Students can Download 2nd PUC Maths Chapter 5 Continuity and Differentiability Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability

### 2nd PUC Maths Continuity and Differentiability One Mark Questions and Answers

Question 1.

Find the derivative of cos (x^{2}) with respect to x.

Answer:

Let y = cos (x^{2})

\(\frac{d y}{d x}\) = -sin(x^{2}).2x.

Question 2.

If tan (2x + 3), find \(\frac{d y}{d x}\)

Answer:

Let y = tan (2x + 3)

\(\frac{d y}{d x}\) = 2.sec^{2}(2x + 3).

Question 3.

The function f(x) = \(\frac{1}{x-5}\) is not continuous at x = 5. Justify the statement.

Answer:

The function is not defined at x = 5.

Question 4.

Sin (x^{2} + 5).

Answer:

Let y = Sin (x^{2} + 5)

Differentiate both sides w.r.t. x, we get

= cos (x^{2} + 5) (2x + 0) = 2x cos (x^{2} + 5).

Question 5.

Cos (sin x).

Answer:

Let y = cos (sin x) .

Differentiate both sides w.r.t. x, we get

= – sin (sin x) cos x = – cos x sin (sin x).

Question 6.

Sin (ax + b).

Answer:

Let y = sin (ax + b)

Differentiate both sides w.r.t. x, we get

= cos (ax + b) {a × 1 + 0} = a cos (ax + b).

Question 7.

Sec (tan √x).

Answer:

Let y = sec (tan √x)

Differentiate both sides w.r.t. x, we get

Question 8.

cos (√x)

Answer:

Let y = cos(√x)

Differentiate both sides w.r.t. x

Question 9.

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

Answer:

Here, f(x) = 5x – 3

Question 10.

Examine the continutiy of the function f(x) = 2x^{2} – 1 at x = 3.

Answer:

Here, f(x) = 2x^{2} – 1

Question 11.

Examine the following functions for continuity:

(a) f(x) = x – 5

(b) f(x) = \(\frac{1}{x-1}\) ≠ 0

(c) f(x) = \(\frac{x^{2}-25}{x+5}\), x ≠ 5

(d) f(x) = |x – 5|.

Answer:

(a) f(x) = x – 5 is a polynomial function, so f(x) is continuous for all values of x.

(b) f(x) = \(\frac{1}{x-5}\) is a quotient function of two polynomial functions, so ,

f(x) is continuous for all values of x provided x ≠ 5.

(c) f(x) = \(\frac{x^{2}-25}{x+5}=\frac{(x+5)(x-5)}{x+5}=x-5\)

∴ f(x) = x – 5 is a polynomial function, so f(x) is continuous at all values of x.

(d)

Also, f(5) = 5 – 5 = 0

∵ LHL = RHL = f(5). therefore, the function is continuous at x = 5.

Question 12.

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

Answer:

Here, f(x) = x^{n}

Thus, f(x) is continuous at x = n, where n is a positive integer.

Question 13.

Differentiate sin (cos (x^{2})) with respect to x.

Answer:

Let y = sin (cos(x^{2}))

\(\frac{d y}{d x}\) = cos(cos(x^{2})). – sin(x^{2}).2x

Question 14.

Differentiate sin (x^{2}) with respect to x.

Answer:

Let y = sin(x^{2})

\(\frac{d y}{d x}\) = cos(x^{2}).2x

Question 15.

Differentiate cos(x^{3}).sin^{2} (x^{5}) with respect tox.

Answer:

Let y = cos(x^{3}).sin^{2} (x^{5})

Question 16.

If 2x + 3y = sin x find \(\frac{d y}{d x}\)

Answer:

Let 2x + 3y = sinx

Differentiating both sides w.r. to x.

2(1) + 3\(\frac{d y}{d x}\) = cos x

\(\frac{d y}{d x}\) = \(\frac{\cos x-2}{3}\)

Question 17.

If 2x + 3y = sin y find \(\frac{d y}{d x}\)

Answer:

Differentiate both sides w.r. t. x.

Question 18.

If ax + by^{2} = cos y find \(\frac{d y}{d x}\)

Answer:

Question 19.

If xy + y^{2} = tan x + y find \(\frac{d y}{d x}\)

Answer:

Question 20.

Differentiate x^{2} + xy + y^{2} = 100 with respect to X.

Answer:

x^{2} + xy + y^{2} = 100

Differentiating with respect to x.

Question 21.

If x^{3} + x^{2}y + xy^{2} + y^{3} = 81 find \(\frac{d y}{d x}\)

Answer:

x^{3} + x^{2}y + xy^{2} + y^{3} = 81

Differentiating w.r. to. x.

Question 22.

If sin^{2}x + cos xy = k find \(\frac{d y}{d x}\)

Answer:

Question 23.

If sin^{2}x + cos^{2}y = 1 find \(\frac{d y}{d x}\).

Answer:

Question 24.

Differentiate w.r. to x : x^{x} + a^{x} + x^{a} + a^{a}

Answer:

Let y = x^{x} + a^{x} + x^{a} + a^{a}

\(\frac{d y}{d x}\) = x^{x}(1 + log x) + a^{x} log_{e}a + ax^{n-1} + 0

Question 25.

Differentiate x^{x} w.r. to x

Answer:

Let y = x^{x}

log y = log x^{x}

log y = x log x.

\(\frac{1}{y} \frac{d y}{d x}\) = x.\(\frac{1}{x}\) + logx(1)

\(\frac{d y}{d x}\) = y(1 + log x) = x^{x} (1 + log x)

Question 26.

Differentiate a^{x} w.r. to x

Answer:

Let y = a^{x}

= x log a

log y = log a^{x}

\(\frac{1}{y} \frac{d y}{d x}\) = x(0) + log a(1)

\(\frac{d y}{d x}\) = y log a = a^{x} log_{e}a

Note : \(\frac{d }{d x}\)(log_{e}x) = \(\frac{1}{x}\)

\(\frac{d }{d x}\)(log_{e}a) = 0 ∵ a is a constant.

Question 27.

If y = x^{5}5^{x} find \(\frac{d y}{d x}\)

Soin.

y = x^{5} .5^{x}

\(\frac{d y}{d x}\) = x^{5} .5^{x} log_{e}5 + 5^{x} .5x^{4}

Question 28.

If y = x^{3}.2^{x} find \(\frac{d y}{d x}\)

Answer:

y = x^{3}.2^{x}

\(\frac{d y}{d x}\) = x^{3}.2^{x} log_{e}2 + 2^{x}(3x^{2})

Question 29.

If y = (log x)^{cos x} find \(\frac{d y}{d x}\)

Answer:

y = (log x)^{cos x}

log y = log(log x)

Question 30.

If y = cosx.cos2x.cos3x find \(\frac{d y}{d x}\).

Answer:

logy = log(cosx.cos2x.cos3x)

= logcosx + logcos2x + logcos3x

cos x cos 2x cos 3x [-tan x – 2 tan 2x – 3 tan 3x]

### 2nd PUC Maths Continuity and Differentiability Two Marks Questions and Answers

Question 1.

Check the continuity of the function f given by f(x) = 2x + 3 at x = 1.

Answer:

Question 2.

Find the derivative of (3x^{2} – 7x + 3)^{5/2} with respect to x.

Answer:

Let y = (3x^{2} – 7x + 3)^{5/2}

\(\frac{d y}{d x}=\frac{5}{2}\) (3x^{2} – 7x + 3)^{5/2} (6x – 7).

Question 3.

If y = (sin^{-1} x)^{x} find \(\frac{d y}{d x}\)

Answer:

Question 4.

Answer:

Question 5.

If y = sin (Iog_{e} x) prove that \(\frac{d y}{d x}=-\sqrt{\frac{y}{x}}\)

Answer:

Question 6.

Find the derivative of x^{x} – 2^{sin x} with respect to x.

Answer:

Let y = x^{x} – 2^{sin x}

Let u = x^{x} and v = 2^{sin x}

∴ y = u – v

Differentiating both sides w.r.t. x, we get

Now, u = x^{x}

Taking log on both sides.

⇒ log u = log x^{x} ⇒ log u = x log x

Question 7.

Find the derivative of \(\frac{e^{x}}{\sin x}\) w.r.t x.

Answer:

Question 8.

Find the derivative of e^{sinx-1} w.r.t. x.

Answer:

y = e^{sinx-1}

Differentiate both sides wor.t. x, we get

Question 9.

Differentiate \(\left(x+\frac{1}{x}\right)^{x}\) W. r. to x. (M. 2015)

Answer:

Question 10.

Answer:

Question 11.

Differentiate (x + 3)^{2} (x + 4)^{3} (x + 5)^{4} w.r. to x.

Answer:

Let y = (x + 3)^{2} (x + 4)^{3} (x + 5)^{4}

log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)

Question 12.

If y = log_{7} (log x) find \(\frac{d y}{d x}\)

Answer:

Question 13.

Differentiate w.r. to x: \(\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^{2}+4}}\)

Answer:

Question 14.

If y = cos ^{-1}(sin x) find \(\frac{d y}{d x}\).

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Differentiate sin^{2} x w.r. t e^{cosx}

Answer:

Question 18.

If y = (3x^{2} – 9x + 5)^{9} find \(\frac{d y}{d x}\)

Answer:

\(\frac{d y}{d x}\) = 9(3x^{2} – 9x + 5)^{8} (6x – 9)

Question 19.

If y = sin^{3}x + cos^{6}x find \(\frac{d y}{d x}\)

Answer:

y = sin^{3}x + cos^{6}x

\(\frac{d y}{d x}\) = 3 sin^{2}x cosx + 6cos^{5}x(-sin x)

Question 20.

If y = (5x)^{3 cos 2x} find \(\frac{d y}{d x}\)

Answer:

y = y = (5x)^{3 cos 2x}

Question 21.

Differentiate w.r. to x. (log x)^{x}

Answer:

Let y = (log x)^{x}

log y = x log (log x)

Question 22.

Differentiate w.r. to x. x^{logx}.

Answer:

Let y = x^{logx}

log y = log x ^{log x}

= log x log x = (log x)^{2}

Question 23.

Differentiate (log x)^{log x} w.r. to x.

Answer:

Let y = (log x)^{log x}

log y = log (log x)

Question 24.

If y = sin^{-1} (x√x) find \(\frac{d y}{d x}\).

Answer:

Question 25.

If x^{y} = y^{x} \(\frac{d y}{d x}\)

Answer:

x^{y} = y^{x}

log x^{y} = log y^{x}

Question 26.

Differentiate : x^{sin x} with respect to x.

Answer:

Let y = x^{sin x}

### 2nd PUC Maths Continuity and Differentiability Three Marks Questions and Answers

Question 1.

Answer:

Question 2.

y = sin^{-1}\(\left(\frac{2 x}{1+x^{2}}\right)\)

Answer:

Given y = sin^{-1}\(\left(\frac{2 x}{1+x^{2}}\right)\)

Putting θ = tan^{-1} x i.e., x = tan θ

Question 3.

Answer:

Substitute tan^{-1} x = θ i.e., x = tan θ

⇒ y = tan^{-1} (tan 3θ) = 3θ = 3 tan^{-1} x

Differentiating both sides w.r.t. x, we get

⇒ 3\(\frac{d}{d x}\)(tan^{-1}) = \(\frac{3}{1+x^{2}}\)

Question 4.

y = cos^{-1}\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1.

Answer:

Let tan^{-1} x = θ i.e., x = tan θ

⇒ y = cos^{-1} (cos 2θ) = 2θ = 2 tan^{-1} x

Differentiating both sides w.r.t. x, we get

Question 5.

y = sin^{-1}\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1.

Answer:

Substitute x = tan θ ⇒ tan^{-1} x = θ

Question 6.

y = cos^{-1}\(\left(\frac{2 x}{1+x^{2}}\right)\), -1 < x < 1.

Answer:

Substitute x = tan θ ⇒ tan^{-1} x = θ

Question 7.

Answer:

Let sin^{-1} x = θ , then x = sin θ

⇒ y = sin^{-1} (2 sin θ cos θ) = sin^{-1}(sin 2θ)

⇒ y = 2θ ⇒ 2 sin^{-1} x

Differentiating both sides w.r.t. x, we get

⇒ \(\frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}\)

Question 8.

Answer:

Let cos^{-1} x = θ, i.e., x = cos θ,

⇒ y sec^{-1} (sec 2θ) ⇒ y = 2θ ⇒ y = 2 cos^{-1} x

Differentiating both sides w.r.t. x, we get

Question 9.

Find \(\frac{d y}{d x}\), if x = at^{2}, y = 2at.

Answer:

Given that x = at^{2}, y = 2at

Question 10.

x = 2at^{2}, y = at^{4}.

Answer:

Given, x = 2at^{2}, y = at^{4}

Differentiating w.r.t. t, we get

Question 11.

x = a cos θ, y = b sin θ.

Answer:

Given, x = a cos θ, y = b sin θ

Question 12.

x = sin t, y = cos 2t.

Answer:

Given, x = sin t, y = cos 2t.

Differentiating w.r.t. t, we get

Question 13.

x = 4t, y = \(\frac{4}{t}\)

Answer:

Given x = 4t, y = \(\frac{4}{t}\)

Question 14.

x = cos θ – cos 2θ, y = sin θ – sin 2θ.

Answer:

Given x = cos θ – cos 2θ, y = sin θ – sin 2θ

Differentiating w.r.t. θ, we get

Question 15.

x = a (θ – sin θ), y = a (1 + cos θ).

Answer:

Given, x = a (θ – sin θ), y = a (1 + cos θ)

Differentiating w.r.t. 0, we get

Question 16.

Answer:

Question 17.

x = a (cos t + log tan\(\frac{t}{2}\)), y = a sint.

Answer:

Given, x = a (cos t + log tan\(\frac{t}{2}\))

Differentiating w.r.t. t, we get

Question 18.

x = a sec θ, y = b tan θ.

Answer:

Given, x = a sec θ, y = b tan θ

Differentiating w.r.t. θ, we get

Question 19.

x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ).

Answer:

Given, x = a (cos θ + θ sin ), y = a (sin θ – θ cos θ)

Differentiating w.r.t. θ, we get

∴ = a {- sin θ + (θ cos θ + sin θ.1)} = a θ cos θ

(Using product rule in \(\frac{d}{d \theta}\) (θ sin θ)

Question 20.

If x = a (θ + sin θ) and y = a (1 – cos θ) prove that \(\frac{d y}{d x}\) = tan\(\left(\frac{\theta}{2}\right)\)

Answer:

We have \(\frac{d x}{d \theta}\) = a(1 + cosθ), \(\frac{d y}{d \theta}\) = a(sin θ)

Question 21.

If a function f(x) is differentiable at x = c prove that it is continuous at x = c.

Answer:

Since f is differentiable at c, we have

Question 22.

Differentiate with respect to x : (sin x)^{x} + sin^{-1} √x

Answer:

Let y =(sin x)^{x} + sin^{-1} √x

y = u + v

Question 23.

Find \(\) given x^{y} + y^{x} = 1

Answer:

Question 24.

Answer:

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

x^{2} + x^{2}y = y^{2} + y^{2}x

x^{2} – y^{2} = y^{2}x – x^{2}y

(x – y)(x + y) = xy (y – x) ⇒ (x – y)(x + y) = -xy (x – y)

∵ x ≠ y

⇒ x + y = -xy

x + y + xy = 0

y + xy = -x

Question 25.

Answer:

Question 26.

If cos y = x cos (a + y) with cos a ≠ ±1

Prove that \(\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}\)

Answer:

Question 27.

If x = a(cos t + t sin t) y = a(sin t – t cos t) find \(\frac{d^{2} y}{d x^{2}}\)

Answer:

x = a(cos t + t sin t)

Question 28.

If x^{y} = e^{x-y} prove that \(\frac{d y}{d x}=\frac{\log _{e} x}{\left(1+\log _{e} e\right)^{2}}\)

Answer:

x^{y} = e^{x-y}

log_{e}x^{y} = log_{e}e^{x-y}

y log_{e}x = (x – y)log_{e}e

y log_{e}x = x – y

y + y log_{e}x = x

### 2nd PUC Maths Continuity and Differentiability Four Marks Questions and Answers

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Define a continuity of a function at a point. Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.

Answer:

Let f(x) be a real valued function on a subset of the real numbers and let c be a point in the domain off. Then f (x) is continuous at x = c if.

Let g (x) = |x| and h (x) = |x + 1|

Now, g (x) = |x| is the absolute valued function, so it is a continuous function for all x ∈ R,

H (x) = |x + 1| is the absolute valued function, so it is a continuous function for all x ∈ R.

Since g (x) and h (x) are both continuous functions for all x ∈ R, so difference of two continuous function is a continuous function for all x ∈ R. Thus f(x) = |x| – |x + 1| is continuous at all points. Hence there is no point at which f(x) is discontinuous.

Question 7.

Answer:

3a + 1 = 3 b+ 3 = 3a + 1

⇒ 3a + 1 = 3b + 3

⇒ 3a = 3b + 3 – 1

⇒ 3a = 3b + 2

⇒ a = b + \(\frac { 2 }{ 3 }\)

Question 8.

Find the points of discontinuity of the function f(x) = x – [x] where [x] indicates the greatest integer not greater than x. Also write the set of values of x where the function is continuous.

Answer:

Let a ∈ Z

∴ f(x) = x – [x] ¡s discontinuous at integral points.

∴ f(x) is continuous at all points.

x ∈ R – Z

Question 9.

Answer:

f(x) is defined at all points on the real line.

Let C be a real number.

Case (1) : Let c < 2.

Question 10.

Define continuity of a function at a point. Find all the points of discontinuity of f defined by f(x) = |x| – |x +1|

Answer:

Let f(x) be a real valued function on a subset of the real numbers and let c be a point in the domain off. Then f(x) is continuous at x = c if.

Let g (x) = |x| and h (x) = |x + 1|

Now, g (x) = |x| is the absolute valued function, so it is a continuous function for all x∈R.

H (x) = |x + 1| is the absolute valued function, so it is a continuous function for all x ∈ R.

Since g (x) and h (x) are both continuous functions for all X ∈ R, so difference of two continuous function is a continuous function for all x ∈ R. Thus f(x) = |x| – |x + 1| is continuous at all points. Hence there is no point at which/(x) is discontinuous.

### 2nd PUC Maths Continuity and Differentiability Five Marks Questions and Answers

Question 1.

If y = A sin x + B cos x, then prove that \(\frac{d^{2} y}{d x^{2}}\) + y = 0.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

If y = 3cos(log x) + 4 sin(log x) show that x^{2}y_{2} + xy_{1} + y = 0.

Answer:

Given, y = 3 cos (log x) + 4 sin (log x)

Differentiating w.r.t. s, we get ……… (i)

Multiplying by x, we get

xy_{1} = -3 sin (log x) + 4 cos (log x) ……….. (ii)

Again differentiating w.r.t. x, we obtain

Multiplying through out by x, we have

x^{2}y_{2} + xy_{1} = – (3 cos (log x) + 4 sin (log x)) [from Eq. (i)]

⇒ x^{2}y_{2} + xy_{1} = -y ⇒ x^{2}y_{2} + xy_{1} + y = 0

Question 5.

If y = 5 cos x – 3 sin x, prove that \(\frac{d^{2} y}{d x}\) + y = 0.

Answer:

Given, y = 5 cos x – 3 sin x

Differentiating twicely w.r.t. x, we get

\(\frac{d y}{d x}\) = -5 sin x – 3 cos x

Question 6.

Answer:

Question 7.

Answer:

Question 8.

If e^{y} (x + 1) = 1, show that \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)

Answer:

Question 9.

y = 500e^{7x} + 600e^{-7x} prove that \(\frac{d^{2} y}{d x^{2}}\) = 49 y.

Answer:

y = 500e^{7x} + 600e^{-7x}

Question 10.

Answer:

Question 11.

Answer:

Question 12.

Answer:

Question 13.

If y^{x} = e^{y-x}, prove that \(\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}\)

Answer:

y^{x} = e^{y-x}

log y^{x} = log e^{y-x}

x log y = (y – x)log e

x + x log y = y

x(1 + log y) = y

Question 14.

Answer:

y = (cos x)^{y}

log y = log (cos x)^{y}

log y = y log cos x

Question 15.

If y = tan^{-1}x find \(\frac{d^{2} y}{d x^{2}}\) im terms of y alone.

Answer:

y = tan^{-1}x

x = tan y