Mathematics Test - 26

Given: sin α + cos α = p

By squaring both the sides, we get

⇒ sin² α + cos² α + 2 sin α cos α = p²

As we know that, sin² x + cos² x = 1 and sin 2x = 2 sin x cos x

⇒ 1 + sin 2α = p² ⇒ sin 2α = p² – 1

⇒ cos² 2α = 1 – sin² 2α = 1 – (p² – 1)²

⇒ cos² 2α = p² (2 – p²)

Given:

A, B, C be subsets of X

Statement 1: A ⊂ C ⇒ (A ∩ B) ⊂ (C ∩ B), (A ∪ B) ⊂ (C ∪ B)

Now, if A ⊂ C then (A ∩ B) ⊂ (C ∩ B), (A ∪ B) ⊂ (C ∪ B) is true.

So, statement 1 is true.

Statement 2: (A ∩ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

Let A = {1, 2, 3}, B = {3, 4, 5} and C = {1, 6, 7, 8}

⇒ A ∩ B = {3} ⊂ (C ∩ B) but A ⊂ C is not true.

Therefore, statement 2 is false.

Statement 3: (A ∪ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

This is true as (A ∪ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

Given,

Total number of cards \(=52\)

Number of king card \(=4\)

Now, 7 cards are drawn from 52 cards.

\(P(3 \text { cards are king })=\frac{{ }^{4} C_{3} \times{ }^{48} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{7} }\)

\(=\frac{4 \times\frac{(48 \times 47 \times 46 \times 45)}{(4 \times 3 \times 2 \times 1)\ }}{\frac{(52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46)}{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} }\)

\(=\frac{4 \times(48 \times 47 \times 46 \times 45) \times(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}{(4 \times 3 \times 2 \times 1) \times(52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46)}\)

\(=\frac{(7 \times 6 \times 5 \times 4 \times 45)}{(52 \times 51 \times 50 \times 49)}\)

\(=\frac{(6 \times 5 \times 4 \times 45)}{(52 \times 51 \times 50 \times 7)}\)

\(=(\frac{(6 \times 4 \times 45)}{(7 \times 52 \times 51 \times 10)}\)

\(=\frac{(6 \times 45)}{(7 \times 13 \times 51 \times 10)}\)

\(=\frac{(6 \times 3)}{(7 \times 13 \times 17 \times 2) }\)

\(=\frac{(3 \times 3) }{(7 \times 13 \times 17) }\)

\(=\frac{9}{1547}\)

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to is to be determined.
• Substitution is done and integral function is then integrated.
• Finally, initial variable t, to be returned.

Let \(\sin x=t \Rightarrow \cos x d x=d t\)

\(\Rightarrow 1=\int \frac{\cos x}{1+\sin ^{2} x} d x\)

\(\Rightarrow 1=\int \frac{1}{1+t^{2}} d t\)

\(\Rightarrow 1=\tan ^{-1} t+c\)

\(\Rightarrow 1=\tan ^{-1}(\sin x)+c\)

Putting the limits,

\(\Rightarrow 1=\left[\tan ^{-1}\left(\sin 90^{\circ}\right)+c\right]-\left[\tan ^{-1}\left(\sin 0^{\circ}\right)+c\right]\)

\(\Rightarrow 1=\tan ^{-1}(1)\)

\(\Rightarrow I=45^{\circ}\) or \(\frac{\pi}{4}\)

Given,\({x}-{iy} =\sqrt{\frac{{a}-{ib}}{{c}-{id}}} \)
\(=\sqrt{\frac{{a}-{ib}}{{c}-{id}} \times \frac{{c}+{id}}{{c}+{id}}}\quad\quad\) [On multiplying numerator and denominator by \(({c}+{id})\)]
\(=\sqrt{\frac{({ac}+{bd})+i({ad}-{bc})}{{c}^{2}+{d}^{2}}}\)
So,\((x-i y)^{2}=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}\)
\(x^{2}-y^{2}-2 i x y=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}\)
On comparing real and imaginary parts, we get
\(x^{2}-y^{2}=\frac{a c+b d}{c^{2}+d^{2}},-2 x y=\frac{a d-b c}{c^{2}+d^{2}}\quad\quad\)........(i)
\(\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}\)
\(=\left(\frac{a c+b d}{c^{2}+d^{2}}\right)^{2}+\left(\frac{a d-b c}{c^{2}+d^{2}}\right)^{2}\quad\quad\) [Using (i)]
\(=\frac{a^{2} c^{2}+b^{2} d^{2}+2 a c b d+a^{2} d^{2}+b^{2} c^{2}-2 a d b c}{\left(c^{2}+d^{2}\right)^{2}}\) \(=\frac{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}{\left(c^{2}+d^{2}\right)^{2}}\)
\(=\frac{a^{2}\left(c^{2}+d^{2}\right)+b^{2}\left(c^{2}+d^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}\) \(=\frac{\left(c^{2}+d^{2}\right)\left(a^{2}+b^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}\)

Given:

\(f: R \rightarrow R\) is a function defined by \(f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi\)

For \(x=n, n \in Z\)

LHL \(=\lim _{ x \rightarrow n ^{-}} f ( x )=\lim _{ x \rightarrow n ^{-}}[ x -1] \cos \left(\frac{2 x -1}{2}\right) \pi=0\)

RHL \(=\lim _{ x \rightarrow n ^{+}} f ( x )=\lim _{ x \rightarrow n ^{+}}[ x -1] \cos \left(\frac{2 x -1}{2}\right) \pi=0\)

\(f( n )=0\)

\(\Rightarrow\) LHL = RHL \(=f(n)\)

\(\Rightarrow f( x )\) is continuous for every real \(x\).

The given equation can be simplified as,

\(\frac{d x}{d y}+\frac{x}{2 y}=\frac{5}{2} y\)

On comparing eqn (i) with standard eqn,\(\frac{d x}{d y}+P x=Q\),we get,

\(P=\frac{1}{2 y}\) and \(Q=\frac{5}{2} y\)

\(\therefore I F=e^{\int P d y}=e^{\int \frac{1}{2 y} d y}\)

\(\Rightarrow I F=e^{\frac{1}{2} \log y}=e^{\log y^{\frac{1}{2}}}\)

\(\Rightarrow I F=\sqrt{y} \cdot\left(\because e^{a \log x}=x^{a}\right)\)

Consider, the slope of the line \((x-2=0)\) is \(m_1\)

So, \(m _1=\infty\).

And, the slope of the line \((\sqrt{3} x-y-2=0)\) is \(m_2\)

So, \(m _2=\sqrt{  3}\).

Now, the angle between the given lines is \(\theta\).

\( \Rightarrow \tan \theta=\left|\frac{m_2-m_1}{1+m_1 \cdot m_2}\right| \)

\( \Rightarrow \tan \theta=\left|\frac{\frac{m_2}{m_1}-1}{\frac{1}{m_1}+m_2}\right| \)

\( \Rightarrow \tan \theta=\frac{1}{\sqrt{3}} \)

\( \Rightarrow \theta=30^{\circ}\)

Given:

\(A =\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]\)

\(A + A ^{T}=1\)

The transpose of matrix \(A\) is given by,

\(A^{T}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\)

As, \(A + A ^{T}=I\)

\(\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]+\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]=I\)

\(\Rightarrow\left[\begin{array}{cc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

As we know that,

If two matrices \(A\) and \(B\) are equal then their corresponding elements are also equal.

\(\therefore 2 \cos 2 \theta=1\)

\(\Rightarrow \cos 2 \theta=\frac{1}{2}\)

\(\Rightarrow \cos 2 \theta=\cos 60^\circ\)

\(\Rightarrow 2 \theta=\frac{\pi}{3}\quad[\because \cos 60^\circ=\frac{\pi}{3}]\)

\(\therefore \theta=\frac{\pi}{6}\)

So, the value of \(\theta\) is \(\frac{\pi}{6}\).

Given, \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x+4=0\).

\(\Rightarrow \alpha+\beta=2\) and \(\alpha \beta=4\)

Consider, \(\alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}-\alpha \beta\right)\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}+2 \alpha \beta-3 a \beta\right)\)

\(\left.\Rightarrow \alpha^{3}+\beta^{3}=(\alpha+\beta)\left[(\alpha+\beta)^{2}-3 a \beta\right)\right]\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(2)\left[(2)^{2}-3(4)\right]\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(2)(-8)\)

\(\Rightarrow \alpha^{3}+\beta^{3}=-16\)

So, if \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x+4=0\), then the value of \(\alpha^{3}+\beta^{3}=-16\).