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Page No 15.17:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.18:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.19:

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 15.20:

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 15.20:

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 15.20:

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 15.20:

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 15.20:

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 15.20:

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 15.8:

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 15.9:

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 15.9:

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 15.9:

Question 7:

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 15.9:

Question 8:

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 15.9:

Question 9:

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 15.9:

Question 10:

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 15.9:

Question 11:

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



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