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**Verify the Rolleâ€™s theorem for each of the functions in Exercises 65 to 69.****Q.65. f (x) = x (x â€“ 1) ^{2} in [0, 1].**

Given that: f(x) = x(x â€“ 1)

= 2x

= 3x

f(0) = 0(0 â€“ 1)

â‡’ f(0) = f(1) = 0

As the above conditions are satisfied, then there must exist at least one point c âˆˆ (0, 1) such that f â€²(c) = 0

âˆ´ f â€²(c) = 3c

â‡’ 3c(c - 1) - 1(c - 1) = 0 â‡’ (c - 1)(3c - 1) = 0

â‡’ c â€“ 1 = 0 â‡’ c = 1

3c â€“ 1 = 0 â‡’ 3c = 1

Hence, Rolleâ€™s Theorem is verified.

Given that:

continuous function in

= 4 sin

= 4 sin x cos x (sin

= â€“ 4 sin x cos x (cos

= â€“ 2.2 sin x cos x.cos 2x

= â€“ 2 sin 2x.cos 2x

= â€“ sin 4x which exists in

So, f (x) is differentiable in

Given that: f(x) = log (x

So, f (x) is differentiable in (â€“ 1, 1).

f(1) = log (1 + 2) - log 3 â‡’ log 3 - log 3 = 0

âˆ´ f(- 1) = f(1) = 0

As the above conditions are satisfied, then there must exist atleast one point c âˆˆ (- 1, 1) such that f â€²(c) = 0.

Hence, Rolleâ€™s Theorem is verified.

Given that: f(x) = x(x + 3) e

âˆ´ f(x) is continuous in [â€“ 3, 0]

So, f (x) is differentiable in (â€“ 3, 0).

f(0) = (0) (0 + 3) e

âˆ´ f(â€“ 3) = f(0) = 0

As the above conditions are satisfied, then there must exist

atleast one point c âˆˆ (- 3, 0) such that

f â€²(c) = 0â‡’

Which gives c = 3, c = â€“ 2 âˆˆ (- 3, 0).

Hence, Rolleâ€™s Theorem is verified.

Given that: f (x) =in [â€“ 2, 2].

âˆ´ f(x) is continuous in [â€“ 2, 2]

So, f â€²(x) is differentiable in (- 2, 2).

So f (â€“ 2) = f (2) = 0

As the above conditions are satisfied, then there must exist

atleast one point c âˆˆ (- 2, 2) such that

f â€²(c) = 0â‡’

Hence, Rolleâ€™s Theorem is verified.

and R.H.L. =

âˆ´ L.H.L. â‰ R.H.L.

So, f(x) is not differentiable at x = 1.

Hence, Rolleâ€™s Theorem is not applicable in [0, 2].

Given that: y = cos x â€“ 1 on [0, 2Ï€]

We have to find a point c on the given curve y = cos x â€“ 1 on [0, 2Ï€] such that the tangent at c âˆˆ [0, 2Ï€] is parallel to x-axis i.e., f â€²(c) = 0 where f â€²(c) is the slope of the tangent.

So, we have to verify the Rolleâ€™s Theorem.

So, it is continuous on [0, 2Ï€].

So, it is differentiable on (0, 2Ï€).

f(0) = cos 0 â€“ 1 = 1 â€“ 1 = 0; f(2Ï€) = cos 2Ï€ â€“ 1 = 1 â€“ 1 = 0

âˆ´ f(0) = f(2Ï€) = 0

As the above conditions are satisfied, then there lies a point

c âˆˆ (0, 2Ï€) such that f â€²(c) = 0.

âˆ´ - sin c = 0 â‡’ sin c = 0

âˆ´ c = nÏ€, n âˆˆ I

â‡’ c = Ï€ âˆˆ (0, 2Ï€)

Hence, c = Ï€ is the point on the curve in (0, 2Ï€) at which the tangent is parallel to x-axis.

Given that: y = x(x â€“ 4), x âˆˆ [0, 4]

Let f(x) = x(x - 4), x âˆˆ [0, 4]

(i) f(x) being an algebraic polynomial, is continuous function everywhere.

So, f(x) = x(x â€“ 4) is continuous in [0, 4].

(ii) f â€²(x) = 2x - 4 which exists in (0, 4).

So, f (x) is differentiable.

(iii) f(0) = 0(0 â€“ 4)

= 0 f(4) = 4(4 â€“ 4) = 0

So f(0) = f(4) = 0

As the above conditions are satisfied, then there must exist at least one point c âˆˆ (0, 4) such that f â€²(c) = 0

âˆ´ 2c - 4 = 0 â‡’ c = 2 âˆˆ (0, 4)

Hence, c = 2 is the point in (0, 4) on the given curve at which the tangent is parallel to the x-axis.

Given that: f (x) =

So, f (x) is differentiable.

As the above conditions are satisfied then there must exist a point

c âˆˆ (1, 4) such that

âˆ´

Hence, Mean Value Theorem is verified.

Given that: f(x) = x

So, f (x) is differentiable.

As the above conditions are satisfied, then there must exist atleast one point c âˆˆ (0, 1) such that

â‡’

â‡’

â‡’ 3c

â‡’ 3c

â‡’ 3c (c - 1) -1 (c - 1) = 0 â‡’ (c - 1) (3c - 1) = 0

â‡’ c â€“ 1 = 0 âˆ´ c = 1

3c â€“ 1 = 0

Hence, Mean Value Theorem is verified.

Given that: f(x) = sin x â€“ sin 2x in [0, Ï€]

So, f (x) is continuous on [0, Ï€].

So, f (x) is differentiable on (0, Ï€).

Since the above conditions are satisfied, then there must exist atleast one point c âˆˆ (0, Ï€) such that

cos c â€“ 2 cos 2c =

â‡’ cos c â€“ 2(2 cos

â‡’ 4 cos

Hence, Mean Value Theorem is verified.

Given that: f (x) =in [1, 5].

â‡’ x

So, f (x) is continuous on [1, 5].

So, f (x) is differentiable in [1, 5].

Since the above conditions are satisfied then there must exist atleast one point c âˆˆ (1, 5) such that

Squaring both sides

â‡’ 2c^{2} = 75 â€“ 3c^{2} â‡’ 5c^{2} = 75 â‡’ c^{2} = 15

âˆ´ c = Â± âˆš15 âˆˆ (1, 5)

Hence, Mean Value Theorem is verified.**Q.77. Find a point on the curve y = (x â€“ 3) ^{2}, where the tangent is parallel to the chord joining the points (3, 0) and (4, 1).**

Given that: y = (x â€“ 3)

Let f(x) = (x â€“ 3)

Hence, by mean value theorem, there must exist a point c on the curve at which the tangent is parallel to the chord joining the points (3, 0) and (4, 1).

âˆ´

âˆ´

If

Hence,is the point on the curve at which the tangent is

parallel to the chord joining the points (3, 0) and (4, 1)

Given that: y = 2x

Let f(x) = 2x

As per the Mean Value Theorem, there must exist a point

c âˆˆ (1, 2) on the curve at which the tangent is parallel to the chord joining the

points A(1, 0) and B(2, 1).

So

â‡’â‡’ 4c = 1 + 5 â‡’ 4c = 6

âˆ´

âˆ´

Hence,is the point on the curve at which the tangent is

parallel to the chord joining the points A(1, 0) and B(2, 1).

Given that:

L.H.L.

â‡’

R.H.L.

For existing the limit

q â€“ 2 â€“ p = 0 â‡’ q â€“ p = 2 ...(i)

If L.H.L. f â€²(1) = R.H.L. f â€²(1) then q = 5.

Now putting the value of q in eqn. (i)

5 â€“ p = 2 â‡’ p = 3.

Hence, value of p is 3 and that of q is 5.

Taking log on both sides

log x

â‡’ log x

â‡’ m log x + n log y = (m + n) log (x + y)

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’Hence proved.

**(ii)** Given that:

Differentiating both sides w.r.t. x

â‡’

Hence,Hence, proved.**Q.81. If x = sint and y = sin pt, prove that****Ans.**

Given that: x = sin t and y = sin pt

Differentiating both the parametric functions w.r.t.t

âˆ´

Again differentiating w.r.t. x,

â‡’

Now we have to prove that

Hence, proved.

Q.82. Find if y = x^{tanx} +**Ans.**

Given that: y =

Let u = x^{tanx }and v =

âˆ´ y = u + v

Differentiating both sides w.r.t. x

...(i)

Now taking u = x^{tan x }

Taking log on both sides log u = log (x^{tan x})

log u = tan x . log x

Differentiating both sides w.r.t. x

â‡’

âˆ´

Taking

Differentiating both sides w.r.t. x

Putting the values ofin eqn. (i)**Objective Type Questions****Q.83. If f (x) = 2x and g (x) =****then which of the following can be a discontinuous function**

We know that the algebraic polynomials are continuous functions everywhere.

âˆ´ f(x) + g(x) is continuous [âˆµ Sum, difference and product

f(x) - g(x) is continuous of two continuous functions is

f(x) . g(x) is continuous also continuous]

is only continuous if g(x) â‰ 0

Here,which is discontinuous at x = 0.

Hence, the correct option is (d).

Given that: f(x) =

For discontinuous function 4x â€“ x

â‡’ x(4 â€“ x

â‡’ x(2 â€“ x)(2 + x) = 0

â‡’ x = 0, x = â€“ 2, x = 2

Hence, the given function is discontinuous exactly at three points.

Hence, the correct option is (c).

Solution.

Given that: f(x) =sinx

Clearly, f(x) is not differentiable at x =

R.H.L. =

Also L.H.L. =

âˆ´ R.H.L. =

So, the given function f(x) is not differentiable at x =

âˆ´ f(x) is differentiable in R -

Hence, the correct option is (b).

Solution.

Given that: f(x) = cot x

â‡’

We know that sin x = 0 if f(x) is discontinuous.

âˆ´ If sin x = 0

âˆ´ x = nÏ€, n âˆˆ nÏ€.

So, the given function f(x) is discontinuous on the set {x = nÏ€; n âˆˆ Z}.

Hence, the correct option is (a).

Solution.

Given that: f(x) =

We know that modulus function is continuous but not differentiable in its domain.

Let g(x) = and t(x) = e

âˆ´ f(x) = got(x) = g[t(x)] =

Since g (x) and t(x) both are continuous at x = 0 but f(x) is not differentiable at x = 0.

Hence, the correct option is (a).

Solution.

Given that: f(x) = x

So, the value of the function f at x = 0, so that f(x) is continuous is 0.

Hence, the correct option is (a).

Solution.

Given that: f(x) = is continuous at x =

L.H.L. ==

R.H.L. =

When f(x) is continuous at x =

âˆ´ L.H.L. = R.H.L.

Hence, the correct option is (c).

(b) f is everywhere continuous but not differentiable at x = nÏ€, n âˆˆ Z.

(c) f is everywhere continuous but not differentiable at x = (2n + 1)

Solution.

Given that: f(x) =|sin x|

Let g(x) = sin x and t(x) =

âˆ´ f(x) = tog(x) = t[g(x)] = t(sin x) =|sin x|

where g(x) and t(x) both are continuous.

âˆ´ f(x) = got(x) is continuous but t(x) is not differentiable at x = 0.

So, f(x) is not continuous at sin x = 0 â‡’ x = nÏ€, n âˆˆ Z.

Hence, the correct option is (b).

Solution.

Given that: y =

â‡’ y = log (1 â€“ x

Differentiating both sides w.r.t. x

Hence, the correct option is (b).

Solution.

Given that: y =

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

Hence, the correct option is (a).

Solution.

Let y = cos

Differentiating both the functions w.r.t. x

â‡’

â‡’

Now

Hence, the correct option is (a).

Solution.

Given that x = t

Differentiating both the parametric functions w.r.t. t

âˆ´

Now differentiating again w.r.t. x

Hence, the correct option is (b).

(b) â€“ 1

(c) 3/2

Solution.

Given that: f(x) = x

We know that if f(x) = x

Theorem in [0 , âˆš3 ] , then

f â€²(c) = 0

â‡’ 3c

âˆ´ c = Â± 1 â‡’ 1 âˆˆ (0, âˆš3 )

Hence, the correct option is (a).

Solution.

Given that: f (x) = x +, x âˆˆ [1, 3]

We know that if f(x) = x +, x âˆˆ [1, 3] satisfies all the

conditions of mean value theorem then

where a = 1 and b = 3

Here c = âˆš3 âˆˆ (1, 3).

Hence, the correct option is (b).

is the function which is continuous everywhere

but fails to be differentiable at x = 0 and x = 1.

We can have more such examples.

Let y = x

Differentiating both the parametric functions w.r.t. xâ‡’

âˆ´

So, the derivative of x

Given that: f (x) = |cosx|

â‡’ f(x) = cos x if

Differentiating both sides w.r.t. x, we get f â€²(x) = - sin x

at

Given that: f (x) = |cosx â€“ sinx |

We know that sin x > cos x if

â‡’ cos x â€“ sin x < 0

âˆ´ f(x) = â€“ (cos x â€“ sin x)

f â€²(x) = - (- sin x - cos x) â‡’ f â€²(x) = (sin x + cos x)

âˆ´

Given that: âˆšx + âˆšy = 1

Differentiating both sides w.r.t. x

â‡’

â‡’

âˆ´

False. Given that f(x) =|x â€“ 1| in [0, 2].

We know that modulus function is not differentiable. So, it is false.

True. We know that modulus function is continuous function on its domain.

So, it is true.

True. We know that the sum and difference of two or more functions is always continuous. So, it is true.

True.

False. Let us take an example: f(x) = sin x and g(x) = cot x

âˆ´ f(x).g(x) = sin x.cot x == cos x which is continuous at x = 0

but cot x is not continuous at x = 0.

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