NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1.

- Continuity and Differentiability Class 12 Ex 5.2
- Continuity and Differentiability Class 12 Ex 5.3
- Continuity and Differentiability Class 12 Ex 5.4
- Continuity and Differentiability Class 12 Ex 5.5
- Continuity and Differentiability Class 12 Ex 5.6
- Continuity and Differentiability Class 12 Ex 5.7
- Continuity and Differentiability Class 12 Ex 5.8

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Continuity and Differentiability |

Exercise |
Ex 5.1 |

Number of Questions Solved |
34 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc 5.1

**Question 1.**

Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

**Solution:**

(i) At x = 0. lim_{x–>0} f (x) = lim_{x–>0} (5x – 3) = – 3 and

f(0) = – 3

∴f is continuous at x = 0

(ii) At x = – 3, lim_{x–>3} f(x)= lim_{x–>-3} (5x – 3) = – 18

and f( – 3) = – 18

∴ f is continuous at x = – 3

(iii) At x = 5, lim_{x–>5} f(x) = lim_{x–>5} (5x – 3) = 22 and

f(5) = 22

∴ f is continuous at x = 5

**Question 2.**

Examine the continuity of the function f(x) = 2x² – 1 at x = 3.

**Solution:**

lim_{x–>3} f(x) = lim_{x–>3} (2x² – 1) = 17 and f(3)= 17

∴ f is continuous at x = 3

**Question 3.**

Examine the following functions for continuity.

(a) f(x) = x – 5

(b) f(x) = , x≠5

(c) f(x) = ,x≠5

(d) f(x) = |x – 5|

**Solution:**

(a) f(x) = (x-5) => (x-5) is a polynomial

∴it is continuous at each x ∈ R.

**Question 4.**

Prove that the function f (x) = x^{n} is continuous at x = n, where n is a positive integer.

**Solution:**

f (x) = x^{n} is a polynomial which is continuous for all x ∈ R.

Hence f is continuous at x = n, n ∈ N.

**Question 5.**

Is the function f defined by continuous at x = 0? At x = 1? At x = 2?

**Solution:**

(i) At x = 0

lim_{x–>0-} f(x) = lim_{x–>0-} x = 0 and

lim_{x–>0+} f(x) = lim_{x–>0+} x = 0 => f(0) = 0

∴ f is continuous at x = 0

(ii) At x = 1

lim_{x–>1-} f(x) = lim_{x–>1-} (x) = 1 and

lim_{x–>1+} f(x) = lim_{x–>1+}(x) = 5

∴ lim_{x–>1-} f(x) ≠ lim_{x–>1+} f(x)

∴ f is discontinuous at x = 1

(iii) At x = 2

lim_{x–>2} f(x) = 5, f(2) = 5

∴ f is continuous at x = 2

**Find all points of discontinuity off, where f is defined by**

**Question 6.**

**Solution:**

at x≠2

**Question 7.**

**Solution:**

**Question 8.**

Test the continuity of the function f (x) at x = 0

**Solution:**

We have;

(LHL at x=0)

**Question 9.**

**Solution:**

**Question 10.**

**Solution:**

**Question 11.**

**Solution:**

At x = 2, L.H.L. lim_{x–>2-} (x³ – 3) = 8 – 3 = 5

R.H.L. = lim_{x–>2+} (x² + 1) = 4 + 1 = 5

**Question 12.**

**Solution:**

**Question 13.**

Is the function defined by a continuous function?

**Solution:**

At x = 1,L.H.L.= lim_{x–>1-} f(x) = lim_{x–>1-} (x + 5) = 6,

R.HL. = lim_{x–>1+} f(x) = lim_{x–>1+} (x – 5) = – 4

f(1) = 1 + 5 = 6,

f(1) = L.H.L. ≠ R.H.L.

=> f is not continuous at x = 1

At x = c < 1, lim_{x–>c} (x + 5) = c + 5 = f(c)

At x = c > 1, lim_{x–>c} (x – 5) = c – 5 = f(c)

∴ f is continuous at all points x ∈ R except x = 1.

**Discuss the continuity of the function f, where f is defined by**

**Question 14.**

**Solution:**

In the interval 0 ≤ x ≤ 1,f(x) = 3; f is continuous in this interval.

At x = 1,L.H.L. = lim f(x) = 3,

R.H.L. = lim_{x–>1+} f(x) = 4 => f is discontinuous at

x = 1

At x = 3, L.H.L. = lim_{x–>3-} f(x)=4,

R.H.L. = lim_{x–>3+} f(x) = 5 => f is discontinuous at

x = 3

=> f is not continuous at x = 1 and x = 3.

**Question 15.**

**Solution:**

At x = 0, L.H.L. = lim_{x–>0-} 2x = 0 ,

R.H.L. = lim_{x–>0+} (0)= 0 , f(0) = 0

=> f is continuous at x = 0

At x = 1, L.H.L. = lim_{x–>1-} (0) = 0,

R.H.L. = lim_{x–>1+} 4x = 4

f(1) = 0, f(1) = L.H.L.≠R.H.L.

∴ f is not continuous at x = 1

when x < 0 f (x) = 2x, being a polynomial, it is

continuous at all points x < 0. when x > 1. f (x) = 4x being a polynomial, it is

continuous at all points x > 1.

when 0 ≤ x ≤ 1, f (x) = 0 is a continuous function

the point of discontinuity is x = 1.

**Question 16.**

**Solution:**

At x = – 1,L.H.L. = lim_{x–>1-} f(x) = – 2, f(-1) = – 2,

R.H.L. = lim_{x–>1+} f(x) = – 2

=> f is continuous at x = – 1

At x= 1, L.H.L. = lim_{x–>1-} f(x) = 2,f(1) = 2

∴ f is continuous at x = 1,

R.H.L. = lim_{x–>1+} f(x) = 2

Hence, f is continuous function.

**Question 17.**

Find the relationship between a and b so that the function f defined by

is continuous at x = 3

**Solution:**

At x = 3, L.H.L. = lim_{x–>3-} (ax+1) = 3a+1 ,

f(3) = 3a + 1, R.H.L. = lim_{x–>3+} (bx+3) = 3b+3

f is continuous ifL.H.L. = R.H.L. = f(3)

3a + 1 = 3b + 3 or 3(a – b) = 2

a – b = or a = b + , for any arbitrary value of b.

Therefore the value of a corresponding to the value of b.

**Question 18.**

For what value of λ is the function defined by

continuous at x = 0? What about continuity at x = 1?

**Solution:**

At x = 0, L.H.L. = lim_{x–>0-} λ (x² – 2x) = 0 ,

R.H.L. = lim_{x–>0+} (4x+ 1) = 1, f(0)=0

f (0) = L.H.L. ≠ R.H.L.

=> f is not continuous at x = 0,

whatever value of λ ∈ R may be

At x = 1, lim_{x–>1} f(x) = lim_{x–>1} (4x + l) = f(1)

=> f is not continuous at x = 0 for any value of λ but f is continuous at x = 1 for all values of λ.

**Question 19.**

Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

**Solution:**

Let c be an integer, [c – h] = c – 1, [c + h] = c, [c] = c, g(x) = x – [x].

At x = c, lim_{x–>c-} (x – [x]) = lim_{h–>0} [(c – h) – (c – 1)]

= lim_{h–>0} (c – h – (c – 1)) = 1[∵ [c – h] = c – 1]

R.H.L. = lim_{x–>c+} (x – [x])= lim_{h–>0} (c + h – [c + h])

= lim_{h–>0} [c + h – c] = 0

f(c) = c – [c] = 0,

Thus L.H.L. ≠ R.H.L. = f (c) => f is not continuous at integral points.

**Question 20.**

Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?

**Solution:**

Let f(x) = x² – sinx + 5,

**Question 21.**

Discuss the continuity of the following functions:

(a) f (x) = sin x + cos x

(b) f (x) = sin x – cos x

(c) f (x) = sin x · cos x

**Solution:**

(a) f(x) = sinx + cosx

**Question 22.**

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

**Solution:**

(a) Let f(x) = cosx

**Question 23.**

Find all points of discontinuity of f, where

**Solution:**

At x = 0

**Question 24.**

Determine if f defined by is a continuous function?

**Solution:**

At x = 0

**Question 25.**

Examine the continuity of f, where f is defined by

**Solution:**

**Find the values of k so that the function is continuous at the indicated point in Questions 26 to 29.**

**Question 26.**

**Solution:**

At x =

L.H.L =

**Question 27.**

**Solution:**

**Question 28.**

**Solution:**

**Question 29.**

**Solution:**

**Question 30.**

Find the values of a and b such that the function defined by

to is a continuous function.

**Solution:**

**Question 31.**

Show that the function defined by f(x)=cos (x²) is a continuous function.

**Solution:**

Now, f (x) = cosx², let g (x)=cosx and h (x) x²

∴ goh(x) = g (h (x)) = cos x²

Now g and h both are continuous ∀ x ∈ R.

f (x) = goh (x) = cos x² is also continuous at all x ∈ R.

**Question 32.**

Show that the function defined by f (x) = |cos x| is a continuous function.

**Solution:**

Let g(x) =|x|and h (x) = cos x, f(x) = goh(x) = g (h (x)) = g (cosx) = |cos x |

Now g (x) = |x| and h (x) = cos x both are continuous for all values of x ∈ R.

∴ (goh) (x) is also continuous.

Hence, f (x) = goh (x) = |cos x| is continuous for all values of x ∈ R.

**Question 33.**

Examine that sin |x| is a continuous function.

**Solution:**

Let g (x) = sin x, h (x) = |x|, goh (x) = g (h(x))

= g(|x|) = sin|x| = f(x)

Now g (x) = sin x and h (x) = |x| both are continuous for all x ∈ R.

∴f (x) = goh (x) = sin |x| is continuous at all x ∈ R.

**Question 34.**

Find all the points of discontinuity of f defined by f(x) = |x|-|x+1|.

**Solution:**

f(x) = |x|-|x+1|, when x< – 1,

f(x) = -x-[-(x+1)] = – x + x + 1 = 1

when -1 ≤ x < 0, f(x) = – x – (x + 1) = – 2x – 1,

when x ≥ 0, f(x) = x – (x + 1) = – 1

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