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JEE Main Practice Test- 14 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 14

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JEE Main Practice Test- 14 - Question 1

Let f(x) be a one-to-one function such that f(1) = 3, f(3) = 1, f '(1) = – 4 and f '(3) = 2. If g = f –1, then the slope of the tangent line to 1/g at x = 1 is

Detailed Solution for JEE Main Practice Test- 14 - Question 1

JEE Main Practice Test- 14 - Question 2

The value of  is equal to

Detailed Solution for JEE Main Practice Test- 14 - Question 2

JEE Main Practice Test- 14 - Question 3

If g (x3 + 1) = x6 + x3 + 2, then the value of g(x2 – 1) is

Detailed Solution for JEE Main Practice Test- 14 - Question 3

g(x3 + 1) = x6 + x3 + 2 = (x3 + 1)2 – x3 + 1
= (x+ 1)– (x+ 1 – 1) + 1 = (x+ 1)– (x+ 1) + 2
Put x3 + 1 = t
So, g(t) = t2 – t + 2
⇒ g(x2 – 1) = (x2 – 1)2 – (x2 – 1) + 2
= x4 – 3x2 + 4. 

JEE Main Practice Test- 14 - Question 4

Suppose that f (0) = 0 and f ' (0) = 2, and let g (x) = f (- x + f (f (x))). The value of g ' (0) is equal to

Detailed Solution for JEE Main Practice Test- 14 - Question 4

g (x) = f (- x + f (f (x))) ;
f (0) = 0; f ' (0) = 2
g ' (x) = f ' (- x + f (f ( x )))· [- 1 + f ' (f (x))· f ' (x )]
g ' (0) = f ' (f (0))· [- 1 + f '(0) · f '(0)]
= f ' (0) [- 1 + (2)(2)]
= (2) (3) = 6 

JEE Main Practice Test- 14 - Question 5

The value of the definite integral,

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JEE Main Practice Test- 14 - Question 6

A line L is perpendicular to the curve  at its point P and passes through (10, –1). The coordinates of the point P are

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only (D) satisfies (1) and (2) both.

JEE Main Practice Test- 14 - Question 7


then the sum of the square of reciprocal of all the values of x where f(x) is non-differentiable, is equal to

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Clearly f(x) is non differentiable at x = 1/9, 1
∴ sum of squares of reciprocals
= 92 + 1 = 82 Ans.

JEE Main Practice Test- 14 - Question 8


h(x) = {x},k(x) = 5log(x + 3) then in [0, 1], Lagranges Mean Value Theorem is NOT applicable to
[Note : where [x] and {x} denote the greatest integer and fractional part function of x respectively]

Detailed Solution for JEE Main Practice Test- 14 - Question 8

f is not differentiable at x = 1/2
g is not continuous in [0, 1] at x = 0 & 1 h is not continuous in [0, 1] at x = 1
k (x) = (x + 3)ln5 = (x + 3)p where 2 < p < 3

JEE Main Practice Test- 14 - Question 9

If the function f (x) = ax e–bx has a local maximum at the point (2, 10), then

Detailed Solution for JEE Main Practice Test- 14 - Question 9

f (2) = 10, hence 2ae–2b = 10
⇒ ae–2b = 5 ....(1)
f ' (x) = a [e–bx – bx e–bx] = 0
f ' (2) = 0
a(e–2b – 2be–2b) = 0
ae–2b (1 – 2b) = 0
⇒ b = 1/2 or a = 0 (rejected)
from (1) if b = 1/2; a = 5e
∴ a = 5e and b = 1/2 Ans.]

JEE Main Practice Test- 14 - Question 10


then the value of x satisfying the equation f (x, 10) = f (x, 11), is

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JEE Main Practice Test- 14 - Question 11

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JEE Main Practice Test- 14 - Question 12

Number of integral solutions of the equation

[Note : where [x] denotes the greatest integer less than or equal to x and sgn x denotes signum function of x.]

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Hence two integral solution will satisfy above equation.

JEE Main Practice Test- 14 - Question 13

The area bounded by the curve y = x2 + 4x + 5 , the axes of co-ordinates & the minimum ordinate is

Detailed Solution for JEE Main Practice Test- 14 - Question 13

y = x2 + 4x + 5 = (x+2)2 + 1


JEE Main Practice Test- 14 - Question 14

Number of roots of the equation x2 –2x–log2 |1 – x | = 3 is

Detailed Solution for JEE Main Practice Test- 14 - Question 14

x2 – 2x – 3 = log2 | 1 – x | 4 points 

JEE Main Practice Test- 14 - Question 15

Let F(x) be the primitive of w.r.t. x. If F(10) = 60 then the value of F(13), is

Detailed Solution for JEE Main Practice Test- 14 - Question 15




given F(10) = 60 = 2 [29 + 1] + C
⇒ C = 0


F(13) = 2 [29 × 2 + 4 × 2]
= 4 × 33 = 132

JEE Main Practice Test- 14 - Question 16


is continuous at x = 0, then
[Note : {x} denotes fractional part of x.]

Detailed Solution for JEE Main Practice Test- 14 - Question 16



∴ k = 0

Note that f (x) is discontinuous at

JEE Main Practice Test- 14 - Question 17


[Note : where C is constant of integration.]

Detailed Solution for JEE Main Practice Test- 14 - Question 17

JEE Main Practice Test- 14 - Question 18

Point 'A' lies on the curve  and has the coordinate  where x > 0. Point B has the coordinates (x, 0). If 'O' is the origin then the maximum area of the triangle AOB is

Detailed Solution for JEE Main Practice Test- 14 - Question 18


JEE Main Practice Test- 14 - Question 19


Which one of the following statement is correct?

Detailed Solution for JEE Main Practice Test- 14 - Question 19

f (x) will be continuous where 3 sin x + a2 – 10a + 30 = 4cos x

or (a – 5)2 + 5 = 4cos x – 3sin x
∴ a = 5 and 4 cos x – 3 sin x = 5

or cos (x + θ) = 1, where tan θ = 3/4

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 20

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

If   then the value of must be


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*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 21

The sum of all the real roots of the equations
|x − 2|2 + |x − 2| − 2 = 0 is …..


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*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 22

If Pn = cosn x + sinn x, then 2.P6 - 3.P4 + 1 = ..


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*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 23

If x = 2 + t3, y = 3t2 and is a constant then the value of 343n must be


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*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 24

The value of 49A + 5B, where A = 1 - log72 and B = - log5 4 is


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JEE Main Practice Test- 14 - Question 25

A transverse wave is travelling along a horizontal string. The first picture shows the shape of the string at an instant of time. This picture is superimposed on a coordinate system to help you make any necessary measurements. The second picture is a graph of the vertical displacement of one point along the string as a function of time. How far does this wave travel along the string in one second?

Detailed Solution for JEE Main Practice Test- 14 - Question 25

From the graphs λ = 9cm
T = 3 sec

JEE Main Practice Test- 14 - Question 26

A cyclic process of an enclosed gas of constant mass is represented by volume (V) against absolute temperature (T) as shown. If P represents pressure, the graph representing the same process can be

Detailed Solution for JEE Main Practice Test- 14 - Question 26

Co mb ina ti on of is ob ori c, is oc hor ic & isothermal.

JEE Main Practice Test- 14 - Question 27

A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:

Detailed Solution for JEE Main Practice Test- 14 - Question 27

4L/5 = λ ⇒ λ = 8cm
hus 2 cm corresponds to Δϕ = z/2
1 cm corresponds to Δϕ = z/4

JEE Main Practice Test- 14 - Question 28

A sound source S and observers O1, O2 are placed as shown. S is always at rest and O1, O2 start moving with velocity v0 at t = 0. At any later instant, let f1 and f2 represent apparent frequencies of sound received by O1 and O2, respectively. The ratio f1/f2 is

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JEE Main Practice Test- 14 - Question 29

Equal masses of three liquids A, B and C have temperatures 10oC, 25oC and 40oC respectively. If A and B are mixed, the mixture has a temperature of 15oC. If B and C are mixed, the mixture has a temperature of 30oC,. If A and C are mixed the mixture will have a temperature of

Detailed Solution for JEE Main Practice Test- 14 - Question 29

msA (15 – 10) = msB (25 – 15) sA = 2sB
msB (30 – 25) = msC (40 – 30)
sB = 2sC ⇒ sA = 4sC
msA (T – 10) = msC (40 – T)
⇒ 4(T – 10) = 40 – T
T = 16°C

JEE Main Practice Test- 14 - Question 30

A steel rod is 4.000 cm in diameter at 30ºC. A brass ring has an interior diameter of 3.992 cm at 30ºC. In order that the ring just slides onto the steel rod, the common temperature of the two should be nearly (αsteel = 11 × 10-6/ºC and αbrass = 19 × 10-6/ºC)

Detailed Solution for JEE Main Practice Test- 14 - Question 30

For ring just slides on to the steel rod the diameter of rod and ring should be equal to each other and suppose due to Δθ increment in temperature the diameter of both are equal then
4 (1+  αs Δθ) = 3.992 (1 + αBrass Δθ) 4 + 4 × 11 × 10-6 × Δθ
= 3.992 + 3.992 × 20 × 10–6 × Δθ
4 + 44 × 10–6 Δθ = 3.992 + 79.84 × 10–6
× Δθ
0.008 = 35.84 × 10–6 Δθ


so if temperature increased by 223°C then ring will start to slide and this temperature will equal to
θ = 30° + Δθ = 30 + 253 = 283°C
θ = 283°C ≈ 280°C

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