Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 9 Limits and Continuity Ex 9.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.5

Question 1.

Prove that f(x) = 2x^{2} +3x – 5 is continuous at all points in R.

Answer:

f(x) = 2x^{2} + 3x – 5

Clearly f(x) is defined for all points of R.

Let x_{0} be an arbitrary point in R. Then

f(x_{0}) = 2x_{0}^{2} + 3x_{0} – 5 ——- (1)

Thus, f(x) is defined at all points of R limit of f(x) exist at all points of R and is equal to the value of the function f (x).

Thus f (x) is continuous at all points of R.

Question 2.

Examine the continuity of the following

(i) x + sin x

Answer:

Let f(x) = sin x

f (x) is defined at all points of R.

Let x_{0} be an arbitrary point in R.

= x_{o} + sin x_{0} …….. (1)

f (x_{o}) = x_{o} + sin x_{o} ……… (2)

From equations (1) and (2) we get

∴ At all points of R, the limit of f (x) exists and is equal to the value of the function.

Thus, f( x) satisfies ail conditions for continuity.

Therefore, f(x) is continuous at all points of f(x).

(ii) X^{2} cos x

Answer:

Let f(x) = x^{2} cos x

f (x) is defined at all points of R.

Let x_{0} be an arbitrary point in R. Then

= x^{2}_{0} cos _{0}

f(x_{0}) = x^{2}_{0} cos _{0}

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

∴ The limit at x = x_{0} exist and is equal to the value of the function f(x) at x = x_{0}.

Since x_{0} is arbitrary, the limit of the function exist and is equal to the value of the function for all points in R.

∴ f( x) satisfies all conditions for continuity. Hence

f (x) is a continuous function in R.

(iii) e^{x} tan x

Answer:

Let f(x) = e^{x} tan x

f (x) is defined at ail points of R.

except at (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.

Let x_{0} be an arbitrary point in R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

∴ Limit at x = x_{0} exist and is equal to the value of the function f(x) at x = x_{0}.

Since x_{0} is arbitrary the limit of the function. f(x) exists at all points in R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z and is equal to the value of the function f (x) at that points.

∴ f (x) satisfies all conditions for continuity. Hence,

f(x) is continuous at all points of R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

(iv) e^{2x} + x^{2}

Answer:

Let f(x) = e^{2x} + x^{2}

Clearly, f(x) is defined for all points in R.

Let x_{0} be an arbitrary point in R.

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

The limit of the function f(x) exist at x = x_{0} and is equal to the value of the function f(x) at x – x_{0}.

Since x_{0} is an arbitrary point in R, the above is true for all points in R. Hence f (x) satisfies all conditions for continuity. Hence f (x) is continuous at all points of R.

(v) x . log x

Answer:

Let f(x) = x log x

The function f(x) is defined in the open interval (0 , ∞) since log x is defined for x > 0. Let x_{0} be an arbitrary point in (0, ∞). Then

= x_{0} log x_{0}

f(x_{0}) = x_{0} log x_{0}

From equation (1) and (2) we have

∴ The limit of the function f(x) exists at x = x_{0} and is equal to the value of the function .

Since x_{0} is an arbitrary point the above is true for all points in (0, ∞).

∴ f(x) is continuous at all points of (0, ∞).

(vi) \(\frac{\sin x}{x^{2}}\)

Answer:

f(x) = \(\frac{\sin x}{x^{2}}\)

f(x) is not defined at x = 0

∴ f(x) is defined for all points of R – {0}

Let x_{0} be an arbitrary point in R – {0}. Then

From equation (1) and (2) we have

∴ The limit of the function f(x) exist at x = x_{0} and is equal to the value of the function f(x) at x = x_{0}.

Since x_{0} is an arbitrary point in R – {0}, the above result is true for all points in R – {0}.

∴ f(x) is continuous at all points of R – {0}.

(vii) \(\frac{x^{2}-16}{x+4}\)

Answer:

Let f(x) = \(\frac{x^{2}-16}{x+4}\)

f(x) is not defined at x = – 4

∴ f(x) is defined for all points of R – {- 4}.

Let x_{0} be an arbitrary point in R – {- 4}. Then

From equation (1) and (2) we have

Thus the limit of the function f (x) exist at x = x_{0} and is equal to the value of the function f (x) at x = x_{0}.

Since x_{0} is an arbitrary point in R – {- 4} the above result is true for all points in R – { – 4}.

∴ f(x) is continuous at all points of R – {- 4}.

(viii) |x + 2| + |x – 1|

Answer:

let f(x) = |x + 2| + |x – 1|

f( x) is defined for all points of R. Let x_{0} be an arbitrary point in R. Then

From equation (1) and (2) we get

Thus the limit of the function f(x) exist at x = x_{0} and is equal to the value of the function at x = x_{0}. Since x = x_{0} is an arbitrary point in R, the above

result is true for all points in R. Hence f (x) is continuous at all points of R.

(ix) \(\frac{|x-2|}{|x+1|}\)

Answer:

Let f(x) = \(\frac{|x-2|}{|x+1|}\)

f(x) is defined for all points of R except at x = – 1.

∴ f (x) is defined for all points of R – { – 1 }.

Let x_{0} be an arbitrary point in R – {- 1}.Then

From equation (1) and (2) we have

Hence the limit of the function f(x) at x = x_{0} exists and is equal to the value of the function at x = x_{0}.

Since x = x_{0} is an arbitrary point in R – { – 1 }, the above result is true for all points in R – {- 1).

∴ f(x) is continuous at all points of R – {- 1}.

(x) cot x + tan x

Answer:

Let f(x) = cot x + tan x

∴ The limit of the function f(x) exists at x = x_{0} and is equal to the value of the function f (x) at x = x_{0}.

Since x_{0} is an arbitrary point , the above result is true for all points of R – \(\left\{\frac{\mathrm{n} \pi}{2}\right\}\), n ∈ z.

∴ f(x) is continuous at all points of R – \(\left\{\frac{\mathrm{n} \pi}{2}\right\}\), n ∈ Z

Question 3.

Find the points of discontinuity of the function f, where

(i)

Answer:

(ii)

Answer:

For the point x_{0} < 2, we have the limit of the function that exists and is equal to the value of the function at that point.

Since x_{0} is an arbitrary point the above result is true for all x < 2.

∴ f(x) is continuous in (-∞, 2).

Let x_{0} be an arbitrary point such that x_{0} > 2 then

∴ For the point x_{0} > 2, the limit of the function exists and is equal to the value of the function.

Since x_{0} is an arbitrary point the above result is true for all x > 2.

∴ The function is continuous at all points of (2, ∞). Hence the given function is continuous at all points of R.

(iii)

Answer:

Clearly, the given function is defined at all points of R.

Case (i) At x = 2

Case (ii) for x < 2

Let x_{0} be an arbitrary point in (- ∞, 2).

∴ f(x)is continuous at x = x_{0} in (- ∞, 2).

Since x_{0} is an arbitrary point in (- ∞, 2), f(x) is continuous at all points. f(-∞, 2).

Case (iii) for x > 2

Let y_{0} be an arbitrary point in (2, ∞). Then

Hence, f(x) is continuous at x = y_{0} in (2, ∞).

Since y_{o} is an arbitrary point of (2, ∞), f (x) is continuous at all points of (2, ∞).

∴ By case (i) case (ii) and case (iii) f(x) is continuous at all points of R.

(iv)

Answer:

Hence, f (x) is continuous at x = x_{0}. Since x_{0} is an arbitrary point of \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), f(x) is continuous at all points of \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).

Hence, f (x) is continuous at all points \(\left[0, \frac{\pi}{2}\right)\),

Question 4.

At the given point x_{0} discover whether the given function is continuous or discontinuous citing the reasons for your answer

(i)

Answer:

(ii)

Answer:

Question 5.

Show that the function

Answer:

Clearly, the given function f(x) is defined at all points of R.

Case (i) Let x_{0} ∈ (- ∞, 1) then

f (x) is continuous at x = x_{0}.

Since x_{0} is arbitrary f(x) is continuous at all points of (-∞, 1).

Case (ii) Let x_{0} ∈ (1, ∞) then

f (x) is continuous at x = x_{0}.

Since x_{0} is an arbitrary point of (1, ∞), f(x) is continuous at all points of (1, ∞).

Case (iii) Let x_{0} = 1 then

Hence, f (x) is continuous at x = 1.

Using all the three cases, we have f (x) is continuous at all the points of R.

Question 6.

Answer:

Question 7.

Graph the function. Show that f(x) continuous on (- ∞, ∞)

Answer:

When x < 0 We have y = 0

When 0 ≤ x < 2 We have y = x^{2}

When x ≥ 2 We have y = 4

Case (i) If x < 0 ie. (-∞, 0) then f (x) = 0

which is clearly continuous in (-∞, 0).

Case (ii) If 0 ≤ x < 2 je. [0 , 2)

Let x_{0} be an arbitrary point in [0, 2)

Hence f(x) is continuous at x = x_{0}. Since x = x_{0} is an arbitrary f(x) is continuous at all points of [0, 2).

Case (iii) x ≥ 2 je. [2, ∞)

f( x) = 4 which is clearly continuous in [2 , ∞)

Case (iv) at x = 2,

f(2) = 4

∴ f(x) is continuous at x = 2.

∴ Using case (j) case (ii) case (iii) and case (iv) we have f(x) is continuous at all points of R.

Question 8.

If f and g are continuous functions with f(3) = 5 and [2f(x) – g(x)] = 4, find g(3)

Answer:

Given f and g are continuous functions.

2 f(3) – g(3) = 4

2 × 5 – g(3) = 4

10 – 4 = g(3)

g(3) = 6

Question 9.

Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.

(i)

Answer:

(ii)

Answer:

Question 10.

A function f is defined as follows:

is the function continuous?

Answer:

Question 11.

Which of the following functions f has a removable discontinuity at x = x_{0}? If the discontinuity is removable, find a function g that agrees with f for x ≠ x_{0} and is continuous on R

(i)

Answer:

f(x) is not defined at x = -2

Redefine the function f(x) as

∴ f (x) has a removable discontinuity at x = -2.

Clearly, g (x) is continuous on R.

(ii)

Answer:

The function f(x) is not defined at x = -4.

Limit the function f (x) exist at x = -4.

∴ The function f (x) has a removable discontinuity at x = -4.

Redefine the function f (x) as

Clearly, the function g(x) is continuous on R.

(iii)

The function f(x) is not defined at x = 9.

∴ Limit of the function f(x) exists at x = 9.

Hence, the function f(x) has a removable discontinuity at x = 9. Redefine the function f(x) as

Clearly, g (x) is defined at all points of R and is continuous on R.

Question 12.

Find the constant b that makes g continuous on (-∞, ∞).

Answer:

Given g is continuous on R.

∴ g (x) is continuous at x = 4.

4^{2} – b^{2} = b × 4 + 20

16 – b^{2} = 4b + 20

b^{2} + 4b + 20 – 16 = 0

b^{2} + 4b + 4 = 0

(b + 2)^{2} = 0

b + 2 = 0 ⇒ b = -2

Question 13.

Consider the function f (x) = x sin \(\frac{\pi}{x}\) What value must we give f (0) in order to make the function continuous everywhere?

Answer:

f(x) = x sin \(\frac{\pi}{x}\)

Define f(x) on R as

∴ f(0) = 0. Then f(x) is continuous on R.

Question 14.

The function f(x) = \(\frac{x^{2}-1}{x^{3}-1}\) is not defined at x = 1. What value must we give f(1) in order to make f(x) continuous at x = 1 ?

Answer:

The function f (x) has a removable discontinuity at x = 1. Redefine f (x) as

∴ f(1) = \(\frac{2}{3}\). Then f(x) will be continuous at x = 1

Question 15.

State how continuity is destroyed at x = x_{0} for each of the following graphs.

Answer:

(a)

The left – hand limit and right hand limit does not coincide at x = x_{0}

(b)

The function f(x) is not defined at x = x_{0} and hence the continuity is destroyed at x = x_{0}

(c)

The limit of f(x) does not exist at x = x_{0}

(d)

The left hand limit and right – hand limit does not coincide at x = x_{0}