Continuity And Differentiability

A function f(x) is said to be continuous at x = c if,

Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) = x³ + 2x – 1 is continuous at x = 1 if -

Clearly,

LHL =

∴ LHL = (1-0)³ + 2(1-0) – 1 = 2 …(1)

Similarly, we proceed for RHL-

RHL =

∴ RHL = (1+0)³ + 2(1+0) – 1 = 2 …(2)

And,

f(1) = (1+0)³ + 2(1+0) – 1 = 2 …(3)

Clearly from equation 1 , 2 and 3 we can say that

∴ f(x) is continuous at x = 1

Given,

…(1)

We need to check its continuity at x = 2

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 2 if -

Clearly,

LHL = {using equation 1}

∴ LHL = (2-0)² = 4 …(2)

Similarly, we proceed for RHL-

RHL =

∴ RHL = 3(2+0) + 5 = 11 …(3)

And,

f(2) = 3(2) + 5 = 11 …(4)

Clearly from equation 2, 3 and 4 we can say that

∴ f(x) is discontinuous at x = 2

TAG:

Given,

…(1)

We need to check its continuity at x = 0

A function f(x) is said to be continuous at x = c if,

Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 0 if -

Clearly,

LHL = {using equation 1}

As we know cos(-θ) = cos θ

⇒ LHL =

∵ 1 – cos 2x = 2sin²x

∴ LHL =

As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)

As we know,

∴ LHL = 2 × 1² = 2 …(2)

Similarly, we proceed for RHL-

RHL =

⇒ RHL =

⇒ RHL =

Again, using sandwich theorem, we get -

RHL = 2 × 1² = 2 …(3)

And,

f (0) = 5 …(4)

Clearly from equation 2, 3 and 4 we can say that

∴ f(x) is discontinuous at x = 0

Given,

…(1)

We need to check its continuity at x = 2

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 2 if -

Clearly,

LHL = {using equation 1}

⇒ LHL =

⇒ LHL =

⇒ LHL =

∴ LHL = 5 -2(0) = 5 …(2)

Similarly we proceed for RHL-

RHL = {using equation 1}

⇒ RHL =

⇒ RHL =

⇒ RHL =

∴ RHL = 5 + 2(0) = 5 …(3)

And,

f(2) = 5 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is continuous at x = 2

Given,

…(1)

We need to check its continuity at x = 4

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 4 if -

Clearly,

LHL = {using equation 1}

⇒ LHL =

∵ h > 0 as defined above.

∴ |-h| = h

⇒ LHL =

∴ LHL = -1/2 …(2)

Similarly, we proceed for RHL-

RHL = {using equation 1}

⇒ RHL =

∵ h > 0 as defined above.

∴ |h| = h

⇒ RHL =

∴ RHL = 1/2 …(3)

And,

f(4) = 0 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is discontinuous at x = 4

Given,

…(1)

We need to check its continuity at x = 0

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 4 if -

Clearly,

LHL = {using equation 1}

∵ h > 0 as defined above.

∴ |-h| = h

⇒ LHL =

As cos (1/h) is going to be some finite value from -1 to 1 as h→0

∴ LHL = 0 × (finite value) = 0 …(2)

Similarly we proceed for RHL-

RHL = {using equation 1}

∵ h > 0 as defined above.

∴ |h| = h

⇒ RHL =

As cos (1/h) is going to be some finite value from -1 to 1 as h→0

∴ RHL = 0 × (finite value) = 0 …(3)

And,

f(0) = 0 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is continuous at x = 0

Given,

…(1)

We need to check its continuity at x = a

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = a if -

Clearly,

LHL = {using eqn 1}

⇒ LHL =

∵ h > 0 as defined above.

∴ |-h| = h

⇒ LHL =

As sin (-1/h) is going to be some finite value from -1 to 1 as h→0

∴ LHL = 0 × (finite value) = 0 …(2)

Similarly we proceed for RHL-

RHL = {using eqn 1}

∵ h > 0 as defined above.

∴ |h| = h

⇒ RHL =

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ RHL = 0 × (finite value) = 0 …(3)

And,

f(a) = 0 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is continuous at x = a

Given,

…(1)

We need to check its continuity at x = 0

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 4 if -

Clearly,

LHL = {using equation 1}

⇒ LHL =

∴ LHL = 0 …(2)

Similarly we proceed for RHL-

RHL = {using equation 1}

⇒ RHL =

⇒ RHL =

∴ RHL = 1 …(3)

And,

f(0) = 0 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is discontinuous at x = 0

Given,

…(1)

We need to check its continuity at x = 1

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 1 if -

Clearly,

LHL = {using equation 1}

∴ LHL = (1-0)²/2 = 1/2 …(2)

Similarly, we proceed for RHL-

RHL = {using eqn 1}

⇒ RHL =

⇒ RHL =

⇒ RHL =

∴ RHL = 2(0)² + 0 + 1/2 = 1/2 …(3)

And,

f(1) = 1²/2 = 1/2 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is continuous at x = 1

Given,

f(x) = |x| + |x – 1| …(1)

We need to check its continuity at x = 1

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = 1 if -

Clearly,

LHL = {using eqn 1}

⇒ LHL =

∵ h > 0 as defined above and h→0

∴ |-h| = h

And (1 – h) > 0

∴ |1 – h| = 1 - h

⇒ LHL =

∴ LHL = 1 …(2)

Similarly we proceed for RHL-

RHL = {using eqn 1}

⇒ RHL =

∵ h > 0 as defined above and h→0

∴ |h| = h

And (1 + h) > 0

∴ |1 + h| = 1 + h

⇒ RHL =

∴ RHL = 1 + 2(0) = 1 …(3)

And,

f(1) = |1|+|1-1| = 1 {using eqn 1} …(4)

Clearly from equation 2 , 3 and 4 we can say that

∴ f(x) is continuous at x = 1

Given,

…(1)

We need to find the value of k such that f(x) is continuous at x = 5.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 5.

∴

As we have to find k so pick out a combination so that we get k in our equation.

In this question we take LHL = f(5)

∴

⇒ {using equation 1}

⇒ 3(5 – 0) – 8 = 2k

⇒ 15 – 8 = 2k

⇒ 2k = 7

∴ k = 7/2

Given,

…(1)

We need to find the value of k such that f(x) is continuous at x = 2.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 2.

∴

As we have to find k so pick out a combination so that we get k in our equation.

In this question we take LHL = f(5)

∴

⇒ {using equation 1}

⇒

⇒

⇒

As the limit can’t be evaluated directly as it is taking 0/0 form.

So, use the formula:

Divide the numerator and denominator by -h to match with the form in formula-

Using algebra of limits, we get,

k =

∴ k =

Given,

…(1)

We need to find the value of k such that f(x) is continuous at x = 0.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 0.

∴

As we have to find k so pick out a combination so that we get k in our equation.

In this question we take LHL = f(0)

∴

⇒ {using eqn 1}

⇒ -1

As we can’t find the limit directly because it is taking 0/0 form.

So, we will rationalize it.

⇒ -1

Using (a+b)(a-b) = a² – b² , we have –

-1

⇒ -1

⇒ -1

⇒ = -1

∴ 2k/2 = -1

∴ k = -1

Given,

…(1)

We need to find the value of k such that f(x) is continuous at x = 0.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 0.

∴

As we have to find k so pick out a combination so that we get k in our equation.

In this question we take LHL = f(0)

∴

⇒ {using equation 1}

∵ cos(-x) = cos x and sin(-x) = - sin x

∴

Also, 1 – cos x = 2 sin² (x/2)

∴

As this limit can be evaluated directly by putting value of h because it is taking indeterminate form(0/0)

So we use sandwich or squeeze theorem according to which –

⇒ 2

Dividing and multiplying by (kh/2)² to match the form in formula we have-

⇒

Using algebra of limits we get –

⇒

Applying the formula-

⇒ 1 × (k²/4) = (1/4)

⇒ k² = 1

⇒ (k+1)(k – 1) = 0

∴ k = 1 or k = -1

Given,

…(1)

We need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now, We need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k

If we show that,

Then there will not be involvement of k in the equation & we can easily prove it.

So let’s take LHL first –

LHL =

⇒ LHL =

⇒ LHL =

∵ h > 0 as defined in theory above.

∴ |-h| = h

∴ LHL =

⇒ LHL =

∴ LHL = …(2)

Now Let’s find RHL,

RHL =

⇒ RHL =

⇒ RHL =

∵ h > 0 as defined in theory above.

∴ |h| = h

∴ RHL =

⇒ RHL =

∴ RHL = …(3)

Clearly form equation 2 and 3,we get

LHL ≠ RHL

Hence,

f(x) is discontinuous at x = 0 irrespective of the value of k.

Given,

…(1)

We need to find the value of a & b such that f(x) is continuous at x = 4.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 4.

∴

As we have to find a & b, so pick out a combination so that we get a or b in our equation.

In this question first we take LHL = f(4)

∴

⇒ {using equation 1}

⇒

∵ h > 0 as defined in theory above.

∴ |-h| = h

⇒

⇒

⇒ a – 1 = a + b

∴ b = -1

Now, taking other combination,

RHL = f(4)

⇒ {using equation 1}

⇒

∵ h > 0 as defined in theory above.

∴ |h| = h

⇒

⇒

⇒ b + 1 = a + b

∴ a = 1

Hence,

a = 1 and b = -1

Given,

f(x) =

To find: Points discontinuity of composite function f(f(x))

As f(x) is not defined at x = -2 as denominator becomes 0,at x = -2.

∴ x = -2 is a point of discontinuity

∵ f(f(x)) =

Clearly f(f(x)) is not defined at x = -5/2 as denominator becomes 0, at x = -5/2.

∴ x = -5/2 is another point of discontinuity

Thus f(f(x)) has 2 points of discontinuity at x = -2 and x = -5/2

Given,

To find: Points discontinuity of function f(t) where t =

As t is not defined at x = 1 as denominator becomes 0,at x = 1.

∴ x = 1 is a point of discontinuity

∵ f(t) =

⇒ f(t) =

Clearly f(t) is not going to be defined whenever denominator is 0 and thus will give a point of discontinuity.

∴ Solution of the following equation gives other points of discontinuities.

-2x² + 5x – 2 = 0

⇒ 2x² – 5x + 2 = 0

⇒ 2x² – 4x – x + 2 = 0

⇒ 2x(x – 2) – (x – 2) = 0

⇒ (2x – 1)(x – 2) = 0

∴ x = 2 or x = 1/2

Hence,

f(t) is discontinuous at x = 1, x = 2 and x = 1/2

Given,

f(x) = |sin x + cos x| …(1)

We need to prove that f(x) is continuous at x = π

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = π if -

Clearly,

LHL =

⇒ LHL {using eqn 1}

∵ sin (π – x) =sin x & cos (π – x) = - cos x

⇒ LHL =

⇒ LHL = | sin 0 – cos 0 | = |0 – 1|

∴ LHL = 1 …(2)

Similarly, we proceed for RHL-

RHL =

⇒ RHL {using eqn 1}

∵ sin (π + x) = -sin x & cos (π + x) = - cos x

⇒ RHL =

⇒ RHL = | - sin 0 – cos 0 | = |0 – 1|

∴ RHL = 1 …(3)

Also, f(π) = |sin π + cos π| = |0 – 1| = 1 …(4)

Clearly from equation 2, 3 and 4 we can say that

∴ f(x) is continuous at x = π …proved

Given,

…(1)

We need to check whether f(x) is continuous and differentiable at x = 2

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).

Mathematically we can represent it as-

Finally, we can state that for a function to be differentiable at x = c

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 2 if -

∴ LHL =

⇒ LHL = {using equation 1}

Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].

E.g. [1.29] = 1; [-4.65] = -4 ; [9] = 9

∵ [2-h] is just less than 2 say 1.9999 so [1.999] = 1

⇒ LHL = (2-0) ×1

∴ LHL = 2 …(2)

Similarly,

RHL =

⇒ RHL = {using equation 1}

∴ RHL = (1+0)(2+0) = 2 …(3)

And, f(2) = (2-1)(2) = 2 …(4) {using equation 1}

From equation 2,3 and 4 we observe that:

∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 2 if -

∴ LHD =

⇒ LHD = {using equation 1}

Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].

E.g. [1.29] = 1 ; [-4.65] = -4 ; [9] = 9

∵ [2-h] is just less than 2 say 1.9999 so [1.999] = 1

⇒ LHD =

⇒ LHD =

∴ LHD = 1 …(5)

Now,

RHD =

⇒ RHD = {using equation 1}

⇒ RHD =

∴ RHD =

⇒ RHD = 0+3 = 3 …(6)

Clearly from equation 5 and 6,we can conclude that-

(LHD at x=2) ≠ (RHD at x = 2)

∴ f(x) is not differentiable at x = 2

Given,

…(1)

We need to check whether f(x) is continuous and differentiable at x = 0

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).

Mathematically we can represent it as-

Finally, we can state that for a function to be differentiable at x = c

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 0 if -

∴ LHL =

⇒ LHL = {using equation 1}

As sin (-1/h) is going to be some finite value from -1 to 1 as h→0

∴ LHL = 0² × (finite value) = 0

∴ LHL = 0 …(2)

Similarly,

RHL =

⇒ RHL = {using equation 1}

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ RHL = (0)²(finite value) = 0 …(3)

And, f(0) = 0 {using equation 1} …(4)

From equation 2,3 and 4 we observe that:

∴ f(x) is continuous at x = 0. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 0 if -

∴ LHD =

⇒ LHD = {using equation 1}

⇒ LHD =

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ LHD = 0×(some finite value) = 0

∴ LHD = 0 …(5)

Now,

RHD =

⇒ RHD = {using equation 1}

⇒ RHD =

As sin (1/h) is going to be some finite value from -1 to 1 as h→0

∴ RHD = 0×(some finite value) = 0

∴ RHD = 0 …(6)

Clearly from equation 5 and 6,we can conclude that-

(LHD at x=0) = (RHD at x = 0)

∴ f(x) is differentiable at x = 0

Given,

…(1)

We need to check whether f(x) is continuous and differentiable at x = 2

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).

Mathematically we can represent it as-

Finally we can state that for a function to be differentiable at x = c

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 2 if -

∴ LHL =

⇒ LHL = {using equation 1}

⇒ LHL =

∴ LHL = (3-h) = 3

∴ LHL = 3 …(2)

Similarly,

RHL =

⇒ RHL = {using equation 1}

⇒ RHL =

∴ RHL = 3+0 = 3 …(3)

And, f(2) = 1 + 2 = 3 {using equation 1} …(4)

From equation 2,3 and 4 we observe that:

∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 2 if -

∴ LHD =

⇒ LHD = {using equation 1}

⇒ LHD =

∴ LHD = 1 …(5)

Now,

RHD =

⇒ RHD = {using equation 1}

⇒ RHD =

∴ RHD = -1 …(6)

Clearly from equation 5 and 6,we can conclude that-

(LHD at x=2) ≠ (RHD at x = 2)

∴ f(x) is not differentiable at x = 2

Given,

f(x) = |x - 5| …(1)

We need to prove that f(x) is continuous but not differentiable at x = 5

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).

Mathematically we can represent it as-

Finally we can state that for a function to be differentiable at x = c

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 5 if -

∴ LHL =

⇒ LHL = {using equation 1}

⇒ LHL =

∴ LHL = |-0| = 0

∴ LHL = 0 …(2)

Similarly,

RHL =

⇒ RHL = {using equation 1}

⇒ RHL =

∴ RHL = |0| = 0 …(3)

And, f(5) = |5-5| = 0 {using equation 1} …(4)

From equation 2,3 and 4 we observe that:

∴ f(x) is continuous at x = 5. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 2 if -

∴ LHD =

⇒ LHD = {using equation 1}

As h > 0 as defined in theory above.

∴ |-h| = h

⇒ LHD =

∴ LHD = -1 …(5)

Now,

RHD =

⇒ RHD = {using equation 1}

As h > 0 as defined in theory above.

∴ |h| = h

⇒ RHD =

∴ RHD = 1 …(6)

Clearly from equation 5 and 6,we can conclude that-

(LHD at x=5) ≠ (RHD at x = 5)

∴ f(x) is not differentiable at x = 5 but continuous at x = 5.

Hence proved.

Given f(x) is differentiable at x = 0 and f(x) ≠ 0

And f(x + y) = f(x)f(y) also f’(0) = 2

To prove: f’(x) = 2f(x)

As we know that,

f’(x) =

as f(x+h) = f(x)f(h)

∴ f’(x) =

⇒ f’(x) = …(1)

As

f(x + y) = f(x)f(y)

put x = y = 0

∴ f(0+0) = f(0)f(0)

⇒ f(0) = {f(0)}²

∴ f(0) = 1 {∵ f(x) ≠ 0 ….given}

∴ equation 1 is deduced as-

f’(x) =

⇒ f’(x) =

⇒ f’(x) = f(x)f’(0) {using formula of derivative}

∴ f’(x) = 2 f(x) …proved {∵ it is given that f’(0) = 2}

Given:

Let Assume

Now, Taking Log on both sides we get,

Now, Differentiate w.r.t x

Now, substitute the value of y

Hence,

We have given

Let us Assume

Now, Taking log on both sides, we get

Since we know,

Since we know, log aⁿ =n log a

Now, Differentiate w.r.t x

Hence,

Given:

To find: Differentiation w.r.t x

we have

Let us Assume

Now, Differentiate w.r.t t

And,

Now, differentiate w.r.t x

Now, using chain rule, we get

Substitute the value of t

Given: log [log (log x⁵)]

To Find: Differentiate the given function w.r.t x

Let Assume y= log [log (log x⁵)]

y= log [log (log x⁵)]

Let log(log x⁵) = u

Let Assume log x⁵=v

Let Assume x⁵=w

Differentiate both side w.r.t x

Now, Bu using chain rule we get, Differentiation of log [log (log x⁵)]

Now, Substitute the value of u , v and w then, we get

Hence, This the differentiation of given function.

Given:

We have

Differentiate w.r.t x

we have sinⁿ (ax² + bx + c)

)

Since, we know, xⁿ=nx^n-1

We have given

Let us Assume

And

So, y= cos v

Now, differentiate w.r.t v

And, v= tan w

Now, again differentiate w.r.t. w

And, we know,

So, differentiate w w.r.t. x we get

Now, using chain rule we get,

Substitute the value of v and w ,

Hence, dy/dy is the differentiation of function.

Let us Assume y= sinx²+sin²x+sin²(x²)

Now, differentiate y w.r.t x

We know,

Hence,

we have given

Let assume

So,

Now, differentiate y w.r.t t

And, differentiate t w.r.t x

Now, using chain we get dy/dx

Substitute the value of t

Hence, this is the differentiation of

Given: (Sin x)^{cos x}

To Find: Differentiate w.r.t x

We have (Sin x)^{cos x}

Let y=(Sin x)^{cos x}

Now, Taking Log on both sides, we get

Log y = cos x.log(sin x)

Now, Differentiate both side w.r.t. x

By using product rule of differentiation

Substitute the value of y , we get

y’ = (Sin x)^{cos x}[cos x.cot x - sin x (log sinx)]

Hence, y’ = (Sin x)^{cos x}[cos x.cot x - sin x (log sinx)]

we have sin^mx.cosⁿx

Taking log both side , we get

Differentiate w.r.t x

Substitute the value of y we get,

We have given, (x + 1)² (x + 2)³ (x + 3)⁴

Let Assume, y=(x + 1)² (x + 2)³ (x + 3)⁴

So,

y=(x + 1)² (x + 2)³ (x + 3)⁴

Taking log both side

Log y=log [(x + 1)² (x + 2)³ (x + 3)⁴]

Log y=log(x + 1)² +log(x + 2)³ +log(x + 3)⁴]

Log y=2log(x + 1) +3log(x + 2) +4log(x + 3)]

Differentiate w.r.t x

= (x+1)(x+2)²(x+3)³[9x²+34x+29]

As we know sin π/4 = cos π/4 = 1/√2

We know cos (a-b) = sin a sin b + cos a cos b

Now,

= - 1

We have

We know,

As the interval is

Differentiate w.r.t. x

We have given tan^-1(sec x + tan x)

Let us Assume (sec x +tan x) =t

SO, y = tan^-1 t

Now, differentiate w.r.t t

Since,

And, t= (sec x+tan x)

Differentiate t w.r.t x

Therefore,

When, substitute the value of t we get

Hence, dy/dx=1/2

We have given

Now, divide by cos x in both numerator and denominator

Since, we know, tan^-1x – tan^-1y =

So,

Hence, -1

We have given,

Let Assume ,

Put x = cos θ then θ=cos^-1x

SO,

And, 4cos³θ-3cosθ=cos3θ

Therefore,

Substitute the value of θ

Now, Differentiate w.r.t x

Since,

Hence,

We have given,

Let us Assume,

Now, put x= a tan θ then θ= tan^-1x/a

So, ,

While, substitute the value of θ

Now, differentiate w.r.t x

Hence,

We have given

Let Assume ,

Put x²= cos 2θ

So,

Differentiate, y w.r.t x

We have given, two parametric equation,

Now, differentiate both equation w.r.t x

We know,

So,

---(i)

and,

---(ii)

Now,

Hence,

we have two equation

Now, differentiate w.r.t θ

So,

---(i)

Also,

---(ii)

We have given, x=3cosθ -2cos³θ , y=3sinθ-2sin³θ

Now, differentiate both the equation w.r.t. x then we get

x=3cosθ -2cos³θ

---(i)

And, for y=3sinθ-2sin³θ

---(ii)

Therefore,

Hence, dy/dx=cot θ

We have given two parametric equation:

Let us Assume t= tan θ

So,

Therefore,

sin x=sin 2θ

x=2θ ---(i)

Also,

Tan y=tan 2θ

y=2θ ---(ii)

from equation (i) and (ii)

y=x

now, differentiate w.r.t x

Hence, dy/dx=1

We have given,

Now, differentiate w.r.t t

---(i)

Also,

---(i)

Now,

Hence,

x = e^cos2t and y = e^sin2t

Now x = e^cos2t,

Taking log on both sides to get,

log x = cos 2t

For y = e^sin2t

Taking log on both sides we get,

log y = sin 2t

∴ cos²2t + sin²2t = (log x)² + (log y)²

1 = (log x)² + (log y)²

Differentiating w.r.t x,

x = asin 2t (1 + cos2t) and y = bcos2t

Differentiate w.r.t t

x = asin 2t (1 + cos2t)

---(i)

Also,

y = bcos2t

Now, for dy/dx

Hence, Proved

x = 3sint – sin3t, y = 3cost – cos 3t

Differentiate w.r.t t in both equation

x = 3sint – sin3t

Now, for y

y = 3cost – cos 3t

At t = π/3

Let us Assume ,

Now, differentiate w.r.t x

----(i)

And, v = sin x

Now,

we have

Let us Assume, p= and q = tan^-1x

And, put x= tan θ

So,

,

Substitute the vaue of θ , we get

Now, differentiate p w.r.t x

-----(i)

Now, for q= tan^-1 x

Differentiate q w.r.t. x

----(ii)

Now, for dp/dq, from equation (i) and (ii)

Hence, dp/dq=1/2

We have,

Use chain rule and quotient rule to get:

Chain Rule

f(x) = g(h(x))

f’(x) = g’(h(x))h’(x)

By Quotient Rule

On differentiating both the sides with respect to x, we get

By product rule:

Multiplying by y² to both the sides, we get

We have,

sec(x + y) = xy

By the rules given below:

Chain Rule

f(x) = g(h(x))

f’(x) = g’(h(x))h’(x)

Product rule:

On differentiating both sides with respect to x, we get

We have,

tan^-1(x² + y²) = a

∴ x² + y² = tan a

On differentiating both sides with respect to x, we get

We have,

(x² + y²)² = xy

By product rule:

Given: ax² + 2hxy + by² + 2gx + 2fy + c = 0 …(i)

Differentiating the above with respect to x, we get

…(ii)

Now, we again differentiate eq (i) with respect to y, we get

…(iii)

Now, multiplying Eq. (ii) and (iii), we get

Hence Proved

Given:

Taking log on both the sides, we get

or ylogx = x

On differentiating above with respect to x, we get

Hence Proved

Given: y^x = e^{y – x}

Taking log on both the sides, we get

log y^x = log e^{y – x}

⇒ x logy = y – x

…(i)

On differentiating both the sides with respect to x, we get

[using (i)]

Hence Proved

Given:

⇒ y = (cos x)^y

Taking log both the sides, we get

log y = y log(cos x)

On differentiating both the sides, we get

Hence Proved

Given: x sin(a + y) + sin a cos(a + y) = 0

⇒ x = - sin a cot (a + y)

Differentiating with respect to y, we get

Hence Proved

Given:

Put x = sin α and y = sin β …(i)

Now, we know that sin²θ + cos² θ = 1

⇒ cos α + cos β = a( sin α – sin β) …(ii)

Now, we use some trigonometry formulas,

So, eq.(ii) become

⇒ 2 cot^-1 a = α – β …(iii)

Now, from eq.(i), we have

α = sin^-1 x and β = sin^-1 y

Now, put value of α and β in eq. (iii), we get

2 cot^-1 a = sin^-1 x – sin^-1 y

or sin^-1 x – sin^-1 y = 2cot^-1 a

On differentiating above with respect to x, we get

Hence Proved

Given: y = tan^-1x …(i)

⇒ tan y = x

On differentiating Eq. (i) with respect to x, we get

Now, again differentiating the above with respect to x, we get

Given: f(x) = x(x – 1)²

⇒ f(x) = x (x² + 1 – 2x)

⇒ f(x) = x³ + x – 2x² in [0,1]

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

On expanding f(x) = x(x – 1)², we get f(x) = x³ + x – 2x²

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R

⇒ f(x) = x³ + x – 2x² is continuous at x ∈ [0,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = x³ + x – 2x²

Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R

⇒ f(x) is differentiable at [0,1]

Hence, condition 2 is satisfied.

Condition 3:

f(x) = x³ + x – 2x²

f(0) = 0

f(1) = (1)³ + (1) – 2(1)² = 1 + 1 – 2 = 0

Hence, f(0) = f(1)

Hence, condition 3 is also satisfied.

Now, let us show that c ∈ (0,1) such that f’(c) = 0

f(x) = x³ + x – 2x²

On differentiating above with respect to x, we get

f’(x) = 3x² + 1 – 4x

Put x = c in above equation, we get

f’(c) = 3c² + 1 – 4c

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

f’(c) = 0

3c² + 1 – 4c = 0

On factorising, we get

⇒ 3c² – 3c – c + 1 = 0

⇒ 3c(c – 1) – 1(c – 1) = 0

⇒ (3c – 1) (c – 1) = 0

⇒ (3c – 1) = 0 or (c – 1) = 0

So, value of

Thus, Rolle’s theorem is verified.

Given: f(x) = sin⁴x + cos⁴x

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

f(x) = sin⁴x + cos⁴x

Since, f(x) is a trigonometric function and trigonometric function is continuous everywhere

⇒ f(x) = sin⁴x + cos⁴x is continuous at

Hence, condition 1 is satisfied.

Condition 2:

f(x) = sin⁴x + cos⁴x

On differentiating above with respect to x, we get

f’(x) = 4 × sin³ (x) × cos x + 4 × cos³ x × (- sin x)

⇒ f’(x) = 4sin³ x cos x – 4 cos³ x sinx

⇒ f’(x) = 4sin x cos x [sin²x – cos² x]

⇒ f’(x) = 2 sin2x [sin²x – cos² x]

[∵ 2 sin x cos x = sin 2x]

⇒ f’(x) = 2 sin 2x [- cos 2x]

[∵ cos² x – sin² x = cos 2x]

⇒ f’(x) = - 2 sin 2x cos 2x

⇒ f(x) is differentiable at

Hence, condition 2 is satisfied.

Condition 3:

f(x) = sin⁴x + cos⁴x

f(0) = sin⁴(0) + cos⁴(0) = 1

Hence, condition 3 is also satisfied.

Now, let us show that c ∈ () such that f’(c) = 0

f(x) = sin⁴x + cos⁴x

⇒ f’(x) = - 2 sin 2x cos 2x

Put x = c in above equation, we get

⇒ f’(c) = - 2 sin 2c cos 2c

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

f’(c) = 0

⇒ - 2 sin 2c cos 2c = 0

⇒ sin 2c cos 2c = 0

⇒ sin 2c = 0

⇒ 2c = 0

⇒ c = 0

Now, cos 2c = 0

Thus, Rolle’s theorem is verified.

Given: f(x) = log (x² + 2) – log3

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

f(x) = log (x² + 2) – log3

Since, f(x) is a logarithmic function and logarithmic function is continuous for all values of x.

⇒ f(x) = log (x² + 2) – log3 is continuous at x ∈ [-1,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = log (x² + 2) – log3

On differentiating above with respect to x, we get

⇒ f(x) is differentiable at [-1,1]

Hence, condition 2 is satisfied.

Condition 3:

∴f(-1) = f(1)

Hence, condition 3 is also satisfied.

Now, let us show that c ∈ (-1,1) such that f’(c) = 0

Put x = c in above equation, we get

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

f’(c) = 0

⇒ 2c = 0

⇒ c = 0 ∈ (-1, 1)

Thus, Rolle’s theorem is verified.

Given: f(x) = x(x + 3)e^-x/2

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

Since, f(x) is multiplication of algebra and exponential function and is defined everywhere in its domain.

is continuous at x ∈ [-3,0]

Hence, condition 1 is satisfied.

Condition 2:

On differentiating f(x) with respect to x, we get

[by product rule]

⇒ f(x) is differentiable at [-3,0]

Hence, condition 2 is satisfied.

Condition 3:

= [9 – 9]e^3/2

= 0

= 0

Hence, f(-3) = f(0)

Hence, condition 3 is also satisfied.

Now, let us show that c ∈ (0,1) such that f’(c) = 0

On differentiating above with respect to x, we get

Put x = c in above equation, we get

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

f’(c) = 0

⇒ (c – 3)(c + 2) = 0

⇒ c – 3 = 0 or c + 2 = 0

⇒ c = 3 or c = -2

So, value of c = -2, 3

c = -2 ∈ (-3, 0) but c = 3 ∉ (-3, 0)

∴ c = -2

Thus, Rolle’s theorem is verified.

Given:

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

Firstly, we have to show that f(x) is continuous.

Here, f(x) is continuous because f(x) has a unique value for each x ∈ [-2,2]

Condition 2:

Now, we have to show that f(x) is differentiable

[using chain rule]

∴ f’(x) exists for all x ∈ (-2, 2)

So, f(x) is differentiable on (-2,2)

Hence, Condition 2 is satisfied.

Condition 3:

Now, we have to show that f(a) = f(b)

so, f(a) = f(-2)

and f(b) = f(2)

∴ f(-2) = f(2) = 0

Hence, condition 3 is satisfied

Now, let us show that c ∈ (0,1) such that f’(c) = 0

On differentiating above with respect to x, we get

Put x = c in above equation, we get

Thus, all the three conditions of Rolle’s theorem is satisfied. Now we have to see that there exist c ∈ (-2,2) such that

f’(c) = 0

⇒ c = 0

∵ c = 0 ∈ (-2, 2)

Hence, Rolle’s theorem is verified

Given:

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

At x = 1

LHL =

RHL =

∵ LHL = RHL = 2

and f(1) = 3 – x = 3 – 1 = 2

∴ f(x) is continuous at x = 1

Hence, condition 1 is satisfied.

Condition 2:

Now, we have to check f(x) is differentiable

On differentiating with respect to x, we get

Now, let us consider the differentiability of f(x) at x = 1

LHD ⇒ f(x) = 2x = 2(1) = 2

RHD ⇒ f(x) = -1 = -1

∵ LHD ≠ RHD

∴ f(x) is not differentiable at x = 1

Thus, Rolle’s theorem is not applicable to the given function.

Given: Equation of curve, y = cos x – 1

Firstly, we differentiate the above equation with respect to x, we get

Given tangent to the curve is parallel to the x – axis

This means, Slope of tangent = Slope of x – axis

⇒ - sin x = 0

⇒ sin x = 0

⇒ x = sin^-1(0)

⇒ x = π ∈ (0, 2π)

Put x = π in y = cos x – 1, we have

y = cos π – 1 = -1 – 1 = -2 [∵ cos π = -1]

Hence, the tangent to the curve is parallel to the x –axis at

(π, -2)

Given: y = x(x – 4)

⇒ y = (x² – 4x)

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

On expanding y = x(x – 4), we get y = (x² – 4x)

Since, x² – 4x is a polynomial and we know that, every polynomial function is continuous for all x ∈ R

⇒ y = (x² – 4x) is continuous at x ∈ [0,4]

Hence, condition 1 is satisfied.

Condition 2:

y = (x² – 4x)

y’ = 2x – 4

⇒ x² - 4x is differentiable at [0,4]

Hence, condition 2 is satisfied.

Condition 3:

y = x² – 4x

when x = 0

y = 0

when x = 4

y = (4)² – 4(4) = 16 – 16 = 0

Hence, condition 3 is also satisfied.

Now, there is atleast one value of c ∈ (0,4)

Given tangent to the curve is parallel to the x – axis

This means, Slope of tangent = Slope of x – axis

⇒ 2x - 4 = 0

⇒ 2x = 4

⇒ x = 2 ∈ (0, 4)

Put x = 2 in y = x² – 4x , we have

y = (2)² – 4(2) = 4 – 8 = -4

Hence, the tangent to the curve is parallel to the x –axis at

(2, -4)

Given:

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Here,

On differentiating above with respect to x, we get

f'(x) = -1 × (4x – 1)^-1-1 × 4

⇒ f’(x) = -4 × (4x – 1)^-2

⇒f’(x) exist

Hence, f(x) is differentiable in (1,4)

We know that,

Differentiability ⇒ Continuity

⇒Hence, f(x) is continuous in (1,4)

Thus, Mean Value Theorem is applicable to the given function

Now,

x ∈ [1,4]

Now, let us show that there exist c ∈ (0,1) such that

On differentiating above with respect to x, we get

Put x = c in above equation, we get

…(i)

By Mean Value Theorem,

⇒

⇒ (4c – 1)² = 45

⇒ 4c – 1 = √45

⇒ 4c – 1 = ± 3√5

⇒ 4c = 1 ± 3√5

but

So, value of

Thus, Mean Value Theorem is verified.

Given: f(x) = x³ – 2x² – x + 3 in [0,1]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

f(x) = x³ – 2x² – x + 3

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R

⇒ f(x) = x³ – 2x² – x + 3 is continuous at x ∈ [0,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = x³ – 2x² – x + 3

Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R

f’(x) = 3x² – 4x – 1

⇒ f(x) is differentiable at [0,1]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function

Now,

f(x) = x³ – 2x² – x + 3 x ∈ [0,1]

f(a) = f(0) = 3

f(b) = f(1) = (1)³ – 2(1)² – 1 + 3

= 1 – 2 – 1 + 3

= 4 – 3

= 1

Now, let us show that there exist c ∈ (0,1) such that

f(x) = x³ – 2x² – x + 3

On differentiating above with respect to x, we get

f’(x) = 3x² – 4x – 1

Put x = c in above equation, we get

f’(c) = 3c² – 4c – 1 …(i)

By Mean Value Theorem,

⇒ f’(c) = -2

⇒ 3c² – 4c – 1 = -2 [from (i)]

⇒ 3c² – 4c -1 + 2 = 0

⇒ 3c² – 4c + 1 = 0

On factorising, we get

⇒ 3c² – 3c – c + 1 = 0

⇒ 3c(c – 1) – 1(c – 1) = 0

⇒ (3c – 1) (c – 1) = 0

⇒ (3c – 1) = 0 or (c – 1) = 0

So, value of

Thus, Mean Value Theorem is verified.

Given: f(x) = sinx – sin2x in [0,π]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

f(x) = sinx – sin 2x

Since, f(x) is a trigonometric function and we know that, sine function are defined for all real values and are continuous for all x ∈ R

⇒ f(x) = sinx – sin 2x is continuous at x ∈ [0,π]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = sinx – sin 2x

f’(x) = cosx – 2 cos2x

⇒ f(x) is differentiable at [0,π]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function

Now,

f(x) = sinx – sin2x x ∈ [0,π]

f(a) = f(0) = sin(0) – sin2(0) = 0 [∵ sin(0°) = 0]

f(b) = f(π) = sin(π) – sin2(π) = 0 – 0 = 0

[∵ sin π = 0 & sin 2π = 0]

Now, let us show that there exist c ∈ (0,1) such that

f(x) = sinx – sin2x

On differentiating above with respect to x, we get

f’(x) = cosx – 2cos2x

Put x = c in above equation, we get

f’(c) = cos(c) – 2cos2c …(i)

By Mean Value Theorem,

[from (i)]

⇒ cos c – 2cos2c = 0

⇒ cos c – 2(2cos² c – 1) = 0 [∵ cos 2x = 2cos²x –1]

⇒ cos c – 4cos² c + 2 = 0

⇒ 4 cos² c – cos c – 2 = 0

Now, let cos c = x

⇒ 4x² – x – 2 = 0

Now, to find the factors of the above equation, we use

[above we let cos c = x]

So, value of

Thus, Mean Value Theorem is verified.

Given:

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

Firstly, we have to show that f(x) is continuous.

Here, f(x) is continuous because f(x) has a unique value for each x ∈ [1,5]

Condition 2:

Now, we have to show that f(x) is differentiable

[using chain rule]

∴ f’(x) exists for all x ∈ (1,5)

So, f(x) is differentiable on (1,5)

Hence, Condition 2 is satisfied.

Thus, mean value theorem is applicable to given function.

Now,

Now, we will find f(a) and f(b)

so, f(a) = f(1)

and f(b) = f(5)

Now, let us show that c ∈ (1,5) such that

On differentiating above with respect to x, we get

Put x = c in above equation, we get

By Mean Value theorem,

Squaring both sides, we get

⇒ 16c² = 24 × (25 – c²)

⇒ 16c² = 600 – 24c²

⇒ 24c² + 16c² = 600

⇒ 40c² = 600

⇒ c² = 15

⇒ c = √15 ∈ (1,5)

Hence, Mean Value Theorem is verified.

Given: Equation of curve, y = (x – 3)²

Firstly, we differentiate the above equation with respect to x, we get

[using chain rule]

Given tangent to the curve is parallel to the chord joining the points (3,0) and (4,1)

i.e.

⇒ 2x – 6 = 1

⇒ 2x = 7

Put in y = (x – 3)² , we have

Hence, the tangent to the curve is parallel to chord joining the points (3,0) and (4,1) at

Given: y = 2x² – 5x + 3 in [1,2]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

y = 2x² – 5x + 3

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R

⇒ y = 2x² – 5x + 3 is continuous at x ∈ [1,2]

Hence, condition 1 is satisfied.

Condition 2:

y = 2x² – 5x + 3

Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R

y’ = 4x – 5 …(i)

⇒ y = 2x² – 5x + 3 is differentiable at [1,2]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function.

Now,

f(x) = y = 2x² – 5x + 3 x ∈ [1,2]

f(a) = f(1) = 2(1)² – 5(1) + 3 = 2 – 5 + 3 = 0

f(b) = f(2) = 2(2)² – 5(2) + 3 = 8 – 10 + 3 = 1

Then, there exist c ∈ (0,1) such that

Put x = c in equation, we get

y’ = 4c – 5 …(i)

By Mean Value Theorem,

⇒ 4c – 5 = 1

⇒ 4c = 6

So, value of

Thus, Mean Value Theorem is verified.

Put in given equation y = 2x² – 5x + 3, we have

⇒ y = 0

Hence, the tangent to the curve is parallel to the chord AB at

Given that, is differentiable at x = 1.

We know that,f(x) is differentiable at x =1 ⇔ Lf’(1) = Rf’(1).

Lf’(1) =

=

= (∵ f(x) = x²+3x+p, if x≤ 1)

=

=

=

=

= 5

Rf’(1) =

=

= (∵ f(x) = qx+2, if x > 1)

=

=

=

=q

Since, Lf’(1) = Rf’(1)

∴ 5 = q (i)

Now, we know that if a function is differentiable at a point,it is necessarily continuous at that point.

⇒ f(x) is continuous at x = 1.

⇒ f(1^-) = f(1⁺) = f(1)

⇒ 1+3+p = q+2 = 1+3+p

⇒ p-q = 2-4 = -2

⇒ q-p = 2

Now substituting the value of ‘q’ from (i), we get

⇒ 5-p = 2

⇒ p = 3

∴ p = 3 and q = 5

We have, x^m.yⁿ = (x+y) ^m+n

Taking log on both sides, we get

log (x^m.yⁿ) = log (x+y) ^m+n

⇒ m log x + n log y = (m+n) log (x+y)

Differentiating both sides w.r.t x, we get

⇒

⇒

⇒

⇒

⇒

⇒

Hence proved.

We have,

_{Differentiating both sides w.r.t x, we get}

_⇒

_⇒

_⇒

_⇒

Hence proved.

_{We have,}

_{x = sin t and y = sin pt}

_⇒

∴

⇒

_{Differentiating both sides w.r.t x, we get}

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Hence proved.

_{We have,}

Putting x^tanx = u and

u = x^tanx

_{Taking log on both sides, we get}

_{log u = tanx log x}

_{Differentiating w.r.t x, we get}

_⇒

_⇒

_⇒ (i)

Now,

⇒

_{Differentiating w.r.t x, we get}

_⇒

_⇒(ii)

Now, y = u + v

⇒

On substituting the values of from (i) and (ii),we get

_⇒

_⇒

We know that if two functions f(x) and g(x) are continuous then {f(x) +g(x)},{f(x)-g(x)}, {f(x).g(x)} and are continuous.

Since, f(x) = 2x and are polynomial functions, they are continuous everywhere.

⇒ {f(x) +g(x)},{f(x)-g(x)}, {f(x).g(x)} are continuous functions.

for,

now, f(x) = 0

⇒ 4x = 0

⇒ x = 0

∴ is discontinuous at x=0.

We have,

Now, f(x) is discontinuous where 4x-x³ = 0.

⇒ x(4x-x²) = 0

⇒ x = 0 or 4x-x² = 0

⇒ x = 0 or x = ±2

⇒ f(x) is discontinuous at exactly three points.

We have, f(x) = |2x – 1| sinx

Now, 2x-1 = 0

⇒

Now we will check the differentiability of f(x) at 1/2.

=

= (∵ f(x) = |2x – 1| sinx)

=

=

=

=

= (∵ f(x) = |2x – 1| sinx)

=

=

=

∴

Hence, f(x) is not differentiable at

We have,

Now, f(x) is discontinuous where sinx=0.

We know that sinx=0 at x = nπ, n ∈ Z

⇒ f(x) = cotx is discontinuous on the set {x = nπ : n ∈ Z}

Given that, f(x) = e^|x|

Let g(x) = |x| and h(x) = e^x

Then, f(x) = hog(x)

We know that, modulus and exponential functions are continuous everywhere.

Since, composition of two continuous functions is a continuous function.

Hence, f(x) = hog(x) is continuous everywhere.

Now, v(x)=|x| is not differentiable at x=0.

Lv’(0) =

=

= (∵ v(x) = |x|)

=

=

=

Rv’(0) =

=

= (∵ v(x) = |x|)

=

=

=

⇒ Lv’ (0) ≠ Rv’(0)

⇒ |x| is not differentiable at x=0.

So, e^|x| is not differentiable at x=0.

Hence, f(x) continuous everywhere but not differentiable at x = 0.

We have, where x ≠ 0.

Given that, the function is continuous at x = 0

⇒

⇒

⇒ f(0) = 0 × (an oscillating number between -1 and 1 )

⇒ f(0) = 0

Hence,the value of the function f at x = 0 is ‘0’.

Given that, is continuous at

Now, LHL =

=

=

RHL =

=

=

=1+n

Since, f(x) is continuous ⇒ LHL = RHL

⇒

⇒

Given that, f(x) = |sinx|

Let g(x) = sinx and h(x) = |x|

Then, f(x) = hog(x)

We know that, modulus function and sine function are continuous everywhere.

Since, composition of two continuous functions is a continuous function.

Hence, f(x) = hog(x) is continuous everywhere.

Now, v(x)=|x| is not differentiable at x=0.

Lv’(0) =

=

= (∵ v(x) = |x|)

=

=

=

Rv’(0) =

=

= (∵ v(x) = |x|)

=

=

=

⇒ Lv’ (0) ≠ Rv’(0)

⇒ |x| is not differentiable at x=0.

⇒ h(x) is not differentiable at x=0.

So, f(x) is not differentiable where sinx = 0

We know that sinx=0 at x = nπ, n ∈ Z

Hence, f(x) is everywhere continuous but not differentiable x = nπ, n ∈ Z

We have,

⇒ y = log(1-x²) – log(1+x²)

⇒

⇒

⇒

⇒

⇒

⇒

We have,

⇒ y² = sinx + y

Differentiating both sides w.r.t x, we get

⇒

⇒

⇒

⇒

Let u = cos^-1(2x² – 1) and v = cos^-1 x

Now, u = cos^-1(2x² – 1)

= cos^-1(2cos²v – 1) [∵v = cos^-1 x ⇒ cos v = x]

= cos^-1(cos2v) [∵ 2cos²x – 1 = cos2x]

⇒ u = 2v

⇒

_{Given that, x = t}², y = t³

_⇒

Now,

⇒

⇒

⇒

⇒

Rolle’s Theorem states that, Let f : [a, b]→R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers.Then there exists some c in (a, b) such that f’(c) = 0.

We have, f(x) = x³ – 3x

Since, f(x) is a polynomial function it is continuous on and differentiable on

⇒

Now, as per Rolle’s Theorem, there exists at least one c ∈ , such that

f’(c) = 0

⇒ 3c² – 3 = 0 [∵ f’(x) = 3x² – 3 ]

⇒ c² = 1

⇒ c = ±1

⇒ c = 1 ∈

Mean Value Theorem states that, Let f : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that

We have,

Since, f(x) is a polynomial function it is continuous on [1,3] and differentiable on (1,3).

Now, as per Mean value Theorem, there exists at least one c ∈ (1,3), such that

⇒

⇒

⇒

⇒ 3(c² – 1) = 2c²

⇒ 3c² – 2c² = 3

⇒ c² = 3

⇒

⇒

Consider, f(x) = |x-1| + |x-2|

Let’s discuss the continuity of f(x).

We have, f(x) = |x-1| + |x-2|

⇒

⇒

When x<1, we have f(x) = -2x+3, which is a polynomial function and polynomial function is continuous everywhere.

When 1≤x<2, we have f(x) = 1, which is a constant function and constant function is continuous everywhere.

When x≥2, we have f(x) = 2x-3, which is a polynomial function and polynomial function is continuous everywhere.

Hence, f(x) = |x-1| + |x-2| is continuous everywhere.

Let’s discuss the differentiability of f(x) at x=1 and x=2.

We have

⇒

⇒

Lf’(1) =

=

= (∵ f(x) = -2x+3, if x< 1)

=

=

Rf’(1) =

= (∵ f(x) = 1, if 1≤ x< 2)

=0

⇒ Lf’(1) ≠ Rf’(1)

⇒ f(x) is not differentiable at x=1.

Lf’(2) =

= (∵f(x) = 1, if 1≤ x< 2 and f(2) = 2×2-3 =1 )

=0

Rf’(2) =

=

= (∵ f(x) = 2x-3, if x≥ 2)

=

=

⇒ Lf’(2) ≠ Rf’(2)

⇒ f(x) is not differentiable at x=2.

Thus, f(x) = |x-1| + |X-2| is continuous everywhere but fails to be differentiable exactly at two points x=1 and x=2.

Let u = x² and v = x³ .

⇒

∴

⇒

Hence, Derivative of x² w.r.t. x³ is .

We have, f(x) = |cos x|

For , cos x > 0

∴ f(x) = cos x

⇒ f’(x) = -sin x

⇒

Hence,

We have, f(x) = | cosx – sinx|

For , sin x > cos x

∴ f(x) = sin x - cos x

⇒ f’(x) = cos x – (-sin x) = cos x + sin x

⇒

Hence,

We have,

On differentiating both sides, we get

⇒

⇒

⇒

⇒

Hence,

False

Rolle’s Theorem states that, Let f : [a, b]→R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers.Then there exists some c in (a, b) such that f’(c) = 0.

We have, f(x) = |x – 1| in [0, 2].

Since, polynomial and modulus functions are continuous everywhere f(x) is continuous.

Now, x-1=0

⇒ x=1

We need to check if f(x) is differentiable at x=1 or not.

We have,

Lf’(1) =

=

= (∵ f(x) = 1-x, if x< 1)

=

=

Rf’(1) =

=

= (∵ f(x) = x-1, if 1< x)

=

⇒ Lf’(1) ≠ Rf’(1)

⇒ f(x) is not differentiable at x=1.

Hence, rolle’s theorem is not applicable on f(x) since it is not differentiable at x=1 ∈ (0,2).

True

Given that, f is continuous on its domain D.

Let a be an arbitrary real number in D. Then f is continuous at a.

⇒

Now,

⇒ [∵ |f|(x) = |f(x)| ]

⇒

⇒

∴ |f| is continuous at x=a.

Since a is an arbitrary point in D. Therefore |f| is continuous in D.

True

Let f be a function defined by f(x) = |1-x + |x|| .

Let g(x) = 1-x + |x| and h(x) = |x| be two functions defined on R.

Then,

hog(x) = h(g(x))

⇒ hog(x) = h(1-x + |x|)

⇒ hog(x) = |(1-x + |x|)|

⇒ hog(x) = f(x) ∀ x ∈ R.

Since (1-x) is a polynomial function and |X| is modulus function are continuous in R.

⇒ g(x) = 1-x + |x| is everywhere continuous.

⇒ h(x) = |x| is everywhere continuous.

Hence, f = hog is everywhere continuous.

True

False

Let f(x) = x and

∴ f(x).g(x) = = 1, which is a constant function and continuous everywhere.

But, is discontinuous at x=0.