SRMJEEE Maths Mock Test - 4 - JEE MCQ

To find the focus of the parabola given by the equation x² − 2x − 8y − 23 = 0, follow these steps:

• Rearrange the equation to match the standard form of a parabola. Start by moving the y-term to the other side:
• x² − 2x = 8y + 23
• Complete the square for the x-terms:
• Take half of the coefficient of x (which is −2), square it, and add it inside the equation:
• x² − 2x + 1 = 8y + 23 + 1
• This simplifies to (x − 1)² = 8y + 24
• Rewrite the equation in the form (x − h)² = 4p(y − k):
• (x − 1)² = 8(y + 3)
• Identify the vertex (h, k):
• The vertex is (1, −3).
• Calculate the value of p where 4p = 8:
• Therefore, p = 2.
• Since the parabola opens upwards, the focus is p units above the vertex:
• The focus is (1, −3 + 2) which simplifies to (1, −1).

To determine where the function f(x) = cot^-1 x + x increases, we need to analyse its derivative. The derivative of cot^-1 x is -1/(1 + x²).

• The derivative of f(x) is: f'(x) = -1/(1 + x²) + 1.
• Simplifying: f'(x) = (x²)/(1 + x²).

Since f'(x) is always non-negative (i.e., f'(x) ≥ 0) for all real numbers:

• The function f(x) increases over the entire real line, (-∞, ∞).

To solve the equation 25cos²θ + 5cosθ - 12 = 0:

• Let x = cosα. The equation becomes 25x² + 5x - 12 = 0.
• Use the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a, where a = 25, b = 5, and c = -12.
• Calculate the discriminant: b² - 4ac = 5² - 4(25)(-12) = 625.
• The roots are x = [−5 ± √625] / 50. This gives x = 3/5 or x = -4/5.

Since (π/2) < α < π, cosα must be negative, so cosα = -4/5.

Calculate sinα using the identity sin²α + cos²α = 1:

• sin²α = 1 - cos²α
• sin²α = 1 - (-4/5)² = 1 - 16/25 = 9/25
• Since π/2 < α < π, sinα is positive, so sinα = 3/5.

Calculate sin2α using the identity sin2α = 2sinαcosα:

• sin2α = 2(3/5)(-4/5)
• sin2α = -24/25

Thus, sin2α is -(24/25).

The given equation of the parabola is (y - 2)² = 20(x + 3).

This is in the form (y - k)² = 4p(x - h), where:

• (h, k) is the vertex of the parabola.
• 4p is the coefficient of x.

From the equation:

• h = -3 and k = 2.
• 4p = 20 gives p = 5.

The parabola opens to the right since p is positive. Therefore, the focus is (h + p, k).

• Calculate the focus: (-3 + 5, 2) = (2, 2).

Thus, the focus of the parabola is (2, 2).

Order = 2, Degree = 4

ANSWER :- c

Solution :- y= sin⁶(x) + cos⁶(x).

y =(sin²x)³ + (cos²x)³

y=(sin²x+cos²x)(sin⁴x+cos⁴x-sin²x.cos²x)

y=(1).[ (sin²x+cos²x)²–3.sin²x.cos²x]

y= 1 - 3 sin²x.cos²x

y = 1 -3(sin x.cos x)²

y= 1- 3(1/2.sin2x)²

y = 1 - (3/4).(sin 2x)²

For minimum value of y , sin2x should be maximum , maximum value of sin 2x =1 or

x=45°.

Minimum value of y= 1 -(3/4).(1)²

= 1 - 3/4 = 1/4

A skew-symmetric matrix is one where the transpose of the matrix is equal to the negative of the matrix itself, i.e., \( A^T = -A \).

For any skew-symmetric matrix of odd order (meaning the number of rows and columns is odd), the determinant is always zero.

This is because:

• The determinant of a matrix is equal to the determinant of its transpose.
• If \( A^T = -A \), then \(\text{det}(A) = \text{det}(-A)\).
• For an odd-ordered matrix, \(\text{det}(-A) = -\text{det}(A)\).

Thus, for odd-ordered skew-symmetric matrices, the only possible solution is \(\text{det}(A) = 0\).

The given differential equation is:

cos(x) cos(y) (dy/dx) = -sin(x) sin(y)

To solve this, separate the variables:

• Rearrange terms to get (dy/dx) = - (sin(x) sin(y)) / (cos(x) cos(y)).
• Simplify to (dy/dx) = - tan(x) tan(y).

Separate the variables:

• dy/tan(y) = - tan(x) dx

Integrate both sides:

• Left side: ∫ csc(y) dy.
• Right side: -∫ tan(x) dx.

This leads to:

• log|sin(y)| = -log|cos(x)| + C

Exponentiate to remove logs:

• sin(y) = C cos(x)

Thus, the solution is:

sin(y) = C cos(x)

Solution:

To find the acute angle between the lines joining the origin to the intersection points of the given curve and line, follow these steps:

• The curve is a circle given by the equation: x² + y² - 2x - 1 = 0.
• Rewriting the equation in standard form, complete the square:
  • (x - 1)² + y² = 2, which represents a circle with centre (1, 0) and radius √2.
• The line is x + y = 1.
• Substitute y = 1 - x into the circle equation to find intersection points:
• (x - 1)² + (1 - x)² = 2, which simplifies to 2x² - 4x = 0.
• Solving gives x = 0 or x = 2. Thus, the points are (0, 1) and (2, -1).
• Find the slopes of lines from the origin to these points:
  • For (0, 1): slope = 1/0 (undefined, vertical line).
  • For (2, -1): slope = -1/2.
• Since the angle is between a vertical line and a slope of -1/2, calculate the angle θ using tan^-1:
  • θ = tan^-1(|-1/2|) = tan^-1(1/2).

Thus, the correct answer is tan^-1(2), which corresponds to option B.

Therefore, the correct option is:  x=cosec θ,y=cot θ

To find the sum of the series 2/3! + 4/5! + 6/7! + ..., let's analyse the pattern:

• The series can be expressed in the form: (2n)/(2n+1)! where n starts from 1.
• This means the terms are: 2/3!, 4/5!, 6/7!, ...
• The series resembles a modified exponential series. The exponential series for e^x is given by:
  • 1 + x/1! + x²/2! + x³/3! + ...
• Here, terms are shifted and modified, resembling:
  • (xⁿ)/(n+1)! with x = 2.
• By substituting x = 1 into the series, it simplifies to:
  • e¹ - 1 (which is e - 1)

Therefore, the sum of the given series is e^-1, which corresponds to option A.

To form a group of 7 with boys in the majority, we need at least 4 boys in the group.

• 4 Boys, 3 Girls:
  • Select 4 boys from 6: \( \binom{6}{4} = 15 \)
  • Select 3 girls from 4: \( \binom{4}{3} = 4 \)
  • Total combinations: \( 15 \times 4 = 60 \)
• 5 Boys, 2 Girls:
  • Select 5 boys from 6: \( \binom{6}{5} = 6 \)
  • Select 2 girls from 4: \( \binom{4}{2} = 6 \)
  • Total combinations: \( 6 \times 6 = 36 \)
• 6 Boys, 1 Girl:
  • Select 6 boys from 6: \( \binom{6}{6} = 1 \)
  • Select 1 girl from 4: \( \binom{4}{1} = 4 \)
  • Total combinations: \( 1 \times 4 = 4 \)

Add up all the combinations for the final result:

• \( 60 + 36 + 4 = 100 \)

The equation of the hyperbola is given by x² - y² = 3. The slope of the tangents that are parallel to the line 2x + y + 8 = 0 is -2 (derived from the line's equation).

To find points on the hyperbola where the tangent has this slope, use the tangent equation for a hyperbola: y = mx ± √(a²m² - b²).

• The slope of -2 implies m = -2.
• Plug into the equation to get 2x - y = 3.
• Now, solve simultaneously with 2x + y + 8 = 0.
• This results in 2x - y = 3 and 2x + y = -8.

By solving these equations, the points of tangency are found to be (2, -1) or (-2, 1), which corresponds to option B.

To differentiate the function:

• Use the formula for the derivative of arctan: if y = arctan(u), then dy/dx = (1/(1+u²))(du/dx).
• Let u = (sin x + cos x) / (cos x - sin x).
• Differentiate u with respect to x:
• Apply the quotient rule to find du/dx.
• Substitute u back into the derivative formula.
• Compute the final expression.

The final result may be undefined.

The line y = 4x - 5 is a tangent to the curve y² = ax³ + b at the point (2, 3). To find a and b, we follow these steps:

• Substitute x = 2 and y = 3 into the curve equation: 3² = a(2)³ + b.
• This simplifies to 9 = 8a + b.
• The slope of the tangent y = 4x - 5 is 4.
• To find the derivative of the curve, differentiate y² = ax³ + b implicitly with respect to x, giving 2y(dy/dx) = 3ax².
• At the point (2, 3), the derivative dy/dx = 4, so substituting gives 2(3)(4) = 3a(2)².
• This simplifies to 24 = 12a, thus a = 2.
• Substitute a = 2 back into 9 = 8a + b to find b = -7.

Therefore, the values are a = 2 and b = -7.

The function y = e^mx represents an exponential curve. To find the differential equation, we first calculate the derivative of y with respect to x.

• Differentiate the function: If y = e^mx, then dy/dx = m * e^mx.
• We can express e^mx in terms of y: Since y = e^mx, it follows that dy/dx = m * y.
• Now, express m in terms of y and x: We know m = log_e(y)/x, which can also be written as log(y)/x.
• Substituting for m in the derivative, we get dy/dx = (y/x) * log(y).

Thus, the differential equation is (dy/dx) = (y/x) * log(y), which corresponds to option C.

ANSWER :- b

Solution :- To form parallelogram we required a pair of line from a set of 4 lines and another pair of line from another set of 3 lines.

Required number of parallelograms = 4C2 x 3C2 = 6×3 = 18

Since a, b, c are in A.P.,
∴ b - a = c - b
Also b - a, c - b, a are in G.P
∴ (c - b)² = (b - a)a
⇒ (b - a)² = (b - a)a
⇒ b - a = a
⇒ b = 2a
Also c = 2b - a = 2(2a) - a = 3a
∴ a : b : c = a : 2a : 3a = 1 : 2 : 3