Comprehensive Mathematics Examination Questions (Linear Programming to Trigonometry)

MathematicsLinear Programming, Calculus, Linear Algebra, Geometry, TrigonometryMedium

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20. Find the maximum value of the objective function $z = 2x + 3y$ under the constraints $x + y \le 6$, $x - y \ge 2$ and $y \ge 0$.

21. If the function $f(x) = 4x - 1$ is defined,

(a) Find the value of $f(2.999)$ and $f(3.001)$

(b) Is function $f(x)$ continuous at the point $x = 3$? Give reason.

22. Solve by matrix method:

$3x + y = 9$ and $2x - y = 1$

23. The angle between a pair of straight lines represented by equation $2x^2 + kxy + 3y^2 = 0$ is $45^\circ$, find the value of $k$.

24. Prove that:

$\frac{1 + \tan\theta}{1 - \tan\theta} = \frac{1 + \sin2\theta}{\cos2\theta}$

25. If $A + B + C = \pi^C$, prove that:

$\cos2A + \cos2B - \cos2C = 1 - 2\sin A \cdot \sin B \cdot \cos C$

26. From a point 45 m and 20 m away from a tower, the angles of elevation of the top of the tower are complementary, find the height of the tower.

Animated Video Solution

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Step by Step Written Solution

1
Step 1

In this problem, we need to find the value of k if the angle between the pair of straight lines represented by the equation two x squared plus k x y plus three y squared equals zero is forty-five degrees.

Angle Between Pair of Lines

$$2x^2 + kxy + 3y^2 = 0$$
$$\theta = 45^\text{o}$$
2
Step 2

First, let's compare our given equation with the general form of a pair of lines passing through the origin, which is a x squared plus two h x y plus b y squared equals zero.

$$ax^2 + 2hxy + by^2 = 0$$
3
Step 3

By comparing coefficients, we find that a equals two, two h equals k, which means h equals k over two, and b equals three.

$$a = 2, \text{ } h = \frac{k}{2}, \text{ } b = 3$$
4
Step 4

Now, we use the formula for the angle theta between such lines, which is tangent theta equals plus or minus two times the square root of h squared minus a b, all divided by a plus b.

$$\tan \theta = \frac{\text{\textpm} 2\text{\textsqrt}{h^2 - ab}}{a + b}$$
5
Step 5

Let's substitute our values into the formula. Tangent of forty-five degrees equals plus or minus two times the square root of k over two squared minus two times three, all over two plus three.

Substitution

$$\tan 45^\text{o} = \frac{\text{\textpm} 2\text{\textsqrt}{(\frac{k}{2})^2 - (2)(3)}}{2+3}$$
6
Step 6

We know that tangent of forty-five degrees is one. On the right side, the denominator becomes five.

7
Step 7

Multiplying both sides by five, we get five equals plus or minus two times the square root of k squared over four minus six.

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About This Question

Subject
Mathematics
Topic
Linear Programming, Calculus, Linear Algebra, Geometry, Trigonometry
Difficulty
Medium
Question Type
Open Ended

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