Mathematics Test 279

Result

Mathematics Test 279

Score
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Rank
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TIME Taken - -

Weekly Quiz Competition

Question 1
4 / -1

A
[x]

B
2[x]

C
1/2 [x]

D
None of these

Solution
Question 2
4 / -1

The area (in sq. units) of the region A = {(x, y) : (x - 1) [x] ≤ y ≤ 2 √x, 0 ≤ x ≤ 2}, where [t] denotes the greatest integer function, is

A

B

C

D

Solution
Question 3
4 / -1

A
1/4

B
1/8

C
1/2

D
1

Solution
Question 4
4 / -1

The maximum and minimum value of f(x) = ab sin x + b cos x + c lie in the interval (assuming |a| < 1, b > 0)

A
[b – c, b + c]

B
(b – c, b + c)

C
[c – b, b + c]

D
None of these

Solution
Question 5
4 / -1

A given right circular cone has a volume p, and the largest right circular cylinder that can be inscribed in the cone has a volume q. Then p : q is

A
9 : 4

B
8 : 3

C
7 : 2

D
None of these

Solution
Question 6
4 / -1

An object is moving in clockwise direction around the unit circle x² + y² = 1. As it passes through the point (1/2, √3/2), its y-coordinate is decreasing at the rate of 3 units per second. The rate at which the x-coordinate changes at this point is (in units per second)

A
2

B
3√3

C
√3

D
2√3
Solution
To find the rate of change of the x-coordinate:
- We know the object is on the unit circle, described by the equation x² + y² = 1.
- The coordinates at the point of interest are (1/2, √3/2).
- The rate of change of the y-coordinate is given as -3 units per second (it is decreasing).
- Differentiate the circle equation with respect to time (t):
- Using implicit differentiation, we get: 2x(dx/dt) + 2y(dy/dt) = 0.
- This simplifies to: x(dx/dt) + y(dy/dt) = 0.
- Substituting the known values:
- x = 1/2, y = √3/2, and dy/dt = -3.
- Thus, the equation becomes: (1/2)(dx/dt) + (√3/2)(-3) = 0.
- Solving for dx/dt:
- We get: (1/2)(dx/dt) - (3√3/2) = 0.
- Rearranging gives: dx/dt = 3√3.
The rate at which the x-coordinate changes at this point is 3√3 units per second.
Question 7
4 / -1

The tangent to the curve y = e^x drawn at the point (c, e^c) intersects the line joining the points (c – 1, e^c^–1) and (c + 1, e^{c + 1})

A
on the left of x = c

B
on the right of x = c

C
at no point

D
at all points
Question 8
4 / -1

The abscissa of the point on the curve 9y²= x³, the normal at which cuts off equal intercepts on the coordinate axes is

A
2

B
4

C
-4

D
-2
Question 9
4 / -1

What is the maximum value of 16 sin θ - 12 sin² θ?

A
3/4

B
4/3

C
16/3

D
4
Solution
To find the maximum value of the expression 16 sin θ - 12 sin² θ, follow these steps:
- Let u = sin θ. The expression becomes 16u - 12u².
- This is a quadratic expression in terms of u: 16u - 12u².
- The standard form is -12u² + 16u. The quadratic opens downwards since the coefficient of u² is negative.
- The maximum value occurs at u = -b/(2a), where a = -12 and b = 16.
- Calculate u = -16 / (2 × -12) = 2/3.
- Substitute u = 2/3 back into the expression:
  
  16(2/3) - 12(2/3)²
  
  Simplify to get 32/3 - 12(4/9)
  
  Further simplify to 32/3 - 16/3 = 16/3
- Therefore, the maximum value is 16/3.
Question 10
4 / -1

f(x) = sin^p θ cos^q θ, (p, q > 0, 0 < θ < π/2) has a point of maxima at

A
θ = tan^-1 √(p/q)

B
θ = tan^-1 √(q/p)

C
θ = tan^-1 (p/q)

D
None of these