Mathematics Test - 36

\(y=\tan x, y=0\) and \(x=\frac{\pi}{4}\)

Let us first draw the graph for above, we get:

\(\therefore\) Area \(=\int_{0}^{\frac{\pi}{4}} y ~d x\)

\(=\int_{0}^{\frac{\pi}{4}} \tan x ~d x\)

\(=[\ln |\sec x|]_{0}^{\frac{\pi}{4}}\)

\(=\ln \left(\sec \frac{\pi}{4}\right)-\ln \left(\sec 0\right)\)

\(=\ln (\sqrt{2})-\ln 1\)

\(=\ln (2)^{(\frac{1}{2})}\)

\(=\frac{\ln (2)}{2}\)

We have \(\overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}-3 \overrightarrow{\mathrm{k}}, \overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{i}}+2 \overrightarrow{\mathrm{j}}-4 \overrightarrow{\mathrm{k}}\) and \(\mathrm{d}=9\)

Using the formula, Distance \(=\frac{|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}|}{|\overrightarrow{\mathrm{n}}|}\).

Distance \(=\frac{|(2 \overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}-3 \overrightarrow{\mathrm{k}}) \cdot(\overrightarrow{\mathrm{i}}+2 \overrightarrow{\mathrm{j}}-4 \overrightarrow{\mathrm{k}})-9|}{|\overrightarrow{\mathrm{i}}+2 \overrightarrow{\mathrm{j}}-4 \overrightarrow{\mathrm{k}}|}\)

\(=\frac{|2-2+12-9|}{\sqrt{1^{2}+2^{2}+(-4)^{2}}}\)

\(=\frac{3}{\sqrt{21}}=\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{7}}=\sqrt{\frac{3}{7}}\)

Let roots of the quadratic equation are \(\alpha, \beta\).

Given, \(\lambda=\frac{\alpha}{\beta}\) and \(\lambda+\frac{1}{\lambda}=1 \Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=1\)

\(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}=1 \text {. }\)

The quadratic equation is, \(3 m^2 x^2+m(m-4) x+2=0\)

\(\therefore \alpha+\beta=\frac{\mathrm{m}(4-\mathrm{m})}{3 \mathrm{~m}^2}=\frac{4-\mathrm{m}}{3 \mathrm{~m}} \text { and } \alpha \beta=\frac{2}{3 \mathrm{~m}^2}\)

Put these values in eq (1)

\(\frac{\left(\frac{4-m}{3 m}\right)^2}{\frac{2}{3 m^2}}=3\)

\(\Rightarrow(m-4)^2=18 \Rightarrow m=4 \pm \sqrt{18}\)

Therefore, least value is

\(4-\sqrt{18}=4-3 \sqrt{2}\)

For a, b in the set of children, we say that a and b are related if a has the same father as b.

Now, we will check for each property one by one.

The relation is obviously reflexive as a has the same father as a.

Therefore, aRa .... (1)

If a has the same father as b, then that means both a and b have the same father.

Therefore, b also has the same father as a. Thus, the relation is symmetric.

Therefore, aRb implies bRa .... (2)

Now, assume that aRb and bRc.

That means a has the same father as b and b has the same father as c. This implies that a, b and c have the same father.

Therefore, a has the same father as c. Thus, the relation is transitive.

Therefore, aRb, bRc implies aRc .... (3)

Thus, from (1), (2) and (3), we conclude that the given relation is an equivalence relation.

Given,

\(2 \tan ^{-1}\left[\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{1} x\right)\right]\)

\(=2 \tan ^{-1}\left[\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(90-\tan ^{-1} x\right)\right]\)

Let \(\tan ^{-1} x=\theta\)

\(=2 \tan ^{-1}[\operatorname{cosec} \theta-\tan (90-\theta)]\)

\(=2 \tan ^{-1}[\operatorname{cosec} \theta-\cot \theta]\)\(\quad\quad(\because \tan(90 - \theta = \cot \theta)\)

\(=2 \tan ^{-1}\left[\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right]\)\(\quad\quad(\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}, \cot \theta = \frac{\cos \theta}{\sin \theta}) \)

\(=2 \tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right]\)

\(=2 \tan ^{-1}\left[\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right]\)

\(=2 \tan ^{-1}\left[\tan \frac{\theta}{2}\right]\)

\(\theta=\tan ^{-1} x\)

Given, \(\frac{{dy}}{{dx}}=2^{{x}-1}\)

\(\Rightarrow \frac{d y}{d x}=\frac{2^{x}}{2}\) \(\Rightarrow 2 d y=2^{x} {dx}\)

Now, variables are separated.

Integrating both sides,

We get 
\(2 \int d y=\int 2^{x} d x \quad\left(\because \int a^{x} d x=\frac{a^{x}}{\operatorname{loga}}\right)\)

\(\Rightarrow 2 y=\frac{1}{\log 2} 2^{x}+c\)

We know, for no real roots, \(D=b^{2}-4 a c<0\)

Given equation is \(\left(a^{2}+b^{2}\right) x^{2}+2(a b+b d) x+c^{2}+d^{2}=0\)

Here, \(a=\left(a^{2}+b^{2}\right), b=2(a b+b d), c=c^{2}+d^{2}\)

\(\left[(a c+b d)^{2}\right]-4\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)<0\)

\(\Rightarrow\left[4\left(a^{2} c^{2}+b^{2} d^{2}+2 a b c d\right)\right]-4\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)<0\)

\(\Rightarrow 4\left[\left(a^{2} c^{2}+b^{2} d^{2}+2 a b c d\right)-\left(a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2}\right)\right]<0\)

\(\Rightarrow 4\left[a^{2} c^{2}+b^{2} d^{2}+2 a b c d-a^{2} c^{2}-a^{2} d^{2}-b^{2} c^{2}-b^{2}\right]<0\)

\(\Rightarrow 4\left[2 a b c d-b^{2} c^{2}-a^{2} d^{2}\right]<0\)

\(\Rightarrow-4\left[a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right]<0\)

\(\Rightarrow-4[a d-b c]^{2}<0\)

Therefore.

\((a d-b c)<0\)

or \({ad}<{bc}\)

\(\therefore a d \neq b c\)

Given: \(f(x)=\log (\sin x)\)

Let's examine the function \(f ( x )\) on the interval \(\left(0, \frac{\pi}{2}\right)\)

So, first let's calculate \(f ^{\prime}( x )\)

\(f^{\prime}(x)=\frac{1}{\sin x} \cdot \frac{d(\sin x)}{d x}=\cot x\)

As we know that \(\cot x>0 \forall x \in \left(0, \frac{\pi}{2}\right)\)

\(f^{\prime}(x)>0 \forall x \in\left(0, \frac{\pi}{2}\right)\)

As we know that for a strictly increasing function \(f^{\prime}(x)>0\) for all \(x \in(a, b)\) So, the given function \(f ( x )\) is a strictly increasing function on \(\left(0, \frac{\pi}{2}\right)\)

\(\therefore\) Option (A) is true

Now let's the function \(f ( x )\) on the interval \(\left(0, \frac{\pi}{2}\right)\)

As we know that, \(f^{\prime}(x)=\cot x\) and \(\cot x<0 \forall x \in\left(0, \frac{\pi}{2}\right)\)

\(f^{\prime}(x)<0 \forall x \in\left(0, \frac{\pi}{2}\right)\)

As we know that for a strictly decreasing function \(f^{\prime}(x)<0\) for all \(x \in(a, b)\) So, the given function \(f ( x )\) is a strictly decreasing function on \(\left(\frac{\pi}{2}, \pi\right)\)

\(\therefore\) Option (B) is true.

Let \(D\) be foot of perpendicular.

The direction ratio of line \(B C\) is:

\(3-4,5-7,3-1=-1,-2,2\)

Equation of \({BC}\),

\(\frac{{x}-4}{3-4}=\frac{{y}-7}{5-7}=\frac{{z}-1}{3-1} \)

\(\Rightarrow \frac{{x}-4}{-1}=\frac{{y}-7}{-2}=\frac{{z}-1}{2}={k}\) (let)

\(\Rightarrow {x}=-{k}+4, {y}=-2 {k}+7\) and \({z}=2 {k}+1\)

General point on \(B C\) be \(D=(-k+4,-2 k+7,2 k+1)\)

Direction ratio of line \(A D\) will be

\(-k+4-1,-2 k+7-0,2 k+1-3\)

\(-k+3,-2 k+7,2 k-2\)

Now, since the line \(B C\) and \(A D\) are perpendicular,

\((-\mathrm{k}+3) \times-1+(-2 {k}+7) \times-2+(2 {k}-2) \times 2=0 \)

\(\Rightarrow {k}-3+4 {k}-14+4 {k}-4=0 \)

\(\Rightarrow 9 {k}=21 \)

\(\Rightarrow {k}=\frac{7}{3} \)

\(\Rightarrow {D}=(-7 / 3+4,-2 \times(7 / 3)+7,2 \times(7 / 3)+1) \)

\(\therefore {D}=\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)

\(\begin{aligned} & \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\ & \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}\end{aligned}\)

the shortest distance between the lines

\(=\left|\frac{(\vec{a}-\vec{b}) \cdot\left(\overrightarrow{d_1} \times \overrightarrow{d_2}\right)}{\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|}\right|\)

\(=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right|\)

\(=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{12 \hat{i}-1 \hat{j}+0 \hat{k}}\right|\)

\(\frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right|\)

3=|λ−4|
λ−4=±3
λ=7,1
Sum of all possible values of λ is =8