VITEEE PCME Mock Test - 1 - JEE MCQ

f(0^-) = f(0) = f(0⁺)


p + 2 = q = 1/2
⇒ p = -(3/2), q = 1/2

The circle x² + y² − 2gx + 2gy + g² = 0 touches the co-ordinate axes.
Now, since it also touches the line xcos α + ysin α = 2,
the perpendicular distance from the center to the line must be equal to the radius.



Hence, option 2 is correct.

The equation of a line perpendicular to the line ax + by + c = 0 is −bx + ay + k = 0.
Hence, the equation of a line perpendicular to the line
3x + 4y + 5 = 0 is 
4x − 3y + k = 0  ...(i)
The line (i)i passes through the point (4, −5), hence the point satisfies the above equation.
Thus, we get
4 × 4 − 3 × (−5) + k = 0
⇒ 16 + 15 + k = 0
⇒ k = −31
Put the value of k in the equation (i) to get the required line as
4x − 3y − 31 = 0.

Given:

cos^-1 x = α, where x lies in the interval (1/√2, 1). This means α = cos^-1(x) lies in (0, π/4).

We are also given the equation:

sin^-1(2x√(1 - x²)) + sec^-1(1 / (2x² - 1)) = 2π/3

Step 1: Express sin^-1(2x√(1 - x²)) in terms of α

• Since x = cos α, then √(1 - x²) = sin α.
• Therefore, 2x√(1 - x²) = 2 cos α sin α = sin 2α.
• Since 2α ∈ (0, π/2), sin^-1(sin 2α) = 2α.

Step 2: Express sec^-1(1 / (2x² - 1)) in terms of α

• 2x² - 1 = 2 cos² α - 1 = cos 2α (using double-angle formula for cosine).
• Thus, 1 / (2x² - 1) = 1 / cos 2α = sec 2α.
• Since 2α ∈ (0, π/2), sec^-1(sec 2α) = 2α.

Step 3: Sum the two inverse functions

Given:

sin^-1(2x√(1 - x²)) + sec^-1(1 / (2x² - 1)) = 2π/3

From the previous steps:

2α + 2α = 4α = 2π/3

Therefore,

4α = 2π/3 ⇒ α = π/6

Step 4: Find tan^-1(2x)

• α = cos^-1(x) = π/6 ⇒ x = cos(π/6) = √3/2.
• Then, 2x = 2 * (√3/2) = √3.
• Hence, tan^-1(2x) = tan^-1(√3) = π/3.

Final answer: tan^-1(2x) = π/3

We have,

Now, we know that 

Therefore, term containing coefficient of x⁵ in  is given by .  
Hence, the coefficient is  ₃₁C⁶ - ₂₁C⁶

∵  A² = A⇒A³ = A⁴ = Aⁿ = A
Also,  (I + A)ⁿ

The shaded region contains elements of A but not in B or C.
Hence, the correct answer is: D) A − (B ∪ C)

We are given an expression:
(2n + r)·r, where n ∈ N and r ∈ N,
and told that it can be written as the sum of k consecutive odd natural numbers.
We are to find the value of k.

Step 1: Understand the sum of k consecutive odd numbers
Let the first odd number in the sum be a.
Then the next consecutive odd numbers will be:
a, a + 2, a + 4, ..., a + 2(k − 1)

This is an arithmetic progression with:

First term = a

Common difference = 2

Number of terms = k

So, the sum is:

S = (k/2) × [2a + 2(k − 1)] = (k/2) × [2(a + k − 1)] = k(a + k − 1)

So,
(2n + r)·r = k(a + k − 1) — (1)

We now want to find k such that this holds true for any n, r ∈ N.

Step 2: Try expressing (2n + r)·r in the form k(a + k − 1)
Let’s write: (2n + r)·r = kr + k(k − 1)
Try expanding RHS of equation (1):
k(a + k − 1) = ka + k(k − 1)

So to match it with LHS: Let’s assume k = r, then:

LHS = (2n + r)·r

RHS = r(a + r − 1)

So,
(2n + r)·r = r(a + r − 1)
Divide both sides by r:

2n + r = a + r − 1 ⇒ a = 2n + 1

Which is odd and natural ⇒ valid!
Hence, when k = r, the sum of r consecutive odd numbers starting from 2n + 1 gives (2n + r)·r

∵ L[R'(0)] ≠ R[f'(0)]
∴ Does not exist.

=

∴ Reading of voltmeter =
⇒ 2r + 8 = 9 ⇒ ​​​​​​​ r = (1/2)Ω 
​​​​​​​Therefore, r = 0.5Ω

As we have the wave number

For hydrogen, wave number is

For helium, wave number is

As area of a point is zero,
∴ ϕ = E (ds) cosθ = Ecosθ × 0 = Zero

In the circuit, the junction diode is in forward bias. In this state, the resistance offered by the diode is zero.

Thus, the total resistance in the circuit is 300 Ω.

Using Ohm’s Law:
I = ΔV / R

Given:
ΔV = (4V - 1V) = 3V, and R = 300 Ω

So,
I = 3V / 300Ω = 10⁻² A

Correct answer: B) 10⁻² A

If the area and temperature of the bodies are same then the emissive power will depend on the nature of surface. The surface which will absorb more will emit more. In all the given colour black will absorb the most so emissive power of C will be maximum.

A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as primary (P) is connected to A.C. supply. The other coil known as Secondary (S) is connected to the load.

• Np​: Number of turns in the primary coil.
• Ns​: Number of turns in the secondary coil.
• Vp​: Voltage across the primary coil (input voltage).
• Vs​: Voltage across the secondary coil (output voltage).

This ratio  N_p / N_s  is called the turn’s ratio.

In Huygens theory, the light is considered as wave. It means all the properties of wave, like interference, diffraction and polarisation can be explained by the Huygens wave theory, but Photoelectric effect shows the particle behaviour of light, so it can't be explained by the wave theory.
Note: Answer given in the book is incorrect. The correct answer is the Photoelectric effect.

= 4.095 × 10³ N/m²

Gross lift on the wing = (p₁ − p₂) × area

= 4.095 × 10³ × 10 × 2

= 81.9 × 10³ N

= 81.9 kN

The magnetic flux linked with the primary coil is given by:

So, voltage across primary coil,

= 4 volts (as φ₀ = constant)
Also, we have
N_P = 50 and N_s = 1500
From relation,

Or V_s = V_p
= 4
= 120 V

Both Para-Cresol and Oleic acid form salt with 10% NaOH, but Para-Cresol salt is soluble whereas Oleic acid salt is insoluble due to very long unsaturated carbon chain.

Let the initial de-Broglie wavelength of the electron (λ₁) = 100 Å.
By applying potential difference, it decreases by 1%.
Thus, λ₂ = 99 Å
Now, 
Thus, 



% Increase in KE 
Thus, the percentage increase in K.E.= 2%

IUPAC name of K₃[Ir(C₂O₄)₃].
The positive counterpart is potassium - will be named as potassium.
The complex is negative and the central atom is Ir(III) - will be named as iridate(III).
Ligand is C₂O₄^-2, which is an anion and the number is three - will be named trioxalato.
So, the name of the compound will be Potassium trioxalatoiridate(III).

(a) In transition metals on moving down the group higher oxidation states are more stable due to smaller size of atoms, due to lanthanide and actinide contractions.
(b) KMnO₄ can oxidise chloride into chlorine, so it will give incorrect results.
(c) Some lanthanoid oxides can be used as phosphors.

Ketones give an addition product having more number of carbon atoms with Grignard reagent, which on hydrolysis gives an alcohol (3°).

Formaldehyde gives primary alcohol with Grignard reagent while any other aldehyde except formaldehyde give secondary alcohol.

Alkyl benzenes are oxidised to benzoic acid using KMnO₄ in dilute H₂SO₄ and heating under reflux.

The observation for this reaction is the decolourisation of purple KMnO₄ and formation of white precipitate benzoic acid.

Maintenance area' is a place where all types of repairing work of memory is done or how much memory is allocated to a particular task is decided. Thus, the correct answer is option 4.

‘Which means “to” is the best choice for the question given.

Here, no punctuation should be appearing between means and “to..”. A non-restrictive relative clause requires the use of ‘which’ and not ‘that’ (as in 3 & 4). Option 2 suggesting a comma between the verb and the quotation is also considered incorrect.

Therefore, this is the correct and best option.