VITEEE PCME Mock Test - 4 - JEE MCQ

Sum of the dice is 7.
S = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
∴ n(S) = 6
Let, E = Event of getting at least three 3 or a die
∴ n(E) = 2
∴ Required probability 

Given

Now, we know that

Therefore,

We know that any function f(x) will be continuous at 
Case I: x = a, where a is rational.
So, 
and

Hence,  function is not continuous for this case.
Case II: x = a, where a is an irrational number.
So, 
and

Hence, function is not continuous for this case.
Hence, the function is discontinuous for all x∈ R.

Given statement:
P(n) : 2ⁿ < n!

The statement is not true for n = 1, 2, 3.

Now, checking for n = 4:
2⁴ < 4!
16 < 24, which is true.

Let P(k) : 2ᵏ < k! be true for some k ≥ 4.

Now, checking for P(k + 1):
2ᵏ⁺¹ < (k + 1)!

From assumption:
2ᵏ < k!

Multiplying both sides by 2:
2ᵏ⁺¹ < 2 × k!

Since (k + 1) > 2, we have:
2 × k! < (k + 1) × k!
which simplifies to:
2ᵏ⁺¹ < (k + 1)!

Thus, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 4.

Correct answer: B) n ≥ 4.

Let the function be f(x) = x² + bx - b.

We are given that the equation of the tangent to the curve at the point P(1, 1) is:

2y = 2x² + 2bx - 2b.

Simplifying this, we get:

y + 1 = 2x ⋅ 1 + b(x + 1) - 2b, which simplifies to y = (2 + b)x - (1 + b).

The tangent meets the coordinate axes at:

• For the x-axis: x_A = (1 + b)/(2 + b)
• For the y-axis: y_B = -(1 + b)

Now, the area of the triangle formed by the tangent and the coordinate axes is given by:

Area of ΔOAB = 1/2 * OA * OB = 2 (given).

Substituting the values of OA and OB:

(1 + b)² + 4(2 + b) = 0
Expanding the equation:

b² + 6b + 9 = 0
Solving this quadratic equation:

(b + 3)² = 0
Hence, b = -3.

Therefore, the value of the y-intercept is:

y = -(1 + b) = -(1 - 3) = 2.

So, the correct answer is C) 2

We have  ...........(i)
On differentiating Eq. (i) w.r.t. x, we get

Put, x = √π/2  in Eq. (ii), we get

Centre≡(0, 3) ≡ (0, 3)
Focus S ≡ (√7, 0)
Radius = BS = 4

∴∴ Its equation is x² + (y−3)² = 4².

x² + y² − 6y − 7 = 0

For an ellipse 
We have a² = 7, b² = 16

Focus: 
Now, for the hyperbola 

Also Focus: 
Given, S and S′ coincide i.e.

Magnetic moment = 10⁴ J/T
B = 4 x 10^-5 T
Work done in moving the magnet in uniform magnetic field,
W = MB(cosθ₂ - cosθ₁)
= 10⁴ x 4 x 10^-5 (cos 0° - cos 60°) = 0.2 J

Induced emf E = IR
Or I = E/R
And
^∴ 
Induced charge, Q = I.dt =
If speed is doubled, then 'dt' decreases. Hence, E and I increase, but Q remains same as it is independent of time and depends on the change in the magnetic flux.

The holes move in the direction of applied electric field and the current density due to them is given as μ_hn_ieE.
The electrons move in direction opposite to that of electric field, and current due to that is in direction opposite to that of movement of electrons, that is, in direction of electric field.
Current density due to movement of electrons is μ_en_ieE.
Hence total current density =

The de-Broglie wavelength is given by
…(i)
where, E is the energy of the electron. The cut-off wavelength λ₀ is given by
…(ii)
From equation (i),
…(iii)
Substituting the value of E from equation (ii), we get

when the balls are warmed, their centre of masse are moving as radii of balls are increasing. The centre of mass of ball A will come down and that of ball B will go up. In case of ball A, the gravitational potential energy decreases. This corresponds to additional heating of ball.

∴ T_A > T_B

In transfer characteristic, the region I is called cutoff region, region II is active region and region III is saturated region.
The transistor work as a switch in cutoff and saturated region, so it will act as a switch in both region I and III.
Transistor work as an amplifier in active region i.e., in region II.

e₀ = NABω. When B and ω are doubled then e₀ becomes four times.

The expression of current in above circuit,

Here in above graph the voltage lags the current. In a capacitor circuit  the voltage lags behind the current by π/2.
Hence circuit contains only capacitor.

At 0°C,
y = A sin(1000t - 3x)

^⇒ T = 277.4 K i.e. K = 4.4°C

Choke (which is an inductor) is preferred over a resistor for limiting current in an AC circuit because it does not dissipate power in the form of heat. Instead, the choke limits the current by creating reactance, and the power loss due to reactance is much lower compared to the power loss in a resistor, which dissipates energy as heat. This makes the choke more efficient in AC circuits, especially in applications like fluorescent lamps or other inductive loads, where energy efficiency is important.

Why not the other options?
A: Choke is cheap: While this may sometimes be true, the main reason for using a choke is its efficiency, not cost.
C: Choke is compact in size: The size of a choke is not the primary reason for preferring it over a resistor.
D: Choke is a good absorber of heat: A choke does not absorb heat, and in fact, power loss in resistors results in heat generation, which is avoided by using a choke.

Therefore, the best answer is B: there is no wastage of power.

From equation (i):

Copper pyrites and Galena are sulphide ores and are concentrated by froth-floatation process.
Cassiterite is concentrated by a magnetic separator.

Given that ρ =2 .75 g cm⁻³
we need to find type of unit cell

= 
∴ Z ≃ 4 So fcc unit cell

Z = 4: FCC (NaCl-type, coordination number 6)

Geometrical isomerism arises due to the restricted rotation in the molecule. In this molecule, a double bond causes restricted rotation and the geometrical isomers are formed. In the given compound, there are three double bonds and each carbon atom is substituted differently. So the number of geometrical isomers will be 2ⁿ = 2³ = 8, where n is the number of double bonds whose each carbon atom is differently substituted.

The only connectable sentence is ‘the fête was held for the first time on a Saturday, usually it is held on a Friday evening.’ Here ‘usually’ has been used as an adverb meaning ‘Under normal conditions; generally’

In a correct clock, the minute hand gains 55 min spaces over the hour hand in 60 min.
To come together again, the minute hand should have gained 60 min spaces over the hour hand.
55 min spaces are gained in 60 min.
60 min spaces will be gained in (60/55) × 60 = 60 (12/11) min
But they come together after 66 min.
Loss in 66 min = 66 -
Loss in one hour = min = 60/121 min

The expression for distance-time relation is:
Distance = Speed × time
In one hour, A and B walk 0.5 km and 0.25 km respectively.
The distance travelled by A in 4 hours = 0.5 × 4 = 2km
The distance travelled by B in 4 hours =0.25 × 4 = 1 km
The expression from Pythagoras theorem is

Since, A walks in the north direction and B walks in the west direction, their path of travel will be at an angle of 90 degrees.
Therefore, from the Pythagoras theorem, the distance (d) between the two will be:

= √5 km
Hence, the first option is the correct one.