**3. **(a) Given: f(x) = x – 5

It is evident that f is defined at every real number k and its value at k is k – 5.

Since therefore, f(x) is continuous at every real number and it is a continuous function.

Since therefore, f(x) is continuous at every point of domain of f and it is a continuous function.

Since therefore, f(x) is continuous at every point of domain of f and it is a continuous function.

(d) Given: f(x) = |x – 5|

Domain of f(x) is real and infinite for all real x

Here f(x) = |x – 5| is a modulus function.

Since, every modulus function is continuous, therefore, f is continuous in its domain R.

**4.** Given: f(x) = x^{n} where n is a positive integer.

And f(n) = n^{n}

Since therefore, f(x) is continuous at x = n.

Here f(x) is defined for x ≤ 2 i.e. on (-∞, 2) and also for x > 2 i.e., on (2, ∞).

∴ Domain of f is (-∞, 2) ∪ (2, ∞) = (-∞, ∞) = R

∴ For all x < 2, f(x) = 2x + 3 is a polynomial and hence continuous and for all x >, f(x) = 2x – 3 is a continuous and hence it is also continuous on R – {2}.

Therefore, does not exist and hence f(x) is discontinuous at x = 2 (only).

Here f(x) is defined for x ≤ -3 i.e., on (-∞, -3) and for -3 < x < 3 i.e., on (-3, 3) and also for x ≥ 3 i.e., on (3, ∞).

∴ Domain of f is = (-∞, -3) ∪ (-3, 3) ∪ (3, ∞) = (-∞, ∞) = R

∴ For all x < -3, f(x) = |x| + 3 = -x + 3 is a polynomial and hence continuous and for all x(-3 < x < 3), f(x) = -2x is a continuous and hence it is also continuous and also for all x > 3, f(x) = 6x + 2. Therefore, f(x) is continuous on R – {-3, 3}.

It is observed that x = -3 and x = 3 are partitioning points of domain R.

Therefore, does not exist and hence f(x) is discontinuous at x = 3 (only).

⇒ f(x) = 1 if x > 0, f(x) = -1 if x < 0 and f(x) = 0 if x = 0

It is clear that domain of f(x) is R as f(x) is defined for x > 0, x < 0 and x = 0.

For all x > 0, f(x) = 1 is a constant function and hence continuous.

For all x < 0, f(x) = -1 is a constant function and hence continuous.

Therefore f(x) is continuous on R – {0}.

Therefore, does not exist and hence f(x) is discontinuous at x = 0 (only).

Since L.H.L. = R.H.L. = f(0)

Therefore, f(x) is a continuous function.

Therefore, f(x) is continuous at x = c < 0

Therefore, f(x) is continuous at x = c > 0

Hence, the function is continuous at all points of its domain.

It is observed that f(x) being polynomial is continuous for x ≥ 1 and x < 1 for all x ∈ R.

And f(1) = 2

Since L.H.L. = R.H.L. = f(1)

Therefore, f(x) is a continuous at x = 1 for all x ∈ R.

Hence, f(x) has no point of discontinuity.

Since L.H.L. = R.H.L. = f(2)

Therefore, f(x) is a continuous at x = 2

Therefore, f(x) is a continuous for all x ∈ R.

Hence the function has no point of discontinuity.

Since L.H.L. ≠ R.H.L.

Therefore, f(x) is discontinuous at x = 1

Therefore, f(x) is a continuous for all x ∈ R – {1}

Hence for all given function x = 1 is a point of discontinuity.

Since L.H.L. ≠ R.H.L.

Therefore, f(x) is discontinuous at

Therefore, f(x) is a continuous for all x ∈ R – {1}

Hence f(x) is not a continuous function.

In the interval 0 ≤ x ≤ 1, f(x) = 3

∴ f is continuous in this interval.

Since L.H.L. ≠ R.H.L.

Therefore, f(x) is discontinuous at x = 1

Since L.H.L. ≠ R.H.L.

Therefore, f(x) is discontinuous at x = 3

Hence, f is discontinuous at x = 1 and x = 3.

Since L.H.L. = R.H.L.

Therefore, f(x) is continuous at x = 0

Since L.H.L. ≠ R.H.L.

Therefore, f(x) is discontinuous at x = 1

When x < 0, f(x) being a polynomial function is continuous for all x < 0.

When x > 1, f(x) = 4x. It is being a polynomial function is continuous for all x > 1.

Hence x = 1 is a point of discontinuity.

Since L.H.L. = R.H.L.

Therefore, f(x) is continuous at x = -1

Since L.H.L. = R.H.L.

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

**18.** Since f(x) is continuous at x = 0.

Here, therefore should be L.H.L. = R.H.L.

⇒ 0 = 1, which is not possible.

∴ for no value of λ, f(x) is continuous at x = 0

Again Since f(x) is continuous at x = 1.

Here, L.H.L. = R.H.L.

∴ for any value of λ, f(x) is continuous at x = 1.

**19. **For any real number we use the symbol [x] to denote the fractional part or decimal part of x. For example,

[3.45] = 0.45

[–7.25] = 0.25

[3] = 0

[–7] = 0

The function g : R → R defined by g(x) = x – [x] ∀x∈∞ is called the fractional part function. It is observed that the domain of the fractional part function is the set R of all real numbers and the range of the set [0, 1).

Hence given function is discontinuous function.

**20. ** Given: f(x) = x^{2} – sin x + 5

And f(π) = π^{2} – sin + 5 = π^{2} + 5

Therefore, f is continuous at x = π

**21.** (a) Let a be an arbitrary real number then

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, f(x) = sin x + cos x is continuous.

(b) Let a be an arbitrary real number then

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, f(x) = sin x – cos x is continuous.

(c) Let a be an arbitrary real number then

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, f(x) = sin x.cos x is continuous.

**22. **(a) Let a be an arbitrary real number then

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, cos x is continuous.

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, f(x) = cos ec x is continuous.

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, f(x) = sec x is continuous.

Therefore, f(x) is continuous at x = a.

Since, a is an arbitrary real number, therefore, f(x) = cot x is continuous.

∴ f is continuous at x = 0.

When x < 0, sin x and x are continuous, then is also continuous.

When x > 0, f(x) = x + 1 is a polynomial, then f is continuous.

Therefore, f is continuous at any point.

Since, therefore, the function f is continuous at x = 0.

Therefore, f is discontinuous at x = 0.

**29.** When x < 5, we have f(x) = kx + 1 which being a polynomial is continuous at each point x < 5.

And, when x > 5,we have f(x) = 3x – 5 which being a polynomial is continuous at each point x > 5.

Now f(5) = 5k + 1 = 3(5 + h) – 5

Since function is continuous, therefore, eq. (i) = eq. (ii)

**30. **For x < 2, function is f(x) = 5, constant, therefore it is continuous.

For 2< x < 10, function f(x) = ax + b, polynomial, therefore, it is continuous.

For x > 10, function is f(x) = 21, constant, therefore it is continuous.

For continuity at

Solving eq.(i) and eq.(ii), we get

a = 2 and b = 1.

**31. **Let f(x) = x^{2} and g(x) = cos x, then

(gof)(x) = g[f(x)] = g(x) = cos x^{2}

Now f and g being continuous it follows that their composite (gof) is continuous.

Hence cos x^{2} is continuous function.

**32. **Given: f(x) = |cos x| ….(i)

f(x) has a real and finite value for all x ∈ R.

∴ Domain of f(x) is R.

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

Since g(x) and h(x) being cosine function and modulus function are continuous for all real x.

Now, (goh)x = g{h(x)} = g(|x|) = cos |x| being the composite function of two continuous functions is continuous, but not equal to f(x)

Again, (hog)x = h{g(x)} = h(cos x) = |cos x| = f(x) [Using eq. (i)]

Therefore, f(x) = |cos x| = (hog)x being the composite function of two continuous functions is continuous.

**33. **Let f(x) = |x| and g(x) = sin|x|, then

(gof)x = g{f(x)} = g(|x|) = sin|x|

Now, f and g being continuous, it follows that their composite, (gof) is continuous.

Therefore, sin|x| is continuous.

Therefore, at x = 0, f(x) is continuous.

Hence, these is no point of discontinuity.

Since R f'(1) ≠ L f'(1)

Therefore, f(x) is not differentiable at x = 1.

Since R f'(1) ≠ L f'(1)

Therefore, f(x) = [x] is not differentiable at x = 1.

Similarly, f(x) = [x] is not differentiable at x = 2.

To simplify the given Inverse Trigonometric function, putting x = tanθ

**70.** Let y = cos x cos 2x cos 3x ….(i)

Taking logs on both sides, we have

**72.** Let y = (log x)^{cos x}….(i)

Taking logs on both sides, we have

**74.** Let y = (x + 3)^{2} (x + 4)^{3} (x + 5)^{4} ….(i)

Taking logs on both sides, we have

From eq.(ii), (iii) and (iv), we can say that value of is same obtained by three different methods.

**87.** Given: u, v and w are functions of x.

= R.H.S Hence proved.

(ii) By Logarithmic differentiation:

Let y = uvw

**116. **Consider f(x) = x^{2} + 2x – 8, x ∈ [-4, 2].

(i) Function is continuous in [-4, 2] as it is a polynomial function and polynomial function is always continuous.

(ii) f'(x) = 2x + 2, f'(x) exists in [-4, 2], hence derivable.

(iii) f(-4) = 0 and f(2) = 0

∴ f(-4) = f(2)

Conditions of Rolle’s theorem are satisfied, hence there exists, at least one c ∈ (-4, 2) such that f'(c) = 0

⇒ 2c + 2 = 0

⇒ c = -1.

**117. **(i) Being greatest integer function the given function is not differentiable and continuous

hence Rolle’s theorem is not applicable.

(ii) Being greatest integer function the given function is not differentiable and continuous hence Rolle’s theorem is not applicable.

(iii) f(x) = x^{2} – 1 ⇒ f(1) = (1)^{2} – 1 = 1 – 1 = 0

f(2) = (2)^{2} – 1 = 4 – 1 = 3 ∴ f(1) ≠ f(2)

Hence, Rolle’s theorem is not applicable.

**118. **For, Rolle’s theorem, if

(i) f is continuous is [a, b]

(ii) f is derivable in [a, b]

(iii) f(a) = f(b)

Then, f'(c) = 0, c ∈ (a, b)

It is given that f is continuous and derivable, but f'(c) ≠ 0

⇒ f(a) ≠ f(b)

⇒ f(-5) ≠ f(5)

**119. **(i) Function is continuous in [1, 4] as it is a polynomial function and polynomial function is always continuous.

(ii) f'(x) = 2x – 4, f'(x) exists in [1, 4], hence derivable. Conditions of MVT theorem are satisfied, hence there exists, at least one c ∈ (1, 4) such that

**120. **(i) Function is continuous in [1, 3] as it is a polynomial function and polynomial function is always continuous.

(ii) f'(x) = 3x^{2} – 10x – 3, it exists in [1, 3], hence derivable.

Conditions of MVT theorem are satisfied, hence there exists, at least one c ∈ (1, 3) such that

Since f(1) ≠ f(3), therefore the value of ‘c’ does not exist such that f(c) = 0.

**121. **Mean Value Theorem states that for a function R, if f : [a, b] → R, if

(i) f is continuous on (a, b)

(ii) f is differentiable on (a, b)

Then there exist some c ∈ (a, b) such that

Therefore, the Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

(i) f(x) = [x] for x ∈ [5, 9]

It is evident that the given function f(x) is not continuous at x = 5 and x = 9.

Therefore,

f(x) is not continuous at [5, 9].

Now let n be an integer such that n ∈ [5, 9]

Since, L.H.L. ≠ R.H.L.,

Therefore f is not differentiable at [5, 9].

Hence Mean Value Theorem is not applicable for for f(x) = [x] for x ∈ [5, 9].

(ii) f(x) = [x] for x ∈ [-2, 2]

It is evident that the given function f(x) is not continuous at x = -2 and x = 2.

Therefore,

f(x) is not continuous at [-2, 2].

Now let n be an integer such that n ∈ [-2, 2].

Since, L.H.L. ≠ R.H.L.,

Therefore f is not differentiable at [-2, 2].

Hence Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [-2, 2]

(iii) f(x) = x^{2} – 1 for x ∈ [1, 2] ……….(i)

Here, f(x) is a polynomial function of degree 2.

Therefore, f(x) is continuous and derivable everywhere i.e., on the real time (-∞, ∞).

Hence f(x) is continuous in the closed interval [1, 2] and derivable in open interval (1, 2).

Therefore, both conditions of Mean Value Theorem are satisfied.

Now, From eq. (i), f'(x) = 2x

∴ f'(c) = 2c

Again, From eq. (i), f(a) = f(1) = (1)^{2} – 1 = 1 – 1 = 0

And From eq. (ii), f(b) = f(2) = (2)^{2} – 1 = 4 – 1 = 3

Therefore, Mean Value Theorem is verified.

which is a constant and is independent of a and b.

**138. ** Given: x = a(cos t + t sin t) and y = a(sin t – t cos t)

Differentiating both sides with respect to t,

**140.** Let p(n) be the given statement in the problem.

Therefore, p(m + 1) is true p(m) if is true but p(1) is true.

∴ By Principal of Induction p(n) is true for all n ∈ N.

**141.** Given: sin(A + B) = sin A cos B + cos A sin B

Consider A and B as function of t and differentiating both sides w.r.t. x,

**142.** Let us consider the function f(x) = |x| + |x – 1|

f is continuous everywhere but it is not differentiable at x = 0 and x = 1.