Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.1

Question 1.

Find the derivatives of the following functions using the first principle.

(i) f(x) = 6

(ii) f(x) = -4x + 7

(iii) f(x) = -x^{2} + 2

Answer:

(i) f(x) = 6

(ii) f(x) = – 4x + 7,

f(x + Δx) = -4(x + Δx) + 7

f(x + Δx) – f(x) = [-4(x + Δx) + 7] – [-4x + 7]

f(x + Δx) – f(x) = [-4(x + Δx) + 7] + 4x – 7

f(x + Δx) – f(x) = -4 Δx

(iii) f(x) = -x^{2} + 2

f (x + Δx) = – (x + Δx)^{2} + 2

f (x + Δx) – f(x) = – [x^{2} + 2x Δx + (Δx)^{2}] + 2 – [- x^{2} + 2]

Question 2.

Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

(i) f (x) = |x – 1|

Answer:

(ii) f (x) = \(\sqrt{1-x^{2}}\)

Answer:

(iii)

Answer:

Question 3.

Determine whether the following function is differentiable at the indicated values.

(i) f(x) = x |x| at x = 0

Answer:

(ii) f(x) = |x^{2} – 1|at x = 1

Answer:

(iii) f(x) = |x| + |x – 1| at x = 0, 1

Answer:

To find the limit at x = 0

First we find the left limit of f(x) at x = 0

When x = 0^{–} |x| = -x and

|x – 1| = -(x – 1)

∴ When x = 0 we have

f(x) = -x – (x – 1)

f(x) = -x – x + 1 = -2x + 1

f(0) = 2 × 0 + 1 = 1

f'(0^{–} = – 2 ……… (1)

∴When x = 0^{+} we have

|x| = x and |x – 1| = – (x – 1)

∴ f(x) = x – (x – 1)

f(x) = x – x + 1

f(x) = 1

f(0) = 1

From equations (1) and (2) , we get

f'(0^{–}) ≠ f’(0^{+})

∴ f(x) is not differentiable at x = 0.

To find the limit at x = 1

First we find the left limit of f (x) at x = 1

When x = 1 , |x| = x and

|x – 1| = -(x – 1)

∴ f(x) = x – (x – 1)

f(x) = x – x + 1 = 1

f(x) = 1

f(1) = 1

When x = 1^{+} , |x| = x and

|x – 1| = x – 1

When x = , |x| = x and

|x – 1| = x – 1

∴ f(x) = x + x – 1 = 2x – 1

f(1) = 2 × 1 – 1 = 2 – 1 = 1

From equations (3) and (4) , we get

f’(1^{–}) ≠ f'(1^{+})

∴ f (x) is not differentiable at x = 1

(iv) f(x) = sin |x| at x = 0

Answer:

First we find the left limit of f (x) at x = 0

When x = 0^{–}, |x| = -x

∴ f(x) = sin (-x) = -sin x

f(0) = -sin 0 = 0

Next we find the right limit of f (x) at x = 0

When x = 0^{+} |x| = x

∴ f(x) = sin x

f(0) = sin 0 = 0

From equations (1) and (2) , we get

f’(0^{–}) ≠ f'(0^{+})

∴ f (x) is not differentiable at x = 0.

Question 4.

Show that the following functions are not differentiable at the indicated value of x

(i)

Answer:

First we find the left limit of f(x) at x = 2

When x = 2, then x ≤ 2

∴ f(x) = -x + 2

f(2) = -2 + 2 = 0

Next we find the right limit of f(x) at x = 2

When x = 2^{+}, then x > 2

∴ f(x) = 2x – 4

f(2) = 2 × 2 – 4 = 4 – 4 = 0

From equation (1) and (2), we get

f’(2^{–}) ≠ f'(2^{+})

∴ f (x) is not differentiable at x = 2

(ii)

Answer:

First we find the left limit of f (x) at x = 0

When x = 0^{–}, then x < 0

∴ f(x) = 3x

f(0) = 3 × 0 = 0

Next we find the right limit of f (x) at x = 0

When x = 0^{+}, then x ≥ 0

∴ f(x) = -4x

f(0) = -4 × 0 = 0

From equations (1) and (2) , we get

f'(0^{–}) ≠ f'((0^{+})

∴ f (x) is not differentiable at x = 0

Question 5.

The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.

Answer:

We know A function f is not differentiable at a point x_{0} belonging to the domain of f if one of the following situations holds

(i) f has a vertical tangent at x_{0}

(ii) The graph of f comes to a point at x_{0} (either a sharp edge ∨ or a sharp peak ∧)

For the given graph f

At x = – 1 , a sharp edge ∨

At x = 8, a sharp peak ∧

At x = 4 , discontinuity

At x = 11, perpendicular tangent

∴ The given graph is not differentiable at

x = – 1, 8, 4, 11

Question 6.

If f(x) = |x + 100| + x^{2}, test whether f’(-100) exists.

Answer:

f(x) = |x + 100| + x^{2}

First let us find the left limit of f( x) at x = – 1100

When x < – 100 ,

f(x) = – (x + 100) + x^{2}

f(- 100) = – (- 100 + 100) + (- 100)^{2}

f(- 100) = 100^{2}

= -1 – 100 – 100

f’ (-100) = -201 ——– (1)

Next let us find the right limit of f( x) at x = -100

when x > – 100

f(x) = x + 100 + x^{2}

f(- 100) = – 100 + 100 + (- 100)^{2}

f(- 100) = 100^{2}

f'(-100^{+}) = – 199 ……… (2)

From equation (1) and (2) , we get

f’(- 100^{–}) ≠ f'(- 100^{+})

∴ f’ (x) does not exist at x = -100

Hence, f'(- 100) does not exist

Question 7.

Examine the differentiability of functions in R by drawing the diagrams. (i) |sin x| (ii) |cos x|

(i) |sin x|

Answer:

At the points, x = 0, π, 2π, 3π, ……….. the graph of the given function has a sharp edge V.

∴ At these points, the function is not differentiable.

∴ The function y = |sin x| is not differentiable at

x = nπ, for all n ∈ Z.

(ii) |cos x|

Answer: