SRMJEE Mock Test - 9 (Engineering) - JEE MCQ

From the mirror formula, we have

^{1/f = 1/v + 1/u} ...(i)
where v is the image distance and u is the object distance.
Differentiating Eq. (i), we get
0 = - 1/v₂ dv - 1/u²du
⇒ dv = - v²/u² du ...(ii)
Given, du = b
∴ dv = - v²/u²  × b
From Eq. (i), we have

⇒
∴ ...(iii)
From Eqs. (ii) and (iii), we get
dv = -b
Hence, size of the image is .

Motion is uniformly accelerated.

Given:
Initial velocity, u = 0
Acceleration, a = a
Velocity at time t = n is v = v

Therefore, the distance traveled in n seconds is given by:

s = ut + (1/2) a t²
Since u = 0, we get:
s = (1/2) a n² ...(i)

Let the distance traveled in (n − 2) seconds be s₁.

Using the equation s = ut + (1/2) a t², we get:
s₁ = 0 + (1/2) a (n − 2)²
Expanding,
s₁ = (1/2) a (n² − 4n + 4) ...(ii)

Now, the distance traveled in the last two seconds is S:

S = s − s₁
Substituting values from (i) and (ii):
S = (1/2) a n² − (1/2) a (n² − 4n + 4)
Simplifying,
S = (1/2) a (4n − 4)
S = 2a (n − 1) ...(iii)

Since the options are given in terms of v, we need to eliminate a from equation (iii).

Using the equation v = u + at, we get:
v = 0 + an
Thus, a = v/n

Substituting this in equation (iii):
S = 2v (n − 1) / n

At critical temperatures, cohesive forces become approximately zero. That's why the surface tension of a liquid at critical temperatures becomes zero.

According to Kepler's law

T² ∝ r³

∴ T₂ = 40 hr

sForinusoidal alternating current or triangualr and square wave alternating current the average value for a period is always zero.

But in case of following currnet peak value will not be equal to rms value.

If AC is the square wave then all these three options are possible. i.e.,

I_rms = I_av = I₀, here I₀ is peak value

Benzaldehyde condenses with N, N-dimethyl aniline in presence of anhydrous ZnCI₂ to give malachite green.

Deoxyribose is a carbohydrate. Hence, the assertion is wrong.
Deoxyribose does not have the formula C_x(H₂O)_y, but is very much a carbohydrate. Hence, the definition of carbohydrates given in the reason is not valid.
The correct definition is that carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones.

Number of moles =
n = 0.00165 (deposited)
n_i = = 0.0375 (initial mole)
n_left = 0.03585 (mole left)
M_left = = 0.143

Neoprene is synthetic rubber. It is a polymer of chloroprene and is resistant to the action of oils, gasoline and other solvents.

When there is a change in the optical rotation due to the change in the equilibrium between two anomers, then the corresponding stereocenters interconvert, and it is called as mutarotation.
When either of glucose's two forms is dissolved in the water, there is a change in the rotation till the equilibrium value of +52.5°.
 

Lead Nitrate Is a mild oxidising agent.
The oxidation of benzyl chloride with lead nitrate gives  benzaldehyde.

Sum of roots = SOR

When m = 0,

α + β = 3, αβ = 1

The system of equations will have a non-trivial solution if and only if Δ = 0 where
.
Thus, Δ = 0, λ ∈ R ⇒ λ = 0

Given: 

Roots: x = 0, 1, -4
Poles: x = -1, 3

Using the wavy curve method:

The sign will not change at x = 1 because the factor (x - 1) has an even power (i.e., 2).
We include the roots and reject the poles in the positive region.
Thus, the solution set is:
x ∈ (-∞, -4] ∪ (-1, 0] ∪ (3, ∞) ∪ {1}

Given equation is: 

Comparing (i) and (ii)

Two triplets:

Let H be the orthocenter.

AH = x₁, BH = x₂, and CH = x₃.

In ΔABE,

∠ABE + ∠BAE + ∠AEB = 180°
⇒ ∠ABE + A + 90° = 180°
⇒ ∠ABE = 90° − A

Similarly, in ΔABD,

∠BAD = 90° − B

Let R be the radius of the circle. Referring to ΔAHB,

Using the sine rule in the triangle, we have:

AH = 2RcosA

Therefore,

Using sine rule in triangle

From equations (i) and (ii),

Similarly, 

In any triangle,

Let the first three terms of the A.P. be a − d, a, a + d.

Given:
Sum of the first three terms:
(a - d) + a + (a + d) = 3a = 33
a = 11

Product of the first three terms:
(a - d) * a * (a + d) = a * (a² - d²) = 1155
11 * (11² - d²) = 1155
11 * (121 - d²) = 1155
121 - d² = 1155 ÷ 11 = 105
d² = 121 - 105 = 16
d = ±4

Case 1: a = 11, d = 4
A.P. = 7, 11, 15, …
The nth term of an A.P. is given by:
Tn = A + (n - 1) * d
For n = 11:
T₁₁ = 7 + 10 * 4 = 47

Case 2: a = 11, d = -4
A.P. = 15, 11, 7, …
For n = 11:
T₁₁ = 15 + 10 * (-4) = -25

Thus, one possible value of the 11th term is −25.

Given, x − |y| − α = 0

 Radius of circle = 2 

Center of the circle ≡ (5, 0)

As given in the question, line does not meet the circle. Therefore, perpendicular distance from the center to the line is greater than the radius

We have,

f′′(x) > 0 ∀ x ∈ R
⇒ f′(x) is an increasing function.

Now,
g(x) = 2f(2x³ − 3x²) + f(6x² − 4x³ − 3)

Differentiating,
g′(x) = (6x² − 6x) ⋅ 2f′(2x³ − 3x²) + (12x − 12x²) ⋅ f′(6x² − 4x³ − 3)

Rewriting,
g′(x) = 12x(x − 1) [ f′(2x³ − 3x²) − f′(6x² − 4x³ − 3) ]

For g(x) to be increasing, g′(x) ≥ 0, i.e.,
12x(x − 1) [ f′(2x³ − 3x²) − f′(6x² − 4x³ − 3) ] ≥ 0

Case I:
When x(x − 1) ≥ 0 ⇒ x ∈ (−∞, 0) ∪ (1, ∞) ...(1)

We must have:
f′(2x³ − 3x²) ≥ f′(6x² − 4x³ − 3)
Since f′(x) is increasing, this means:
2x³ − 3x² > 6x² − 4x³ − 3

Simplifying,
6x³ − 9x² + 3 > 0
2x³ − 3x² + 1 > 0
(x − 1)²(2x + 1) > 0

Since (x − 1)² is always non-negative except at x = 1,
we get 2x + 1 > 0 ⇒ x > −1/2

Thus, x ∈ (−1/2, ∞) − {1} ...(2)

From (1) and (2), we conclude:
x ∈ (−1/2, 0) ∪ (1, ∞)

Case II:
When x(x − 1) ≤ 0 ⇒ x ∈ [0,1] ...(3)

We must have:
f′(2x³ − 3x²) − f′(6x² − 4x³ − 3) ≤ 0

Again, simplifying,
(x − 1)²(2x + 1) < 0

Since (x − 1)² is always non-negative except at x = 1,
we get 2x + 1 < 0 ⇒ x < −1/2

Thus, x ∈ (−∞, −1/2) ...(4)

From (3) and (4), we get x ∈ ϕ (empty set).

Conclusion:
g(x) is increasing for x ∈ (−1/2, 0) ∪ (1, ∞).

The centre and radius of the circle x² + y² + 2gx + 2fy + c = 0 are respectively (-g, -f) and √(g² + f² - c).

Let, for the circle S₁ : x² + y² = 4, centre C₁ (0, 0), r₁ = 2 units.

And, for the circle S₂ : x² + y² + 6x + 8y - 24 = 0, centre C₂ (-3, -4), r₂ = √(3² + 4² + 24) = 7 units.

The distance between the points (x₁, y₁) and (x₂, y₂) is √((x₁ - x₂)² + (y₁ - y₂)²).

Hence, the distance between the centres of the circles is:
C₁C₂ = √((-3 - 0)² + (-4 - 0)²) = 5 units.

Also, |r₁ - r₂| = 5 units.

Since, C₁C₂ = |r₁ - r₂|, hence both the circles touch each other internally.

Therefore, the common tangent is the same as the radical axis of the two circles, i.e., S₁ - S₂ = 0.

Thus, the common tangent is:
x² + y² - 4 - (x² + y² + 6x + 8y - 24) = 0

⇒ -6x - 8y + 20 = 0

⇒ 3x + 4y - 10 = 0

Clearly, the point (6, -2) lies on the common tangent.

We have,

f(−x) = f(x); [∵ f is an even function]
g(−x) = −g(x); [∵ g is an odd function]

Now,
(fog)(−x) = f[g(−x)]
= f[−g(x)]
= f[g(x)]
= (fog)(x) ∀x ∈ R.

∴ (fog) is an even function.

Area of square ABCD = 64 cm²
CD × CD = 64
CD = 8 cm
InΔDCE,
Area of ΔDCE = 1/2× DC × CE = 1/2 × 8 × CE = 4 CE ... (i)
Also,
tan ∠EDC = tan 45 =
Or CE = 8 × tan 45^o
CE = 8 (As tan 45^o = 1)
From (i),
Area of DCE = 4 × 8 = 32 sq. cm

 If 10 sin θ^{_{= 6,}}
sinθ = 6/10 = 3/5

sinθ =
Base = = 4k
tanθ = 3/4
cotθ = 4/3
tanθ+ cotθ = 3/4 + 4/3 =