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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3

Queston 1.

Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.

Solution:

f(x) is not continuous at x = 0. So Rolle’s Theorem is not applicable.

(ii) f(x) = tan x, x ∈ [0, π]

The function tan x is not continuous on [0, π] because it is not defined at \(\frac { π }{ 2 }\)

The function is not differentiable on (0, π), because it is not continuous at \(\frac { π }{ 2 }\)

and f(0) = tan (0) = 0

f(π) = tan π = 0

f (0) = f(π) = 0

Since ‘tan x’ is not continuous in [0, π] and not differentiable in (o, π), Rolle’s theorem is not applicable.

(iii) f(x) = x – 2 log x, x ∈ [2, 7]

f(x) is continuous on [2, 7] and

f(x) is differentiable on (2, 7)

f(2) = 2 – 2 log 2, f (7) = 7 – 2 log 7

f(2) ≠ f(7)

Hence, Rolle’s theorem is not applicable.

Question 2.

Using Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:

Solution:

Question 3.

Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals :

Solution:

The function is not continuous at x = 0. So Lagrange’s mean value theorem is not applicable in the given interval.

(ii) f(x) = |3x + 1|, x ∈ [-1, 3]

3x + 1 = 0

x = –\(\frac { 1 }{ 3 }\)

f(x) is not differentiable at x = –\(\frac { 1 }{ 3 }\)

Hence, Lagrange’s mean value theorem is not applicable.

Question 4.

Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:

(i) f(x) = x^{3} – 3x + 2, x ∈ [-2, 2]

(ii) f(x) = (x – 2)(x – 7), x ∈ [3, 11]

Solution:

(i) f(x) = x^{3} – 3x + 2

Here a = -2, b = 2

Question 5.

Show that the value in the conclusion of the mean value theorem for

Solution:

(i)

Question 6.

A racecar driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours?

Solution:

Here the interval is [0, 2] and f(0) = 20, f(2) = ?

f(b) – f(a) ≤ (b – a)f’c)

here f (a) = 20

⇒ f(b) – 20 ≤ 150(2 – 0)

⇒ f(b) ≤ 320

(i.e) f(2) = 320 km.

Question 7.

Suppose that for a function f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) – f(1) ≤ 3.

Solution:

Question 8.

Does there exist a differentiable function f(x) such that f(0) = -1, f(2) = 4 and f'(x) ≤ 2 for all x. Justify your answer.

Solution:

f(0) = -1, f(2) = 4, f(x) ≤ 2

Here a = 0, b = 2

So this is not possible

Question 9.

Show that there lies a point on the curve where the tangent is drawn is parallel to the x-axis.

Solution:

⇒ There lies a point in [-3,0], where the tangent is parallel to the x-axis.

Question 10.

Using mean value theorem prove that for, a > 0, b > 0, |e^{-a} – e^{-b}| < |a – b|

Solution:

### Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3 Additional Questions Solved

Question 1.

Verify Rolle’s theorem for the following:

(i) f(x) = x^{3} – 3x + 3, 0 ≤ x ≤ 1

(ii) f(x) = tan x, 0 ≤ x ≤ π

(iii) f(x) = | x |, – 1 ≤ x ≤ 1

(iv) f (x) = sin^{2} x, 0 ≤ x ≤ π

(v) f(x) = e^{x} sin x, 0 ≤ x ≤ π

(vi) f(x) = x(x – 1) (x – 2), 0 ≤ x ≤ 2

Solution:

(i) f(x) = x^{3} – 3x + 3, 0 ≤ x ≤ 1. f is continuous on [0, 1] and differentiable in (0, 1)

f(0) = 3 and f(1) = 1 ∴ f(a) ≠ f(b)

∴ Rolle’s theorem, does not hold, since f(a) = f(b) is not satisfied.

Also note that f’ (x) = 3x^{2} – 3 = 0 ⇒ x^{2} = 1 ⇒ x = ±1

There exists no point c ∈ (0, 1) satisfying f’ (c) = 0.

(ii) f(x) = tan x, 0 ≤ x ≤ π.

f ‘(x) is not continuous in [0, π] as tan x tends to + ∞ at x = \(\frac{\pi}{2}\),

∴ Rolle’s theorem is not applicable.

(iii) f (x) = | x |, -1 ≤ x ≤ 1

f is continuous in [-1, 1] but not differentiable in (-1, 1) since f'(0) does not exist.

∴ Rolle’s theorem is not applicable.

(iv) f (x) = sin^{2}x, 0 ≤ x ≤ π

f is continuous in [0, π] and differentiable in (0, π). f(0) = f(π) = 0

(i.e.,) f satisfies hypothesis of Rolle’s theorem.

f’ (x) = 2 sin x cos x = sin 2x

(v) f(x) = e^{x}, sin x, 0 ≤ x ≤ π

e^{x} and sin x are continuous for all x, therefore the product e^{x} sin x is continuous in 0 ≤ x ≤ π.

f’ (x) = e^{x} sin x + e^{x} cos x = e^{x} (sin x + cos x) exist in 0 < x < π ⇒ f'(x) is differentiable in (0, π)

f (0) = e° sin 0 = 0

f (π) = e^{π} sin π = 0

∴ f satisfies hypothesis of Rolle’s theorem.

Thus there exists c ∈ (0, π) satisfying f'(c) = 0 ⇒ e^{c} (sin c + cos c) = 0

⇒ e^{c} = 0 or sin c + cos c = 0

⇒ e^{c} = 0 ⇒ c = -∞ which is not meaningful here.

(vi) f(x) = x (x – 1) (x – 2), 0 ≤ x ≤ 2

f is not continuous in [0, 2] and differentiable in (0, 2)

f(0) = 0 = f(2), satisfying hypothesis of Rolle’s theorem.

Now f'(x) = (x – 1) (x – 2) + x(x – 2) + x(x – 1) = 0

Note: There could exist more than one such ‘c’ appearing in the statement of Rolle’s theorem.

Question 2.

Suppose that f(0) = -3 and f'(x) ≤ 5 for all values of x, how large can f(2) possibly be?

Solution:

Since by hypothesis f is differentiable, f is continuous everywhere. We can apply Lagrange’s Law of the mean on the interval [0, 2], There exist atleast one ‘c’ ∈ (0, 2) such that

f(2) – f(0) = f'(c)(2 – 0)

f(2) = f(0) + 2f'(c) = -3 + 2f'(c)

Given that f'(x) ≤ 5 for all x.

In particular we know that f'(c) ≤ 5. Multiplying both sides of the inequality by 2, we have

2f'(c) ≤ 10

f(2) = -3 + 2f'(c) < -3 + 10 = 7

i.e., the largest possible value of f(2) is 7.

Question 3.

Using Rolle’s theorem find the points on the curves = x^{2} +1, -2 ≤ x ≤ 2 where the tangent is parallel to X-axis

Solution:

a = -2,

b = 2

f(x) = x^{2} + 1

f(a) = f(-2) = 4 + 1 = 5

f(b) = f(2) = 4 + 1 = 5

So, f(a) = f(b)

f'(x) = 2x

f'(x) = 0 ⇒ 2x = 0

x = 0 where 0 ∈ (-2, 2)

at x = 0, y = 0 + 1 = 1

So, the point is (0, 1) at (0, 1) the tangent drawn is parallel to X-axis

Question 4.

Find ‘C’ of Lagrange’s mean value theorem for the function f(x) = 2x^{3} + x^{2} – x – 1, [0, 2]

Solution:

f(x) = 2x^{3} + x^{2} – x – 1

a = 0,

b = 2

f(a) = f(0) = -1

f(b) = f(2) = 2(8) = 4 – 2 – 1 = 16 + 4 – 2 – 1 = 17

By Lagrange’s mean value theorem, we get a constant c ∈ (a, b) such that

Question 5.

Find ‘C’ of Lagrange’s mean value theorem for the function f(x) = x^{3} + x^{2} – 3x in [1, 3]

Solution:

f(x) = x^{3} + x^{2} – 3x

a = 1,

b=3

f(a) = f(1) = 1 – 5 – 3 = -7

f(b) = f(3) = 27 – 5(9) – 3(3)

= 27 – 45 – 9 = -27

f'(x) = 3x^{2} – 10x – 3

f'(x) = 3c^{2} – 10c – 3

From (1) and (2),

3c^{2} – 10c – 3 = -10

3c^{2} – 10c – 3 + 10 = 0

3c^{2} – 10c + 7 = 0

3c^{2} – 3c – 7c + 7 = 0

So, Lagrange’s mean value theorem is true with c = \(\frac{7}{3}\)