WBJEE Maths Test - 1 - JEE MCQ

To find the coefficient of x³ in (√x⁵ + 3/√x³)⁶, rewrite the expression as:

(x^5/2 + 3x^-3/2)⁶.

Using the binomial theorem, each term can be expressed as:

binom(6, k) (x^5/2)^{6 - k} (3x^-3/2)^k = binom(6, k) 3^k x^{15 - 4k}.

To find the coefficient of x³, set the exponent equal to 3:

• 15 - 4k = 3 gives k = 3.

Now, calculate the coefficient:

• binom(6, 3) × 3³ = 20 × 27 = 540.

We are given two expressions and need to find the relationship between a and b such that their coefficients match.

1. For the expression [ax² + 1/(bx)]¹¹, we use the binomial theorem:

• The general term is ➸(11, k) (ax²)^11-k (1/(bx))^k.
• Simplifying, each term's exponent of x is 2(11 - k) - k = 22 - 3k.
• Setting this equal to 7: 22 - 3k = 7 → k = 5.
• The coefficient is ➸(11, 5) a⁶ / b⁵.

2. For the expression [ax - 1/(b x²)]¹¹, similarly:

• The general term is ➸(11, m) (ax)^11-m (-1/(b x²))^m.
• Simplifying, each term's exponent of x is (11 - m) - 2m = 11 - 3m.
• Setting this equal to -7: 11 - 3m = -7 → m = 6.
• The coefficient is ➸(11, 6) a⁵ / b⁶ (since (-1)⁶ = 1).

3. Equating the coefficients:

➸(11, 5) a⁶ / b⁵ = ➸(11, 6) a⁵ / b⁶

• Since ➸(11, 5) = ➸(11, 6), we have:
• ➸(11, 5) a⁶ / b⁵ = ➸(11, 6) a⁵ / b⁶
• Simplifying: ab = 1.

Thus, the relationship is ab = 1.

To determine when the third term in the expansion of (x² - 1/x³)ⁿ is independent of x, we use the binomial theorem. The general term is given by:

T_k+1 = C(n, k) · (x²)^{n - k} · (-1/x³)^k.

For the third term (when k = 2):

T₃ = C(n, 2) · (x²)^{n - 2} · (-1/x³)².

Simplifying this, we have:

T₃ = (n(n - 1)/2) · x^{2n - 4} · x^-6 = (n(n - 1)/2) · x^{2n - 10}.

For T₃ to be independent of x, the exponent must be zero:

2n - 10 = 0 ⇒ n = 5.

The correct equation is:

x² + y² - x - 2y = 0.

The standard form of the circle's equation is:

(x - 2)² + (y - 2)² = r².

Substituting the point (4,5), we find:

r² = 13.

Expanding and converting to general form gives:

x² + y² - 4x - 4y - 5 = 0, which is option B.

The degree of a differential equation is the power to which the highest derivative is raised. In this case, the term d²y/dx²  is the highest derivative

• The power of this is 1.
• Therefore, the degree of the differential equation is 1.

Let u = x³. Then, y = u², so:

• dy/du = 2u = 2x³.

Thus, the differential of x⁶ with respect to x³ is:

• 2x³.

The given differential equation is:

2x ^dy/_dx - y = 3.

To solve this, we first rewrite it in standard linear form:

^dy/_dx - 1/2x y = 3/2x.

The integrating factor (IF) is calculated as:

IF = e^{∫ -1/(2x) dx} = e^{-1/2 ln x} = 1/√x.

Multiplying both sides by the IF gives:

1/√x ^dy/_dx - 1/2x · 1/√x y = 3/2x · 1/√x.

Simplifying, we get:

^d/_dx ( y · 1/√x ) = 3/2x^3/2.

Integrating both sides with respect to x gives:

y · 1/√x = -3 · 1/√x + C.

Multiplying through by √x gives:

y = -3 + C√x.

Rewriting this as:

y + 3 = C√x.

Squaring both sides results in:

(y + 3)² = C² x.

This is the equation of a parabola. Therefore, the solution represents a parabola.

Given that |z| = 4 and arg z = (5π/6), we can express the complex number z in its polar form:

z = r (cos θ + i sin θ)

• r = |z| = 4
• θ = arg z = (5π/6)

Calculating the trigonometric functions:

• cos((5π/6)) = -√3/2
• sin((5π/6)) = 1/2

Substituting these values into the polar form:

z = 4(-√3/2 + i(1/2))

z = 4(-√3/2) + 4(1/2)i

z = -2√3 + 2i

Thus, the correct answer is option C.

Given x = a cos³θ and y = a sin³θ, we find dy/dx by first differentiating both equations with respect to θ.

• dx/dθ = -3a cos²θ sinθ
• dy/dθ = 3a sin²θ cosθ

Thus, we can express dy/dx as follows:

dy/dx = (dy/dθ) / (dx/dθ) = [3a sin²θ cosθ] / [-3a cos²θ sinθ] = -tanθ.

Next, we compute √[1 + (dy/dx)²]:

• (dy/dx)² = tan²θ
• So, 1 + tan²θ = sec²θ

Taking the square root gives:

√(sec²θ) = |secθ|.

Since arc length is non-negative, the absolute value ensures positivity. Therefore, the correct answer is D, |secθ|.

 9x² + 25y² = 225   [ send this 225 on left side in divide ]
9/225 x² +25/225 y² = 1
 x² +  y² = 1
25      9
after sending 225 in divide in lhs

e=√25-9
       5
e=4/5

The given parametric equations are:

• x = sin² t
• y = 2cos t

To identify the type of graph, we can eliminate the parameter t. Using the trigonometric identity:

sin² t + cos² t = 1,

we can express cos t from the second equation:

cos t = y/2.

Substituting this into the first equation gives:

x = 1 - (y/2)².

Rearranging this leads to:

y² = 4(1 - x).

This is the standard form of a parabola that opens to the left. Therefore, the graph represents a full parabola, not just a portion.

We start with the given parametric equations:

• x = a (coshθ + sinhθ)
• y = b (coshθ - sinhθ)

Using the identities for hyperbolic functions, we know:

• coshθ + sinhθ = e^θ
• coshθ - sinhθ = e^-θ

Substituting these into the equations gives:

• x = a e^θ implies e^θ = x/a
• y = b e^-θ implies e^-θ = y/b

Since e^-θ = 1/e^θ, we substitute:

• y/b = a/x implies xy = ab.

This is the equation of a hyperbola. Therefore, the correct answer is A.

We have , f x = x 2 + k x + 1
⇒ f ′ x = 2 x + k . Also , f ″ x = 2
Now , f ″ x = 2, ∀ x ∈ [ 1,2 ]
⇒ f ″ x > 0, ∀ x ∈ [ 1,2 ]
⇒ f ′ x is an increasing function in the interval [ 1,2 ] .
⇒ f ′ 1 is the least value of f′(x) on [ 1,2 ]
But f ′ x > 0 ∀ x ∈ [ 1,2 ]
[ ∵ f x is increasing on [1,2] ]
∴ f ′ 1 > 0, ∀ x ∈ [ 1,2 ] ⇒ k > − 2
Thus, the least value of k is -2

To find the multiplicative inverse of z², we start by calculating z². Given z = 1 + i:

We can calculate z² as follows:

• z² = (1 + i)² = 1² + 2(1)(i) + i²
• = 1 + 2i - 1 = 2i.

Next, we find the multiplicative inverse of 2i. The multiplicative inverse of a complex number w is given by:

Inverse formula: 1/w = w̅/|w|2,

where w̅ is the conjugate of w and |w| is its modulus.

For w = 2i:

• The conjugate w̅ = -2i.
• The modulus squared |w|² = (0)² + (2)² = 4.

Thus, the inverse of 2i is:

1/(2i) = -2i/4 = -1/2i.

Therefore, the multiplicative inverse of z² is -i¹/2, which corresponds to option C.

Given that a, b, c, and d are positive real numbers with the condition abcd = 1, we aim to minimise (1 + a)(1 + b)(1 + c)(1 + d). By applying the AM-GM inequality, we find that when all variables are equal (i.e., a = b = c = d = 1), the product is minimised.

Substituting these values yields:

• (2)⁴ = 16

This confirms that the minimum value is 16.

To find the equation of the parabola with vertex at (2, -1) and focus at (2, -3):

1. The vertex form of a parabola that opens upward or downward is (x - h)² = 4p(y - k), where (h, k) is the vertex and p is the directed distance from the vertex to the focus.
2. Here, the vertex is at (2, -1), so:
   • h = 2
   • k = -1
   The focus is at (2, -3), which is 2 units below the vertex, so p = -2.
3. Substituting these values into the equation: (x - 2)² = 4(-2)(y + 1). Simplifying gives: (x - 2)² = -8(y + 1).
4. Expanding and rearranging terms to convert it into the general form: x² - 4x + 4 = -8y - 8. Bringing all terms to one side results in: x² - 4x + 8y + 12 = 0.

Thus, the correct equation is option B.

To solve the expression sin[(1/2)cos^-1(4/5)], we can use trigonometric identities.

Let θ = cos^-1(4/5). This implies that:

• cosθ = 4/5
• θ is in the range [0, π]

We need to find sin(θ/2). Using the half-angle identity for sine:

sin(θ/2) = √[(1 - cosθ)/2]

Substituting cosθ = 4/5 into the formula gives:

sin(θ/2) = √[(1 - 4/5)/2] = √[(1/5)/2] = √(1/10) = 1/√10

Since θ/2 is in [0, π/2], sin(θ/2) is positive. Therefore:

sin[(1/2)cos^-1(4/5)] = 1/√10

The correct answer is option B: 1/√10.

Step 1: Recall the Polar-Pole Relationship

For the parabola y² = 4ax, the polar of a point (x₁, y₁) is given by:
y * y₁ = 2a(x + x₁)

In this case, the given parabola is y² = 4x, which means 4a = 4, so a = 1.
Substitute into the polar equation:
y * y₁ = 2(x + x₁)
or rearranged:
2x - y₁ * y + 2x₁ = 0

Step 2: Compare with the Given Line

The given line is:
2x + 3y - 4 = 0

Now match this with the polar equation:
2x - y₁ * y + 2x₁ = 0

To represent the same line, their coefficients must be proportional. That gives us the following proportion:
2 / 2 = -y₁ / 3 = 2x₁ / -4

Simplify the left side:
1 = -y₁ / 3 = 2x₁ / -4

Step 3: Solve for x₁ and y₁

From 1 = -y₁ / 3, we get:
-y₁ = 3, so y₁ = -3

From 1 = 2x₁ / -4, we get:
2x₁ = -4, so x₁ = -2

Step 4: Identify the Pole

So the coordinates of the pole are (-2, -3)

Final Answer:

The pole of the line 2x + 3y - 4 = 0 with respect to the parabola y² = 4x is (-2, -3).

Correct option: (b)

Given word has 9 letters out which I, U, A and E are wovels.

So if we club them together, we are left with 5 letters and 1 group of letters.

N, S, R, N, C and IUAE.

In this, N comes twice.

So number of arrangements will be,

6! / 2! = 360.

Also, the group of wovels has 4 distinct letters.

So, it can be arranged in,

4! = 24 ways.

Therefore, total number of arrangements,

= 360 * 24

= 8640.