Mathematics Test-16

Given:

Simplify the given function.

\(\frac{1}{x-\sqrt{x}}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}\)

\(\sqrt{x}-1=t\)

Differentiating both sides with respect to \(x\)

\(\frac{1}{2} \cdot x^{\frac{1}{2}-1}-0=\frac{d t}{d x}\)

\(\left[\therefore\left(x^{n}\right)^{\prime}=n x^{n-1}\right]\)

\(\frac{1}{2} \cdot x^{-\frac{1}{2}} =\frac{d t}{d x}\)

\(d x =\frac{2 d t}{x^{-\frac{1}{2}}}\)

\(d x =2 \sqrt{x} d t\)

Integrating the function

\(\int \frac{1}{x-\sqrt{x}} \cdot d x\)

\(=\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} \cdot d x\)

putting \(t=\sqrt{x}-1\) and \(d x=2 \sqrt{x} d t\)

\(=\int \frac{1}{\sqrt{x}(t)} \cdot 2 \sqrt{x} \cdot d x\)

\(=\int \frac{2}{t} \cdot d t\)

\(=2 \int \frac{1}{t} \cdot d t\)

\(=2 \log |t|+C\)

\([\because\) \(\int \frac{1}{x} \cdot d x=\log |x|\)\(]\)

\(=2 \log |\sqrt{x}-1|+C\)

\([\because\) \(t=\sqrt{x}-1]\)

Given:

\(\vec{a}=\hat{i}+\hat{j}-\hat{k}\)

\(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)

We know that

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)

where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\)

Finding \(|\vec{a}|,|\vec{b}|\) and \(\vec{a} \cdot \vec{b}\)

Magnitude of \(\vec{a}=\sqrt{1^2+1^2+(-1)^2}\)

\(|\vec{a}|=\sqrt{1+1+1}=\sqrt{3}\)

Magnitude of \(\vec{b}=\sqrt{1^2+(-1)^2+1^2}\)

\(|\vec{b}|=\sqrt{1+1+1}=\sqrt{3}\)

Finding \(\vec{a} \cdot \vec{b}\)

\(\vec{a} \cdot \vec{b} =(1 \hat{i}+1 \hat{j}-1 \hat{k}) \cdot(1 \hat{i}-1 \hat{j}+1 \hat{k})\)

\(=1 \cdot 1+1 \cdot(-1)+(-1) 1\)

\(=1-1-1\)

\(=-1\)

Now,

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)

Putting values

\(-1=\sqrt{3} \times \sqrt{3} \times \cos \theta\)

\(-1=3 \cos \theta\)

\(\cos \theta=\frac{-1}{3}\)

\(\theta=\cos ^{-1}\left(\frac{-1}{3}\right)\)

Therefore, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\cos ^{-1}\left(\frac{-1}{3}\right)\)

Here, we first have to select \(9\) students from \(17\) students.

Tip:

SELECT \(=\) Combination \(={ }^{{n}} {C}_{{r}}=\frac{{n} !}{{r} !({n}-{r}) !}\)

SELECT and ARRANGE \(=\) Permutation \(={ }^{{n}} {P}_{{r}}=\frac{{n} !}{({n}-{r}) !}\)

Arrange \(9\) students in circle \(=9-1=8 !\) ways \(17-9=8\) students remain.

Arrange remaining \(8\) students in another circle \(=8-1=7 !\) ways

\(\therefore\) Total ways to arrange \(17\) students in \(2\) circles\(=\)\({ }^{17} {C}_{9} \times 8 ! \times 7 !\)

Given radius \(=5 \mathrm{~cm}\) and \(\mathrm{OM=4~cm}\)

Now by Pythagoras Theorem,

\(r^2=O M^2+P M^2 \)

\(5^2=4^2+PM^2 \)

\(25=16+PM^2 \)

\(PM^2=25-16=9 \)

\(\mathrm{PM}=\sqrt{9}=3\)

Now length of chord \(\mathrm{PN}=3+3=6 \mathrm{~cm}\)\(~~~~~~\)(Since the angular bisector divides the chord into equal parts.)

Given,

\(\frac{d^{2} y}{d x^{2}}+3\left(\frac{d y}{d x}\right)^{2}=x^{2} \log \left(\frac{d^{2} y}{d x^{2}}\right)\)

For the given differential equation the highest order derivative is \(2\).

The given differential equation is not a polynomial equation because it involved a logarithmic term in its derivatives so, its degree is not defined.

Let \(f(x)=\frac{x^{2}}{1+x^{2}}\)

Clearly domain (f) =R

Let \(y=f(x) \Rightarrow y=\frac{x^{2}}{1+x^{2}}\)

\(\Rightarrow y+x^{2} y=x^{2} \Rightarrow x^{2}-x^{2} y=y\)

\(\Rightarrow x^{2}(1-y)=y \Rightarrow x=\pm \sqrt{\frac{y}{1-y}}\)

Clearly x will take real values,

If \(\frac{y}{1-y} \geq 0 \Rightarrow \frac{y-0}{y-1} \leq 0\)

⇒ 0 ≤ y < 1 ⇒ y ∈ [0, 1)

We know that,

\(\sec ^{2} \theta-\tan ^{2} \theta=1\)

\(\Rightarrow \sec ^{2} \theta=1+\tan ^{2} \theta\)

\(\Rightarrow \sec \theta=\sqrt{\left(1+\tan ^{2} \theta\right)}\)

Given:cot 35° = m

Find: The value ofsec 55°

\(\Rightarrow \sec 55^{\circ}=\sqrt{\left(1+\tan ^{2} 55^{\circ}\right)}\)

\(\Rightarrow \sec 55^{\circ}=\sqrt{\left(1+\tan ^{2}\left(90^{\circ}-35^{\circ}\right)\right.}\)

\(\Rightarrow \sec 55^{\circ}=\sqrt{\left(1+\cot ^{2} 35^{\circ}\right)}\)

\(\Rightarrow \sec 55^{\circ}=\sqrt{\left(1+\mathrm{m}^{2}\right)}\)

Formula used

\(A =\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|\)

\(\operatorname{det}( A )= a _1\left|\begin{array}{ll}b_2 & b_3 \\ c_2 & c_3\end{array}\right|- a _2\left|\begin{array}{ll}b_1 & b_3 \\ c_1 & c_3\end{array}\right|+ a _3\left|\begin{array}{ll}b_1 & b_2 \\ c_1 & c_2\end{array}\right|\)

\(=a_1\left(b_2 c_3-b_3 c_2\right)-a_2\left(b_1 c_3-b_3 c_1\right)+a_3\left(b_1 c_2-b_2 c_1\right)\)

Given

\(\left|\begin{array}{ccc}\cos C & \tan A & 0 \\ \sin B & 0 & -\tan A \\ 0 & \sin B & \cos C\end{array}\right|\)

\(=\cos C(0+\tan A \sin B)-\tan A(\sin B \cos C)\)

\(=\tan A \sin B \cos C-\tan A \sin B \cos C\)

\(=0\)

Given:

\(P\) divides \(\overrightarrow{A B}\) in the ratio \(3: 4\)

\(\Rightarrow \overrightarrow{O P}=\frac{3 \overrightarrow{O B}-4 \overrightarrow{O A}}{3-4}\)

\(=\frac{3 \vec{b}-4 \vec{a}}{-1}\)

\(\overrightarrow{O P}=4 \vec{a}-3 \vec{b}\)

\(Q\) divides \(\overrightarrow{B C}\) in the ratio \(2: 1\) both externally.

\(\Rightarrow\overrightarrow{O Q}=\frac{2 \overrightarrow{O C}-\overrightarrow{O B}}{2-1}\)

\(=2 \vec{c}-\vec{b}\)

\(\therefore \overrightarrow{P Q}= \overrightarrow{OQ}-\overrightarrow{OP}\)

\(=2 \vec{c}-\vec{b}-(4 \vec{a}-3 \vec{b})\)

\(=2 \overrightarrow{ c }+2 \overrightarrow{ b }-4 \overrightarrow{ a }\)

\(=2(\vec{c}+\vec{b}-2 \vec{a})\)