Mathematics Test - 16

Given,

\(\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x\)

Let \(I=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x\)

Using \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\), we get

\(\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x\)

\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x\)

\(\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x\)

\(\Rightarrow I=\int_{0}^{\frac{\pi}{4}}(\log 2) d x-\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x\)

\(\Rightarrow I=(\log 2) \int_{0}^{\frac{\pi}{4}} 1 d x-I\)

\(\Rightarrow 2 \mathrm{I}=(\log 2)[\mathrm{x}]_{0}^{\frac{\pi}{4}}\)

\(\Rightarrow 2 \mathrm{I}=(\log 2)\left(\frac{\pi}{4}\right)\)

\(\Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2\)

\(\therefore \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x=\frac{\pi}{8} \log 2\)

Given,

\([\vec{a} \vec{b} \vec{c}]=1\)

Take L.H.S. \(\frac{\vec{a} \cdot \vec{b} \times \vec{c}}{\vec{c} \times \vec{a} \cdot \vec{b}}+\frac{\vec{b} \cdot \vec{c} \times \vec{d}}{\vec{a} \times \vec{b} \cdot \vec{c}}+\frac{\vec{c} \cdot \vec{a} \times \vec{b}}{\vec{b} \times \vec{d} \cdot \vec{a}}\)

\(=\frac{[\vec{a} b \vec{c}]}{[\vec{d} \vec{a} \vec{b}]}+\frac{\mid \vec{b} \overrightarrow{c a}]}{[\vec{a} \vec{b} \vec{c}]}+\frac{[\vec{c} a \vec{b}]}{[\vec{b} \vec{c} d]} \)

\(=1+1+1 \)

\(=3\)

Given: \(I = ∫ e^x cos x dx\)

Integrating by ILATE rule,

\(I = cos x ∫ e^x dx - ∫ (-sin x ∫ e^x dx) dx\)

\(I = e^x cos x + ∫ e^x sin x dx\)

\(I = e^x cos x + sin x ∫ e^x dx - ∫ (cos x ∫ e^x dx) dx\)

\(I = e^x cos x + e^x sin x - ∫ e^x cos x dx\)

\(I = e^x cos x + e^x sin x - I\)

\(2I = e^x cos x + e^x sin x\)

\(I=\frac{e^{x} \cos x+e^{x} \sin x}{2}\)

Given:

\(\frac{\operatorname{cosec} \theta+\sin \theta}{\operatorname{cosec} \theta-\sin \theta}=\frac{21}{15}\)

Using componendo & dividendo

\(\frac{\operatorname{cosec} \theta+\sin \theta+\operatorname{cosec}\theta-\sin \theta}{\operatorname{cosec} \theta+\sin \theta-\operatorname{cosec} \theta+\sin \theta}=\frac{21+15}{21-15}\)

\(\Rightarrow \frac{\operatorname{cosec} \theta}{\sin \theta}=\frac{36}{6}=6\)

\(\Rightarrow \operatorname{cosec}^{2} \theta=6\)

We know that,

\(1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\)

\(\cot ^{2} \theta=6-1=5\)

\(\tan \theta=\frac{1}{\sqrt{5}}\)

Given that,

\(\frac{2}{1+\cot ^{2} \theta}+\frac{4}{1+\tan ^{2} \theta}+2 \sin ^{2} \theta\)

We know that,

\(\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)

So,

\(\frac{2}{\operatorname{cosec}^{2} \theta}+\frac{4}{\sec ^{2} \theta}+2 \sin ^{2} \theta\)

\(\Rightarrow 2 \sin ^{2} \theta+4 \cos ^{2} \theta+2 \sin ^{2} \theta\)

\(\Rightarrow 4 \sin ^{2} \theta+4 \cos ^{2} \theta=4\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=4\)

\(\therefore\) The numerical value of \(\frac{2}{1+\cot ^{2} \theta}+\frac{4}{1+\tan ^{2} \theta}+2 \sin ^{2} \theta\) is \(4\).

Given:

\(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+2\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)-\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0\)

For the given differential equation the highest order derivative is 3.

Now, the power of the highest order derivative is 1.

We know that, the degree of a differential equation is the power of the highest derivative.

So, the degree of the differential equation is 1.

Given,

\(y=(\tan x)^{\tan x^{\tan x} \cdots(1)}\)

As we know,

\(\log _{b}\left(M^{p}\right)=p \log _{b}(M)\)

\(\log _{b}(M N)=\log _{b}(M)+\log _{b}(N)\)

Taking log on both sides, we get

\(\log y=\log (\tan x)^{\tan x^{\tan x}}\)

\(\log y=(\tan x)^{\tan x} \log (\tan x)\)

Again, taking log on both sides, we get

\(\log (\log y)=\log (\tan x)^{\tan x}+\log (\log (\tan x))\) \(\log (\log y)=\tan x \log (\tan x)+\log (\log (\tan x))\)

As we know,

\(\frac{d}{d x}[f(x) g(x)]=f(x) \frac{d}{d x}[g(x)]+g(x) \frac{d}{d x}[f(x)]\)

\(\frac{d}{d x} \tan x=\sec ^{2} x\)

\(\frac{d}{d x} \log x=\frac{1}{x}\)

Now, differentiating with respect to \(x\), we get

\(\frac{1}{\log y} \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\tan x\left(\frac{\sec ^{2} x}{\tan x}\right)+\sec ^{2} x \log (\tan x)+\frac{1}{\log (\tan x)} \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\)

\(\frac{d y}{d x}=y \log y\left[\tan x\left(\frac{\sec ^{2} x}{\tan x}\right)+\sec ^{2} x \log (\tan x)+\frac{1}{\log (\tan x)} \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\right]\)

At \(x=\frac{\pi}{4}\),

From equation (1), we get

\(y=\left(\tan \frac{\pi}{4}\right)^{\tan \frac{\pi}{4} \tan \frac{\pi}{4}}\)

\(y=1^{1^{1}}=1\)

\(\log y=\log 1=0\)

So, \(\frac{d y}{d x}\left(x=\frac{x}{4}\right)=0\)

We know that equation of circle is,

\(x^{2}+y^{2}+2 g x+2 f y+c=0 \text {...(1) }\)

whose, centre \(=(-g,-f)\) and radius \(=\sqrt{g^{2}+f^{2}-c}\).

Given,

\(x^{2}+y^{2}-x-3 y+\frac{5}{2}=0\)

On comparing above equation with equation (1), we get

\(2 g x=-x, 2 f y=-3 y\)

\(g=-\frac{1}{2}, f=-\frac{3}{2}, c=\frac{5}{2}\)

Then, Centre \(=\left(\frac{1}{2}, \frac{3}{2}\right)\)

\(\text { Radius }=\sqrt{\left(-\frac{1}{2}\right)^{2}+\left(-\frac{3}{2}\right)^{2}-\left(\frac{5}{2}\right)} \)

\(=\sqrt{\frac{1}{4}+\frac{9}{4}-\frac{5}{2}} \)

\(=\sqrt{\frac{1+9-10}{4}} \)

\(=0\)

Thus, we can say that a circle whose radius is zero represents a point.

If a line makes \(\alpha, \beta, \gamma\) angle with the co-ordinate axis. Then, \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)

As we know, \(\cos ^{2} \alpha=1-\sin ^{2} \alpha\)

\(\left(1-\sin ^{2} \alpha\right)+\left(1-\sin ^{2} \beta\right)+\left(1-\sin ^{2} \gamma\right)=1 \)

\(\Rightarrow 3-\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=1 \)

\(\Rightarrow\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=2\)

\(\therefore\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=2\)