VITEEE: Subject Wise and Full Length VITEEE PCME Mock Test - 9 Free Online

Let O be one of the vertex of a cube at the origin and three edges through point O be the co-ordinate axes.
The four diagonals are OP, AA', BB' and CC'. Let 'a' be the length of
each edge. Then, the co-ordinates of P, A and A' are (a, a, a), (a, 0, 0), (0, a, a).
The direction ratios of OP are a, a, a.
The direction cosines of OP are
i.e.
Similarly, direction cosines of AA' are

Let θ be the angle between the diagonals OP and AA'.
Then,
= =

or,

Let f(x) = bx² + ax + c
Since, f(0) = 0 ⇒ c = 0
And f(1) = 0  ⇒ a + b = 1
∴ f(x) = ax + (1 − a)x²
Also, f′(x) > 0 for x ∈ (0, 1)
⇒ a + 2(1 − a)x > 0 ⇒  a(1 − 2x) + 2x > 0
⇒ 0 < a < 2
Since, x ∈ (0, 1)
∴ f(x) = ax + (1 − a)x²; 0 < a < 2
For a = 1, f(x) = x (not second degree polynomial)

A linear programming deals with the optimization (minimization or maximization) of a linear function (objective function) of a number of variables (decision variables) subject to a number of conditions on the variables, in the form of linear equations or inequations in variables involved.

Since, ω is a complex cube root of unity
Now, 
= ω + ω² = −1

The distance from the point P(3,1) to the required line will be maximum if the line drawn from (3,1) to the required line is perpendicular at point Q(1,2).

Since the required line passes through Q(1,2), we have the slope of line PQ = (1−2) / (3−1) = -1/2.

⇒ Slope of the required line is -1 / (-1/2) = 2.

Using the point-slope form of a straight line:

(y−2) = 2(x−1)
⇒ y - 2 = 2x - 2
⇒ 2x - y = 0
⇒ y = 2x.

Given,

The given function is undefined when its denominator vanishes.

⇒ (x² − 1)(x + 3) = 0

⇒ x = ±1, −3

Hence, the function is discontinuous at those points.

Let I = 

(∵|sinx| is periodic with period π and  . If T is the period of the function f(x).

Given, (2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0


On comparing the corresponding components, we have
6μ – 27λ = 0, –2μ + 27 = 0, 2λ – 6 = 0

Hence, λ = 3 and μ = 27/2

Family of circles passing through the points of intersection of the given curves x²/4 + y² = 1 and x²/a² + y² = 1 is given by:

If the above equation represents the required circle, then coefficient of x² = coefficient of y²

On substituting , we get

So, the required circle is x² + y² = 1

Given that, 
Curve 
Differentiate both sides w.r.t. x, we get,

Slope of tangent at point (α, β):

Equation of tangent is:

Compare the equation with given equation

⇒ α = a and β = b.

Given the curve y = k sin x,

The required area is = 

Hence, the area is 2|k| square units.

Given that α, β are the roots of the equation:

(x − a)(x − b) + c = 0

Expanding:
x² − (a + b)x + ab + c = 0

From this, we identify:
α + β = (a + b)
αβ = ab + c

Now, consider the given equation:
(x − α)(x − β) − c = 0

Expanding:
x² − (α + β)x + αβ − c = 0

Substituting α + β = a + b and αβ = ab + c:
x² − (a + b)x + ab = 0

Thus, the roots of this equation are a, b.

We evaluate the radioactive emission magnitude by the amount of atoms per unit of time converted into new atoms. The standard unit employed for evaluating radioactivity is Curie. 
One Curie is equivalent to 3.7×10¹⁰ radioactive decays per second, approximately the number of decays occurring in 1 gm of radium per second, and is equal to 3.7×10¹⁰ Bq becquerels. 

Germanium is a tetravalent atom. When the crystal of Germanium is doped with Gallium, which is a trivalent atom, a hole is created due to lack of one valence electron in gallium. Thus, impurity of lower group creates a positively charged hole. The type of extrinsic semiconductor obtained is p-type semiconductor. On applying potential difference, p-holes conduct electric current.

In β-decay, mass number is unaffected. Atomic number increase by one.

Given,
Current in the breakdown region, Iₓ = 20 mA
Applied voltage, Vₓ = 80 V

So, the Zener impedance, Zₙ = Vₓ / Iₓ
Zₙ = (80 × 10⁻³) / (20 × 10⁻³)
Zₙ = 4 Ω

Initial kinetic energy of the body:


Let, at height h, the kinetic energy reduces to half, i.e., it becomes 4m. This is also equal to the potential energy. Hence,
mgh = 4 m
Or,

(voltage lead, so Circuit is RL)
X_L = R
ωL = R
100 L = R

For Daniell cell,
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Applying the Nernst equation, we get:

E_cell = 1.10 - 0.0295 = 1.0705 V

Acidic strength is the tendency to give H⁺ ions. The correct order of acidic strength of given acids is
CH₃COOH<C₆H₅COOH<HCOOH
This order depends on pKa value of these acids and stability of conjugate base formed by these acids.
Acidic character is directly proportional to 
−Inductive effect and −esonance effect.

According to the Aufbau principle electrons are filled in the increasing order of their energies in the orbitals. Atomic orbitals which are lower in energies are filled before the higher energy orbitals.
If this rule is not followed then the first f-block atom will have the atomic number 47.

Acidic character of oxides depends upon the electronegativity and oxidation state of the central atom.
Higher the electronegativity and oxidation state of the central atom, more is the tendency to attract electrons (using Lewis concept). Hence, more is the acidic character of the oxides.
The oxidation state of the centric atom in all the given oxides is +5 and nitrogen is the most electronegative among all. Hence, N₂O₅ is the most acidic.

NO₂  is a very strong electron-withdrawing group (EWG), thereby deactivating the ring strongly. The amount of electron donation into the ring (activating the ring towards electrophilic aromatic substitution) therefore decreases.

Mass of electron (m) = 9.1×10⁻³¹ Kg

Uncertainty in momentum (△p)=1.0×10⁻¹⁰

According to Heisenberg's principle,

=0.527 × 10⁻²⁴
△x = 5.27 × 10⁻²⁵ 

The balanced equation:

Where 'x'  is the degree of dissociation.

This indicates that the degree of dissociation is very less and thus: [I₂(g)] > [I(g)]

Given that the profit of shopkeeper T in December 2009 is 5.31 and in January 2010 is 5.69.
Therefore, the difference is:
⇒ (5.69−5.31) × 1000
⇒ 0.38 × 1000
⇒ ₹380
Hence, the correct answer is ₹380