SRMJEEE Maths Mock Test - 4 - JEE MCQ

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).

The sum of the products of elements in any row of a determinant A with the same row is equal to the determinant of A, denoted as |A|.

This property holds because:

• In a determinant, multiplying a row by itself involves calculating the dot product of the row with itself.
• This operation does not change the value of the determinant, hence the result is |A|.

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)

The standard deviation is a measure of how spread out numbers are in a set of data. It is particularly useful in probability experiments like this one involving a die.

• Consider a single die with six faces. Each face has an equal probability of showing up, which is 1/6.
• A success occurs when a 3 is thrown, so the probability of success (p) is 1/6.
• The probability of failure (q) is 5/6, since it's the complement of success.
• For 8 throws, the number of successes follows a binomial distribution.
• The formula for the standard deviation (σ) in a binomial distribution is:
  σ = √(n × p × q)
• Substitute the values: n = 8, p = 1/6, and q = 5/6.

• Calculation:

  σ = √(8 × 1/6 × 5/6)

  σ = √(40/36) = √(10/9)

  σ = 1.054

The correct answer, rounded to two decimal places, is approximately 4/3.

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.

The domain of the inverse sine function, denoted as sin^-1 x, is the range of values for which the function is defined.

• The domain of sin^-1 x is [-1, 1].
• This means sin^-1 x is defined for all x where -1 ≤ x ≤ 1.

The given differential equation is:

(dy/dx) = (y/x) + (φ(y/x)/φ'(y/x))

To solve this, note that the equation is in the form of a homogeneous differential equation. Let's introduce a substitution:

• Let v = y/x. Thus, y = vx.
• Differentiating, we have: dy/dx = v + x(dv/dx).

Substitute back into the original equation:

• v + x(dv/dx) = v + φ(v)/φ'(v)
• Cancel v from both sides to get: x(dv/dx) = φ(v)/φ'(v)

Separate the variables:

• φ'(v) dv = φ(v) dx/x

Integrate both sides:

• ∫φ'(v) dv = ∫φ(v) dx/x
• Integration gives: φ(v) = k ln(x), where k is the constant of integration.

Substituting v = y/x back, we get:

• φ(y/x) = k ln(x)

To simplify, we can express this as a power of x:

• φ(y/x) = k x

Thus, the solution is:

φ(y/x) = k x

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.

To find the point on the curve y = x² + 4x + 3 closest to the line y = 3x + 2, follow these steps:

• Consider a point (x, y) on the curve, where y = x² + 4x + 3.
• The perpendicular distance from this point to the line y = 3x + 2 is given by the formula:
• d = |y - 3x - 2| / √(32 + 1)
• Substitute y = x² + 4x + 3 into the distance formula:
• d = |x2 + 4x + 3 - 3x - 2| / √10
• Simplify the expression:
• d = |x2 + x + 1| / √10
• Minimise d by minimising the expression x² + x + 1.
• Find the derivative: f'(x) = 2x + 1 and set it to zero:
• 2x + 1 = 0
• x = -1/2
• Substitute x = -1/2 back into y = x² + 4x + 3:
• y = (-1/2)2 + 4(-1/2) + 3
• y = 1/4 - 2 + 3
• y = 1.25
• The closest point is (-1/2, 1.25), which corresponds to option B.

To find the equation of the line passing through the intersection of the lines 3x - 2y - 1 = 0 and x - 4y + 3 = 0, and the point (π, 0), follow these steps:

• Find the point of intersection of the given lines:
• Express these equations in a system and solve them simultaneously.
• For example, solve for x and y using substitution or elimination.
• Once the intersection point is found, use it along with the point (π, 0) to determine the equation of the desired line.
• The line equation can be written in the form y = mx + c, where m is the slope.
• Calculate m as the slope between the intersection point and (π, 0).
• Substitute one point into the line equation to find c.
• After determining m and c, rewrite the equation in the desired form.

The solution is the equation x - y = π(1 - y), which matches option A.

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