f'c=0.

​Hence, Rolle's theorem is verified.


(ix)

 The given function isfx=e1-x2.

Since exponential function  is everywhere continuous and differentiable, e1-x2 is continuous on -1, 1 and differentiable on -1, 1.

Also,
 f1=f-1=1

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c-1, 1 such that f'c=0.

We have
fx=e1-x2f'x=-2xe1-x2

f'x=0-2xe1-x2=0x=0

Thus, c=0-1, 1 such that f'c=0.

​Hence, Rolle's theorem is verified.


(x)

 The given function is fx=logx2+2-log3, which can be rewritten as fx=logx2+23.

Since logarithmic function is differentiable and so continuous in its domain, fx=logx2+23 is continuous on -1, 1 and differentiable on -1, 1.

Also,
 f1=f-1=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c-1, 1 such that f'c=0.

We have
fx=logx2+23f'x=32xx2+2=6xx2+2

f'x=06xx2+2=0x=0

Thus, c=0-1, 1 such that f'c=0.

​Hence, Rolle's theorem is verified.


(xi)

 The given function isfx=sinx + cosx.
Since sinx and cosx are everywhere continuous and differentiable, fx=sinx + cosx is continuous on 0, π2 and differentiable on 0, π2.

Also,
 fπ2=f0=1

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π2 such that f'c=0.

We have
fx=sinx+cosxf'x=cosx-sinx

f'x=0cosx-sinx=0tanx=1x=π4

Thus, c=π40, π2 such that f'c=0.

​Hence, Rolle's theorem is verified.


(xii)
The given function isfx=2sinx +sin2x.

Since sinx & sin2x are everywhere continuous and differentiable, fx  is continuous on 0, π and differentiable on 0, π.

Also,
 fπ=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=2sinx+sin2xf'x=2cosx+2cos2x

 f'x=02cosx+2cos2x=0cosx+cos2x=0cosx+2cos2x-1=02cos2x+cosx-1=0cosx+1 2cosx-1=0cosx=-1, cosx=12cosx=cosπ, cosx=π3x=π, π3

Thus, c=π30, π such that f'c=0.

​Hence, Rolle's theorem is verified.


(xiii)


The given function isfx=x2-sinπx6.

Since sinx & x2 are everywhere continuous and differentiable, fx  is continuous on -1, 0 and differentiable on -1, 0.

Also,
 f-1=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c-1, 0 such that f'c=0.

We have
fx=x2-sinπx6f'x=12-π6cosπx6

 f'x=012-π6cosπx6=0cosπx6=3πx=-6πcos-13π

Thus, c=-6πcos-13π-1, 0 such that f'c=0.

​Hence, Rolle's theorem is verified.

(xiv)


The given function isfx=6xπ-4sin2x.

Since sin2x & x are everywhere continuous and differentiable, fx  is continuous on 0, π6 and differentiable on 0, π6.

Also,
 fπ6=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π6 such that f'c=0.

We have
fx=6xπ-4sin2xf'x=6π-8 sinx cosx

 f'x=06π-8sinxcosx=0sin2x=32πx=12sin-132π

Thus, c=12sin-132π0, π6 such that f'c=0.

​Hence, Rolle's theorem is verified.



(xv)

The given function isfx=4sinx.

Since sine function and exponential function are everywhere continuous and differentiable, fx  is continuous on 0, π and differentiable on 0, π.

Also,
 fπ=f0=1

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=4sinxf'x=4sinxcosxlog4

 f'x=04sinxcosxlog4=04sinxcosx=0cosx=0x=π2

Thus, c=π20, π such that f'c=0.

​Hence, Rolle's theorem is verified.


(xvi)  

According to Rolle’s theorem, if(x) is a real valued function defined on [ab] such that it is continuous on [ab], it is differentiable on (ab) and f(a) = f(b), then there exists a real number c ∈(ab) such that f(c) = 0.

Now, f(x) is defined for all x ∈[1, 4].
At each point of [1, 4], the limit of f(x) is equal to the value of the function. Therefore, f(x) is continuous on [1, 4].

Also, exists for all x ∈(1, 4).

So, f(x) is differentiable on (1, 4).

Also,
 f(1) = f(4) = 0

Thus, all the three conditions of Rolle’s theorem are satisfied.

Now, we have to show that there exists c ∈(1, 4) such that.

We have
 

f'x=02x-5=0x=52

[Since  ∈(1, 4) such that]

Hence, Rolle’s theorem is verified.


(xvii)

The given function is fx=sin4x + cos4x.
Since sinx and cosx are everywhere continuous and differentiable, fx=sin4x + cos4x is continuous on 0, π2 and differentiable on 0, π2.

Also,
 fπ2=f0=1

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π2 such that f'c=0.

We have
fx=sin4x+cos4xf'x=4sin3xcosx-4cos3xsinx

f'x=04sin3xcosx-4cos3xsinx=0sin3xcosx-cos3xsinx=0tan3x-tanx=0tanxtan2x-1=0tanx=0, tan2x=1tanx=0, tanx=±1x=0, x=π4, 3π4

Thus, c=π40, π2 such that f'c=0.

​Hence, Rolle's theorem is verified.


(xviii)

The given function is fx=sinx -sin2x.

Since sinx and sin2x are everywhere continuous and differentiable, fx  is continuous on 0, π and differentiable on 0, π.

Also,
 fπ=f0=0

Thus, fx satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists c0, π such that f'c=0.

We have
fx=sinx-sin2xf'x=cosx-2cos2x

 f'x=0cosx-2cos2x=0cosx-2cos2x+1=02cos2x-cosx-1=0cosx-1 2cosx+1=0cosx=1, cosx=-12cosx=cosπ2, cosx=2π3x=π2, 2π3

Thus,  c=π2, 2π30, π such that f'c=0.

​Hence, Rolle's theorem is verified.​

Page No 14.9:

Question 4:

Using Rolle's theorem, find points on the curve y = 16 − x2, x ∈ [−1, 1], where tangent is parallel to x-axis.

Answer:

The equation of the curve is 
y=16-x2.          ...(1)

Let Px1,y1 be a point on it where the tangent is parallel to the x-axis.

Then,
 dydxP=0       ...(2)

Differentiating (1) with respect to x, we get

dydx=-2xdydxP=-2x1-2x1=0   from 2x1=0

Px1, y1 lies on the curve y=16-x2.

 y1=16-x12

When x1=0,
 y1=16

Hence, 0, 16 is the required point.

Page No 14.9:

Question 5:

At what points on the following curves, is the tangent parallel to x-axis?
(i) y = x2 on [−2, 2]
(ii) y = e1-x2 on [−1, 1]
(iii) y = 12 (x + 1) (x − 2) on [−1, 2].

Answer:

(i) Let fx=x2
Since fx is a polynomial function, it is continuous on -2, 2 and differentiable on -2, 2.

Also, f2=f-2=4

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c-2, 2 for which f'c=0.

But f'c=02c=0c=0

 fc=f0=0

By the geometrical interpretation of Rolle's theorem, 0, 0 is the point on y=x2, where the tangent is parallel to the x-axis.

(ii) Let fx=e1-x2
Since fx is an exponential function, which is continuous and derivable on its domain, fx is continuous on -1, 1 and differentiable on -1, 1.

Also, f1=f-1=1

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c-1, 1 for which f'c=0.

But f'c=0-2ce1-c2=0c=0     e1-c20

 fc=f0=e

By the geometrical interpretation of Rolle's theorem, 0, e is the point on y=e1-x2 where the tangent is parallel to the x-axis.

(iii) Let fx=12x+1x-2    ...(1)

fx=12x2-x-2
fx=12x2-12x-24

Since fx is a polynomial function, fx is continuous on -1, 2 and differentiable on -1, 2.

Also, f2=f-1=0

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c-1, 2 for which f'c=0.

But f'c=024c-12=0c=12

 fc=f12=-123232=-27    (using (1))

By the geometrical interpretation of Rolle's theorem, 12,-27 is the point on y=12x+1x-2​ where the tangent is parallel to the x-axis.

Page No 14.9:

Question 6:

If f : [−5, 5] → R is differentiable and if f' (x) doesnot vanish anywhere, then prove that f (−5) ± f (5).

Answer:

It is given thatis a differentiable function.
Every differentiable function is a continuous function. Thus,
(a) f is continuous in [−5, 5].
(b) is differentiable in (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that

It is also given thatdoes not vanish anywhere.


Hence proved.

Page No 14.9:

Question 7:

Examine if Rolle's theorem is applicable to any one of the following functions.
(i) f (x) = [x] for x ∈ [5, 9]
(ii) f (x) = [x] for x ∈ [−2, 2]
Can you say something about the converse of Rolle's Theorem from these functions?

Answer:

By Rolle’s theorem, for a function, if

(a) f is continuous on [ab],

(b) f is differentiable on (ab) and

(c) (a) = f (b),

then there exists some c ∈ (ab) such that .

Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

(i)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = 5 and = 9.

Thus, f (x) is not continuous on [5, 9].

The differentiability of f on (5, 9) is checked in the following way.

Let be an integer such that n ∈ (5, 9).

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, is not differentiable on (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on.

(ii)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = −2 and = 2.

Thus, f (x) is not continuous on [−2, 2].

The differentiability of f on (−2, 2) is checked in the following way.

Let be an integer such that n ∈ (−2, 2).

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, is not differentiable on (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on.

Page No 14.9:

Question 8:

It is given that the Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, x [1, 2] at the point x = 43. Find the values of b and c.

Answer:

As, the Rolle's theorem holds for the function f(x) = x3 + bx2 + cxx  [1, 2] at the point x = 43

So, f1=f213+b12+c1=23+b22+c21+b+c=8+4b+2c3b+c+7=0                   .....iAnd f'43=03432+2b43+c=0                As, f'x=3x2+2bx+c163+8b3+c=08b+3c+16=0                  .....iiii-i×3, we ge8b-9b+16-21=0-b-5=0b=-5Substituting b=-5 in i, we get3-5+c+7=0-15+c+7=0c=8



View NCERT Solutions for all chapters of Class 15