Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

JEE Advanced Maths Continuity Previous Year Questions with Solutions

Continuity JEE Advanced previous year questions with solutions are given on this page. These solutions are prepared by our subject experts. A function is said to be continuous if its graph is a single unbroken curve. Checking continuity at a given point, algebra of continuous functions, continuity of composite functions, etc are the important topics in continuity. Learning these solutions will definitely help you to easily crack the problems from continuity, for the JEE Advanced exam.

Question 1: Let [x] be the greatest integer less than or equal to x. Then at which of the following point(s) the function f(x) = x cos (π(x + [x])) is discontinuous?

(a) x = 2

(b) x = 0

(c) x = 1

(d) x = -1

Answer: a, c, d

Solution:

Given f(x) = x cos (π(x + [x]))

At x = 2

limx→ 2-x cos (π(x + [x])) = 2 cos (π+2π)

= 2 cos 3π

= -2

limx→ 2+x cos (π(x + [x])) = 2 cos (2π+2π)

= 2 cos 4π

= 2

LHL ≠ RHL

So f(x) is discontinuous at x = 2.

At x = 0

limx→ 0-x cos (π(x + [x])) = 0 cos (-π+0)

= 0

limx→ 0+x cos (π(x + [x])) = 0

LHL = RHL

So f(x) is continuous at x = 0.

At x = 1

limx→ 1-x cos (π(x + [x])) = cos (π)

= -1

limx→ 1+x cos (π(x + [x])) = cos 2π

= 1

LHL ≠ RHL

So f(x) is discontinuous at x = 1.

At x = -1

limx→ -1-x cos (π(x + [x])) = -cos (-3π)

= 1

limx→ -1+x cos (π(x + [x])) = -cos 2π

= -1

LHL ≠ RHL

So f(x) is discontinuous at x = 1.

The function is discontinuous at x = 2, 1, -1

Question 2: For every pair of continuous functions f, g: [0, 1] → R such that max {f(x): x∈[0, 1] = max {g(x) : x∈ [0, 1]}, the correct statements is (are)

(a) (f(c))2 + 3f(c) = (g(c))2 + 3g(c) for some c ∈ [0,1]

(b) (f(c))2 + f(c) = (g(c))2 + 3g(c) for some c ∈ [0,1]

(c) (f(c))2 + 3f(c) = (g(c))2 + g(c) for some c ∈ [0,1]

(d) (f(c))2 = (g(c))2 for some c∈ [0,1]

Answer: a, d

Solution:

Let f and g be maximum at c1 and c2 respectively.

c1, c2 ∈ [0, 1]

Then, f(c1) = g(c2)

Let h(x) = f(x) – g(x)

Then h(c1) = f(c1) – g(c1) > 0

And h(c2) = f(c2) – g(c2) < 0

So h(x) = 0 has atleast one root in [c1, c2]

i.e f(c) = g(c) for c∈ [[c1, c2]

which shows that options a and d are correct.

Question 3: Suppose that f(x) is a differentiable function such that f ‘(x) is continuous f ‘(0) = 1 and f ” (0) does not exist. If g(x) = x f ‘ (x). Then,

(a) g'(0) does not exist

(b) g'(0) = 0

(c) g'(0) = 1

(d) g'(0) = 2

Answer: c

Solution:

Given f ‘(0) = 1

g(x) = x f ‘ (x)

g’(x) = f’(x) + x f’’(x)

g’(0) = f’(0) + 0 f’’(0)

So g’(0) = 1

Hence option (c) is the answer.

Question 4: If a function f is such that f(0) = 2, f(1) = 3, f(x+2) = 2f(x) – f(x+1) for x ≥ 0, then f(5) is equal to

(a) -7

(b) -3

(c) 17

(d) 13

Answer: d

Solution:

Given f(0) = 2, f(1) = 3

f(x+2) = 2f(x) – f(x+1)

Put x = 0,

Then f(2) = 2f(0) – f(1)

= 4-3

= 1

Put x = 1

f(3) = 2f(1) – f(2)

= 6-1

= 5

Put x = 2

f(4) = 2f(2) – f(3)

= 2 – 5

= -3

f(5) = 2f(3) – f(4)

= 10 – -3

= 13

Hence option d is the answer.

Question 5: Let [t] denote the greatest integer ≤ t and limx→ 0 x[4/x] = A. Then the function, f(x) = [x2] sin πx is discontinuous, when x is equal to

(a) √(A+1)

(b) √(A+5)

(c) √(A+21)

(d) √A

Answer: a

Solution:

limx→0 x[4/x] = A

limx→ 0 x[4/x – {4/x}] = A

=> limx→0 4 – x{4/x} = A

=> 4-0 = A

As, f(x) = [x2]sin (πx) will be discontinuous at non-integers

And when x = √(A+1)

=> x = √5, which is not an integer.

Hence f(x) is discontinuous when x is equal to √(A+1).

Hence option a is the answer.

Question 6: Let f: [0, 2] → R be a function which is continuous on [0, 2] and its differentiable on (0, 2) with f(0) = 1. Let F(x) =

\(\begin{array}{l}\int_{0}^{x^{2}}f(\sqrt{t})dt\end{array} \)
for x∈ [0, 2]. If F(x) = f’(x) for all x belongs to (0, 2), then F(2) equals

(a) e2 – 1

(b) e4 – 1

(c) e – 1

(d) e4

Answer: b

Solution:

F(x) =

\(\begin{array}{l}\int_{0}^{x^{2}}f(\sqrt{t})dt\end{array} \)

F’(x) = f(x) 2x

F’(x) = f’(x) for all x belongs to (0, 2)

=> f(x) 2x = f’(x)

=> f’(x)/f(x) = 2x

On integrating w.r.t.x,we get

ln f(x) = x2 + c

f(x) =

\(\begin{array}{l}e^{x^{2}+c}\end{array} \)

=

\(\begin{array}{l}e^{x^{2}}.e^{c}\end{array} \)

Since f(0) = 1

=> ec = 1

=> f(x) =

\(\begin{array}{l}e^{x^{2}}\end{array} \)

Hence F(x) =

\(\begin{array}{l}\int_{0}^{x^{2}}e^{x}dx\end{array} \)

=

\(\begin{array}{l}e^{x^{2}}-1\end{array} \)

F(2) = e4 – 1

Hence option b is the answer.

Question 7: Let [x] be the greatest integer less than or equal to x. Then at which of the following points the function f(x) = x cos(π(x + [x]) is discontinuous?

(a) x = -1

(b) x = 0

(c) x = 1

(d) x = 2

Answer: a, c, d

Solution:

Given f(x) = x cos (π(x + [x])

Let x = n be any integer not equal to zero.

Then limx→ n cos (π(x + [x])) = n cos (π(n + n-1))

= n cos (2n-1)π

= -n

And limx→ n+x cos (π(x + [x])) = n cos(π(n + [n]))

= n cos 2nπ

= n

So LHL ≠ RHL

=> f is discontinuous at x = -1, 1, 2

At x = 0, LHL = RHL = 0 = f(0)

So f is continuous at x = 0.

Question 8: If f(x) = cos x cos 2x cos 4x cos 8x cos 16x, then f ‘(π/4) is equal to

(a) √2

(b) 1/√2

(c) 0

(d) √3/2

Answer: a

Solution:

f(x) = cos x cos 2x cos 4x cos 8x cos 16x

Multiply and divide by 2 sin x

f(x) = (2 sin x cos x cos 2x cos 4x cos 8x cos 16x)/2 sin x

= 2 sin 2x cos 2x cos 4x cos 8x cos 16x)/22 sin x

= 2 sin 4x cos 4x cos 8x cos 16x)/23 sin x

= 2 sin 8x cos 8x cos 16x)/24 sin x

= 2 sin 16 x cos 16x)/25 sin x

= sin 32x/25 sin x

f’(x) = (1/25)[ 32 sin x cos 32 x – sin 32x cos x]/sin2 x

f ‘(π/4) = (1/25)[ 32 sin π/4 cos 8π – sin 8π cos π/4]/sin2 π/4

= (1/25)[ (32×1/√2 – 0)/(½) ]

= √2

Hence option (a) is the answer.

Question 9: If f(x) = [x] – [x/4], x∈R, where [x] denotes the greatest integer function, then

(a) f is continuous at x = 4

(b) limx→ 4+ f(x) exists but limx→ 4- f(x) does not exist

(c) Both limx→ 4-f(x) and limx→ 4+f(x) exist but are not equal

(d) limx→ 4- f(x) exists but limx→ 4+ f(x) does not exist

Answer: a

Solution:

limx→ 4- ([x] – [x/4]) = 3-0 = 3

limx→ 4+- ([x] – [x/4]) = 4-1 = 3

RHL = LHL

f(4) = [4] – [4/4]

= 3

LHL = RHL = f(4)

So f(x) is continuous at x = 4.

Hence option a is the answer.

Question 10: Let f be a composite function of x defined by f(u) = 1/(u2+u-2), u(x) = 1/(x-1). Then the number of points x where f is discontinuous is

(a) 4

(b) 3

(c) 2

(d) 1

Answer: b

Solution:

Given f(u) = 1/(u2+u-2)

= 1/(u+2)(u-1), is discontinuous at x = -2, 1

u(x) = 1/(x-1), is discontinuous at x = 1

When u = -2,

Then 1/(x-1) = -2

=> x = -½

When u = 1, then 1/(x-1) = 1

=> x = 2

Hence the given composite function is discontinuous at three points x = 1, ½ and 2.

Hence option b is the answer.

Recommended video

Also Read

JEE Main Maths Limits, Continuity and Differentiability Previous Year Questions With Solutions

Leave a Comment

Your Mobile number and Email id will not be published.

*

*