Which is the root of the equation x4-9×3+6×2+2?

A. 1
B. -1
C. -2
D. None of these

Answer

Correct Answer: D. None of these 


Solution

The roots of equation x^4 – 9x^3 + 6x^2 + 2 cannot be determined exactly by factoring or using the rational root theorem, as there are no rational roots.

However, we can use numerical methods or a computer algebra system to find the approximate roots. Using a graphing calculator or a computer algebra system, we can plot the function y = x^4 – 9x^3 + 6x^2 + 2 and find its zeros, which are the roots of the equation.

The approximate roots are:

x ≈ -2.5061 x ≈ -0.2213 x ≈ 0.1156 x ≈ 8.6118

Therefore, none of the given options A, B, C, or D is the root of the equation x^4 – 9x^3 + 6x^2 + 2.

The value of the polynomial 4x^4-3x^3+2x^2-x+5 at x=3 is:

A. 21
B. 263
C. 313
D. 150

Answer

Correct Answer: B. 263


Solution

The value of the polynomial 4x^4 – 3x^3 + 2x^2 – x + 5 at x = 3 is obtained by replacing x with 3 in the expression for the polynomial and then simplifying:

4(3)^4 – 3(3)^3 + 2(3)^2 – 3 + 5 = 4(81) – 3(27) + 2(9) – 3 + 5 = 324 – 81 + 18 – 3 + 5 = 263

Therefore, the value of the polynomial 4x^4 – 3x^3 + 2x^2 – x + 5 at x = 3 is 263.

If S1 is the solution set of equation f(x,y)=0 and S2 is the solution set of g(x,y)=0 then the solution set of the system f(x,y)=0 and g(x,y)=0 is:

A. S1-S2
B. S1∩S2
C. S1US2
D. S1+S2

Answer

Correct Answer: B. S1∩S2


Solution

The solution set of the system f(x,y)=0 and g(x,y)=0 is the intersection of S1 and S2, which consists of all the solutions that satisfy both equations simultaneously.

In other words, the solution set of the system f(x,y)=0 and g(x,y)=0 is the set of all (x,y) pairs that satisfy both equations. This set can be written as:

S1 ∩ S2 = {(x,y) : f(x,y) = 0 and g(x,y) = 0}

So the solution set of the system is the set of all common solutions of the two equations.

Sign of the cos(-885) is ?

A. +ve
B. -ve
C. Both (-ve) & (+ve)
D. 0

Answer

Correct Answer: B. -ve


Solution

We can find the sign of cos(-885) using the fact that cosine is an even function, which means that:

cos(-x) = cos(x)

for any value of x.

Therefore, we have:

cos(-885) = cos(885)

To determine the sign of cos(885), we need to know the quadrant in which the angle 885 degrees lies.

Starting from the positive x-axis and rotating counterclockwise, we have:

  • 0 to 90 degrees: First quadrant
  • 90 to 180 degrees: Second quadrant
  • 180 to 270 degrees: Third quadrant
  • 270 to 360 degrees: Fourth quadrant

Since 885 degrees is more than 360 degrees, we can subtract 360 from it until we get an angle in the range of 0 to 360 degrees:

885 – 360 = 525 525 – 360 = 165

Therefore, 885 degrees is equivalent to an angle of 165 degrees, which lies in the second quadrant.

In the second quadrant, the cosine function is negative, so we have:

cos(885) = cos(165) < 0

Therefore, the sign of cos(-885) is negative.

1° = ____ radians.

A. 0.1745
B. 0.01745
C. 0.0001745
D. 0.001745

Answer

Correct Answer: B. 0.01745


Solution

To convert degrees to radians, we use the formula:

radians = (π/180) * degrees

where π (pi) is the constant ratio of the circumference to the diameter of a circle and is approximately equal to 3.14159.

Therefore, to convert 1 degree to radians, we have:

radians = (π/180) * 1 = 0.017453 radians (rounded to six decimal places)

So 1 degree is approximately equal to 0.017453 radians.

The ratio between the area of the sector to the length of arc of a circle is ?

A. 2r/1
B. r/2
C. 2/r
D. 1/2r

Answer

Correct Answer: B. r:2

Solution

The ratio between the area of the sector to the length of arc of a circle is given by:

θ/360°

where θ is the central angle of the sector measured in degrees.

The length of the arc intercepted by the central angle θ is given by:

L = (θ/360°) * 2πr

The area of the sector defined by the central angle θ is given by:

A = (θ/360°) * πr^2

The ratio of the area of the sector to the length of the arc is then:

A/L = [(θ/360°) * πr^2] / [(θ/360°) * 2πr] = r/2

So the ratio between the area of the sector to the length of the arc of a circle is simply half the radius of the circle.

 

cot^–1(x) = ?

A. Cos-1x
B. -Tan-1x
C. 1/tan-1x
D. tan-1(1/x)

Answer

Correct Answer: D. tan-1(1/x)

Solution

To find cot⁻¹(x), we need to find the angle whose cotangent is x. That is, we need to find θ such that:

cot(θ) = x

Using the definition of cotangent in terms of the adjacent and opposite sides of a right triangle, we can draw a right triangle with an angle θ such that the adjacent side has length x and the opposite side has length 1. This gives us:

cot(θ) = x = adjacent/opposite = x/1

Using the Pythagorean theorem, we can find the length of the hypotenuse:

adjacent^2 + opposite^2 = hypotenuse^2

x^2 + 1^2 = hypotenuse^2

hypotenuse^2 = x^2 + 1

Taking the square root of both sides, we get:

hypotenuse = ±√(x^2 + 1)

However, since cot(θ) is positive, the angle θ must be in the first or fourth quadrant, where the hypotenuse is positive. Therefore, we have:

hypotenuse = √(x^2 + 1)

Now, we can use the definition of the inverse tangent function to find the angle that has a tangent of x:

tan(θ) = 1/x

θ = tan⁻¹(1/x)

Therefore, the solution is:

cot⁻¹(x) = tan⁻¹(1/x)

Therefore, cot⁻¹(x) = tan⁻¹(1/x).

Range of Tan^–1(x) is ?

A. [-π/2,π/3]
B. [0,π]
C. (-π/2,π/2)
D. R

Answer

Correct Answer: C. (-π/2,π/2)


Solution

The range of the tangent function is all real numbers, except for odd multiples of π/2, i.e., (-∞, -π/2), (-π/2, π/2), and (π/2, ∞). Therefore, the range of the arctangent function, denoted as tan⁻¹(x), will be a subset of (-π/2, π/2).

To find the exact range of tan⁻¹(x), we can use the fact that the tangent function is an odd function, i.e., tan(-θ) = -tan(θ) for any angle θ. This implies that the arctangent function is an odd function as well, i.e., tan⁻¹(-x) = -tan⁻¹(x). Therefore, the range of tan⁻¹(x) must be symmetric about the origin.

Since tan(π/4) = 1, we know that tan⁻¹(1) = π/4. Similarly, tan(-π/4) = -1, which implies that tan⁻¹(-1) = -π/4. These are the two endpoints of the range of tan⁻¹(x). Therefore, the range of tan⁻¹(x) is:

(-π/2, π/2)

which is the same as the range of the arctangent function.