Mathematics Test - 40

The equation of the given parabola is, y² = 5x ……….(1)

Differentiating both sides of (1) with respect to y, we get,

2y = 5(dx/dy)

Or dx/dy = 2y/5

Take any point P((5/4)t², (5/2)t). Then, the normal to the curve (1) at P is,

-[dx/dy]_P = -(2*5t/2)/5 = -t

By the question, slope of the normal to the curve (1) at P is tan45°.

Thus, -t = 1

Or t = -1

So, the required equation of normal is,

y – 5t/2 = -t(x – 5t²/4)

Simplifying further we get,

4(x – y) = 15

Given, y² = 3x ……….(1) and y = 2x + 4 ……….(2)

Differentiating both sides of (1) with respect to y we get,

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]_P = -2y₁/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y₁/3*2 = -1

Since the slope of the line (2) is 2

Or y₁ = 3/4

Since the point P(x₁, y₁) lies on (1) hence,

y₁² = 3x₁

As, y₁ = 3/4, so, x₁ = 3/16

Therefore, the required equation of the normal is

y – y₁ = -(2y₁)/3*(x – x₁)

Putting the value of x₁ and y₁ in the above equation we get,

16x + 32y = 27.

The equation of any circle through the points of intersection of the given circle is,

x² + y² + 2x + 4y + 1 + k(x² + y² – 1) = 0

x² + y² + 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0

Clearly, the co-ordinates of the center of the circle (1) are, (-1/(1 + k), -2/(1 + k)) and its radius,

= √[(1/(1 + k))² + (2/(1 + k))² – ((1 – k)]/(1 + k))

= √(4 + k²)/(1 + k)

Clearly, the line x + 2y + 5 = 0 is tangent to the circle (1), hence, the perpendicular distance of the line from the center of the circle = radius of the circle

± (-1/(1 + k))– 2(2/(1 + k)) + 5/ √(1² + 2²) = √(4 + k²)/(1 + k)

Or ±(5k/√5) = √(4 + k²)

Or 5k² = 4 + k²

Or 4k² = 4

Or k = 1 [as, k ≠ -1]

Putting k = 1 in (1), equation of the given circle is,

x² + y² + x + 2y = 0

We have, x²/a + y²/b = 1 ……….(1)

and

x²/c + y²/d = 1 ……….(2)

Let, us assume curves (1) and (2) intersect at (x₁, y₁). Then

x₁²/a + y₁²/b = 1 ……….(3)

and

x₁²/c + y₁²/d = 1 ……….(4)

Differentiating both side of (1) and (2) with respect to x we get,

2x/a + 2y/b(dy/dx) = 0

Or dy/dx = -xb/ya

Let, m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x₁, y₁); then,

m₁ = [dy/dx]_{(x1, y1)} = -(bx₁/ay₁) and m₂ = [dy/dx]_{(x1, y1)} = -(dx₁/cy₁)

By question as the curves (1) and (2) intersects at right angle, so, m₁m₂ = -1

Or -(bx₁/ay₁)*-(dx₁/cy₁) = -1

Or bdx₁² = -acy₁² ……….(5)

Now, (3) – (4) gives,

bdx₁²(c – a) = acy₁²(d – b) ……….(6)

Dividing (6) by (5) we get,

c – a = d – b

Or a – b = c – d.

The above is a condition for a symmetric relation.

For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}

1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.

Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. Hence, they are symmetric.

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

The equation is cos^-1 x + cos^-1 y + cos^-1 z = 3π

This means cos^-1 x = π, cos^-1 y = π and cos^-1 z = π

This will be only possible when it is in maxima.

As, cos^-1 x = π so, x = cos π = -1 similarly, y = z = -1

Therefore, x + y + z = -1 -1 -1

So, x + y + z = -3.

• (A) Unit Vector → (I) A vector with a magnitude of 1, often used to specify direction.
• (B) Position Vector → (II) A vector representing a point’s position in space with respect to the origin.
• (C) Zero Vector → (III) A vector with zero magnitude and no particular direction.
• (D) Collinear Vectors → (IV) Vectors that are parallel or lie along the same straight line.

Thus, the correct answer is A (A) - (I), (B) - (II), (C) - (III), (D) - (IV).

• (A) Skew-Symmetric Matrix: A skew-symmetric matrix satisfies the condition Aᵀ = -A, corresponding to (I).
• (B) Diagonal Matrix: In a diagonal matrix, all off-diagonal elements are zero, which matches (II).
• (C) Orthogonal Matrix: An orthogonal matrix satisfies A * Aᵀ = I, corresponding to (III).
• (D) Scalar Matrix: A scalar matrix has equal diagonal elements, corresponding to (IV).

• (A) Derivative of log(x) → (II) The derivative of the natural logarithm function, log(x), is 1/x.
• (B) Derivative of e^x → (I) The rate of change of the function y = e^x with respect to x is itself.
• (C) Derivative of sin(x) → (IV) The rate of change of the sine function with respect to x is cos(x).
• (D) Second-Order Derivative → (III) The second derivative of a function gives information about the concavity of the graph.

Thus, the correct answer is (1) (A) - (II), (B) - (I), (C) - (IV), (D) - (III).