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Trigonometric Functions

Multiple Choice Questions

31.

In a AABC, if $∠$ C = 90°, r and R are the inradius and circumradius of the $∆$ ABC respectively, then 2(r + R) is equal to

b + c
c + a
a + b
a + b + c

32.

Let $α and β$ be two distinct roots of $acos (θ) + bsin (θ) = c$ where a, b, c are three real constants and $θ \in [0, 2 π]$ . Then, $α + β$ is also a root of the same equation, if

a + b = c
b + c = a
c + a = b
c = a

33.
If $\cos (x) and \sin (x)$ are solutions of the differential equation
$a_{0} \frac{d^{2} y}{d x^{2}} + a_{1} \frac{d y}{d x} + a_{2} y = 0$
where a₀, a₁ and a₂ are real constants, then which of the following is/are always true?

$Acos (x) + Bsin (x)$ is a solution, where A and B are real constants

$Acos (x + \frac{π}{4})$ is a solution, where A is a real constant

$Acos (x) \sin (x)$ is a solution, where A is a real constant

$Acos (x + \frac{π}{4}) + Bsin (x - \frac{π}{4})$ is a souton, where A and B are real constants

$Acos (x) + Bsin (x)$ is a solution, where A and B are real constants

$Acos (x + \frac{π}{4})$ is a solution, where A is a real constant

$Acos (x + \frac{π}{4}) + Bsin (x - \frac{π}{4})$ is a souton, where A and B are real constants

(a) Let $f (x) = \cos (x) and g (x) = \sin (x)$

Consider the Wronskian of f(x) and g(x),

$W = |\begin{matrix} f (x) & g (x) \\ f' (x) & g' (x) \end{matrix}| = |\begin{matrix} \cos (x) & \sin (x) \\ - \sin (x) & \cos (x) \end{matrix}| = \cos^{2} (x) + \sin^{2} (x) = 1 \neq 0$

Thus, the functions are linearly independent. So, the general solution of given differential equation is given by $y = Acos (x) + Bsin (x)$ , where A and B are real constants.

[ $∵$ if y₁ and y₂ are linearly independent solutions of the differential equation ay'' + by' + c = 0, then the general solution is y = c₁y₁ + c₂y₂, where c₁ and c₂ are constants]

Hence, option (a) is true.

(b) Let $y = Acos (x + \frac{π}{4})$

$= A (\cos (x) . \cos (\frac{π}{4}) - \sin (x) . \sin (\frac{π}{4})) [∵ \cos (A + B) = \cos (A) . \cos (B) - \sin (A) . \sin (B)] = \frac{A}{\sqrt{2}} (\cos (x) - \sin (x)) = \frac{A}{\sqrt{2}} \cos (x) + (- \frac{A}{\sqrt{2}}) \sin (x)$

which is in the form of general solution.

Hence, option (b) is true

(d) Let y $= Acos (x + \frac{π}{4}) + Bsin (x - \frac{π}{4})$

$= Acos (x + \frac{π}{4}) + Bsin (x - \frac{π}{4}) = A (\cos (x) . \frac{1}{\sqrt{2}} - \sin (x) . \frac{1}{\sqrt{2}}) + B (\sin (x) . \frac{1}{\sqrt{2}} - \cos (x) . \frac{1}{\sqrt{2}}) = \cos (x) . (\frac{A}{\sqrt{2}} - \frac{B}{\sqrt{2}}) + \sin (x) . (\frac{B}{\sqrt{2}} - \frac{A}{\sqrt{2}})$

which is in the form of general solution.

Hence, option (d) is true.

34.

Which of the following statements is /are correct for $0 < θ < \frac{π}{2}$

${(\cos (θ))}^{1 / 2} \leq \cos (\frac{θ}{2})$
${(\cos (θ))}^{3 / 4} \geq \cos (\frac{3 θ}{4})$
$\cos (\frac{5 θ}{6}) \geq {(\cos (θ))}^{5 / 6}$
$\cos (\frac{7 θ}{8}) \leq {(\cos (θ))}^{7 / 8}$

35.

The value of $\tan (\frac{π}{2}) + 2 \tan (\frac{2 π}{5}) + 4 \cot (\frac{4 π}{5})$ is

$\cot (\frac{π}{5})$
$\cot (\frac{2 π}{5})$
$\cot (\frac{4 π}{5})$
$\cot (\frac{3 π}{5})$

36.

The range of the function y = $3 \sin (\sqrt{\frac{π^{2}}{16} - x^{2}})$ is

$[0, \sqrt{3 / 2}]$
[0, 1]
$[0, 3 / \sqrt{2}]$
$[0, \infty]$

37.

In a $∆ ABC$ , a, b, c are the sides of the triangle opposite to the angles A, B, C, respectively. Then, the value of a³sin(B - C) + b³sin(C - A) + c³sin(A - B) is equal to

38.

$\cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$

is equal to zero
lies between 0 and 3
is a negative number
lies between 3 and 6

39.

The minimum value of $2^{\sin (x)} + 2^{\cos (x)}$ is

$2^{1 - 1 / \sqrt{2}}$
$2^{1 + 1 / \sqrt{2}}$
$2^{\sqrt{2}}$
2

40.

If $p = (\begin{matrix} \cos (\frac{π}{4}) & - \sin (\frac{π}{4}) \\ \sin (\frac{π}{4}) & \cos (\frac{π}{4}) \end{matrix}) and X = (\begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix})$ . Then, p³X is equal to

$(\begin{matrix} 0 \\ 1 \end{matrix})$
$(\begin{matrix} - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix})$
$(\begin{matrix} - 1 \\ 0 \end{matrix})$
$(\begin{matrix} - \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} \end{matrix})$