Mathematics Test - 38

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x² – 24x + 45

3x² – 24x + 45 = 0

Or x² – 8x + 15 = 0

Or(x – 3)(x – 5) = 0

Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5

Therefore, f’(x) = 0 for x = 3 and x = 5.

If h be a positive quantity, however small, then,

f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.

f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.

Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.

So, f(x) has minimum at 5.

Putting, x = 5 in (1)

Thus, its maximum value is,

f(3) = 5³ – 12*5² + 45*5 + 8 = 58.

Assume that the velocity of the particle at time t second is vcm/sec.

Then, v = dx/dt = 4t³/12 – 6t²/3 + 6t/2 + 1

So, v = dx/dt = t³/3 – 2t²/ + 3t + 1

Thus, dv/dt = t² – 4t + 3

And d²v/dt² = 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t² – 4t + 3 = 0

Or (t – 1)(t – 3) = 0

Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3

Now, [d²v/dt²]_{t = 3} = 2*3 – 4 = 2 > 0

Thus, v is minimum at t = 3.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x² – 24x + 45

3x² – 24x + 45 = 0

Or x² – 8x + 15 = 0

Or (x – 3)(x – 5) = 0

Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5

Therefore, f’(x) = 0 for x = 3 and x = 5.

If h be a positive quantity, however small, then,

f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.

f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.

Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.

So, f(x) has maximum at 3.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x² – 24x + 45

3x² – 24x + 45 = 0

Or x² – 8x + 15 = 0

Or (x – 3)(x – 5) = 0

Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5

Therefore, f’(x) = 0 for x = 3 and x = 5.

If h be a positive quantity, however small, then,

f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.

f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.

Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.

So, f(x) has minimum at 5.

The given function is f(x) = ∣x − 1∣, x ∈ R.

It is known that a function f is differentiable at point x = c in its domain if both

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.

As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.

Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.

Therefore, f(x) has a local minima at x = 1.

Assume that the velocity of the particle at time t second is vcm/sec.

Then, v = dx/dt = 4t³/12 – 6t²/3 + 6t/2 + 1

So, v = dx/dt = t³/3 – 2t²/ + 3t + 1

Thus, dv/dt = t² – 4t + 3

And d²v/dt² = 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t² – 4t + 3 = 0

Or (t – 1)(t – 3) = 0

Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3

Now, [d²v/dt²]_{t = 3} = 2*3 – 4 = 2 > 0

Thus, v is minimum at t = 3.

Putting t = 3 in (1) we get,

3³/3 – 2(3)²/ + 3(3) + 1

= 1 cm/sec.

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.