Mathematics Test-13

Let \(\tan ^{-1} \sqrt{3}=x .\)

Then,

tan \(x=\sqrt{3}=\tan \frac{\pi}{3}\).

\(\therefore \tan ^{-1} \sqrt{3}=\frac{\pi}{3}\)

Let \(\sec ^{-1}(-2)=y .\)

Then,

\(\sec y=-2=-\sec \left(\frac{\pi}{3}\right)=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \frac{2 \pi}{3}\)

\(\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}\)

Therefore, \(\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}\)

According to the question,

\(a+6 d=\frac{1}{5}\)......\((1)\)

\(a+4 d=\frac{1}{7}\).......\((2)\)

Subtracting the two equations we get,

\(2 d=\frac{1}{5}-\frac{1}{7}=\frac{2}{35}\)

\(\Rightarrow d=\frac{1}{35}\)

Substituted inequation\((1)\) to get as:

\(a=\frac{1}{5}-6 d\)

\(\Rightarrow a=\frac{1}{5}-\frac{6}{35}\)

\(\Rightarrow a=\frac{1}{35}\)

Sum of first \(n\) terms = \(\frac{n}{2}[2d+(n-1)d]\)

Sum of first \(13\) terms = \(\frac{13}{2}\left[2×\ \frac{1}{35}+(12)× \frac{1}{35}\right]\)

\(=3\left[\frac{1}{35}+\frac{6}{35}\right]\)

\(=13× \frac{7}{35}\)

=\(\frac{13}{5}=2.6\)

Given,

The system of equations

\(2 x+y-3 z=5\)

\(3 x-2 y+2 z=5 \text { and }\)

\(5 x-3 y-z=16\)

So, \(A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 3 & -2 & 2 \\ 5 & -3 & -1\end{array}\right]\)

\(\operatorname{det}(A)=|A|=2 \times\{(-2 \times-1)-(-3 \times 2)\}-1 \times\{(3 \times-1)-(2 \times 5)\}+(-3) \times\{(3 \times-3)-(5 \times-2)\}\)

\(\Rightarrow|A|=2 \times(2+6)-1 \times(-3-10)-3 \times(-9+10)\)

\(\Rightarrow|A|=16+13-3=26\)

\(\therefore|A| \neq 0\)

So, system is consistent having unique solution.

Given,

Quadratic equations \(3 x^{2}+ax+1=0\) and \(2 x^{2}+bx+1=0\) have a common root.

Let \(\alpha\) be the common root.

\(3 \alpha^{2}+\mathrm{a} \alpha+1=0 \ldots(1)\)

\(2 \alpha^{2}+\mathrm{b} \alpha+1=0 \ldots(2)\)

Multiplying \((1)\) by \(2\) and \((2)\) by \(3 .\) Further solving those two equations for \(\alpha\) gives.

\(\alpha=\mathrm{b}-\mathrm{a} \ldots(3)\)

Substituting \((3)\) in \((1)\), we get

\(\Rightarrow 2(\mathrm{~b}-\mathrm{a})^{2}+\mathrm{b}(\mathrm{b}-\mathrm{a})+1=0\)

\(\Rightarrow 3 b^{2}+2 \mathrm{a}^{2}-5 \mathrm{ab}+1=0\)

\(\therefore 5 a b-3 b^{2}-2 a^{2}=1\)

The equation of the line in intercept form is given by:

\(\frac{x}{a}+\frac{y}{b}=1\)

It is given that both the intercepts are equal so we have \(a = b\).

Therefore, the equation of the line is given as follows:

\(\frac{x}{a}+\frac{y}{a}=1\)

\(\Rightarrow\frac{x+y}{a}=1\)

\(\Rightarrow x+y=a\)

\(\Rightarrow y=-x+a\)

Therefore, comparing with the slope-intercept form \(y=mx+c\)we observe that the slope of the line is -1.

We know that slope is given by \(m=tan \theta\)

Where, \(\theta\)is the angle made with the positive direction of the \(X\)-axis.

Therefore, in the given case we get,

\(\tan \theta=-1\)

\(\Rightarrow \theta=135^{\circ}\)

Hence the correct option is (D).

Let \((r+1)^{\text {th }}\) term be the greatest term in the expansion of \((1+x)^{2 n} \cdot\)

Then,

\(T_{r+1}={ }^{2 n} C_{r} x^{2 n-r}\)

The coefficient of \(T_{r+1}\) is \({ }^{2 n} C_{r}\).

Clearly, \({ }^{2 n} C_{r}\) will be greatest,

if \(r=\frac{2 n}{2}=n\)

Thus, \(T_{n+1}={ }^{2 n} C_{n} x^{n}\) has the greatest coefficient.

Now, \(T_{n+1}\) will be the greatest term.

\(\Rightarrow T_{n+1}>T_{n} \quad\)

\(\Rightarrow \frac{T_{n+1}}{T_{n}}>1\) and \(\frac{T_{n+2}}{T_{n+1}}<1\)

\(\Rightarrow \frac{{ }^{2 n} C_{n} x^{n}}{2 n_{n-1} x^{n-1}}>1\) and \(\frac{2 \mathrm{n} C_{n+1} x^{n+1}}{2 n C_{n} x^{n}}<1\)

\(\Rightarrow \left(\frac{n+1}{n}\right) x>1 \) and \(\left(\frac{n}{n+1}\right) x<1\)

\(\Rightarrow x>\frac{n}{n+1} \) and \(x<\frac{n+1}{n}\)

\(\Rightarrow \frac{n}{n+1}\)

Let \(I=\int \log \left(x+\frac{1}{x}\right) d x\)

\(\Rightarrow I=\int \log \left(\frac{x^{2}+1}{x}\right) d x\)

\(\Rightarrow I=\int \log \left(x^{2}+1\right) d x-\int \log x d x\left(\because \log \frac{a}{b}=\log a-\log b\right)\)

\(\Rightarrow I=\log \left(x^{2}+1\right) \cdot x-\int \frac{2 x}{x^{2}+1} \cdot x d x-(x \log x-x)+C\) (Integrating by parts.)

\(\Rightarrow I=x \log \left(x^{2}+1\right)-2 \int \frac{x^{2}}{x^{2}+1} d x-x \log x+x+C\)

\(\Rightarrow I=x \log \left(x^{2}+1\right)-2 \int\left(1-\frac{1}{x^{2}+1}\right) d x-x \log x+x+C\)

\(\Rightarrow I=x \log \left(x^{2}+1\right)-2 x+2 \tan ^{-1} x-x \log x+x+C\)

\(\Rightarrow I=x \log \left(\frac{x^{2}+1}{x}\right)-x+2 \tan ^{-1} x+C\)

\(\Rightarrow I=x \log \left(x+\frac{1}{x}\right)-x+2 \tan ^{-1} x+C\)

Any plane passing through \((0,7,-7)\) is \(a(x-0)+b(y-7)+c(z+7)=\) \(0 \ldots\) (i)

If (i) contains the given line then it must pass through the point \((-1,3,-2)\) and must be parallel to the given line.

If (i) passes through \((-1,3,-2)\) then,

\(a(-1-0)+b(3-7)+c(-2+7)=0\)

⇒ \(a+4 b-5 c=0 \ldots\) (ii)

If (i) is parallel to the given line then,

\((-3) a+2 b+1 . c=0\)

⇒ \(-3 a+2 b+c=0 \ldots\) (iii)

By cross multiplication of (ii) and (iii), we get

\(\frac{a}{(4+10)}=\frac{b}{(15-1)}=\frac{c}{(2+12)}\)

⇒ \(\frac{\mathrm{a}}{14}=\frac{\mathrm{b}}{14}=\frac{\mathrm{c}}{14}\)

⇒ \(\frac{\mathrm{a}}{1}=\frac{\mathrm{b}}{1}=\frac{\mathrm{c}}{1}=\mathrm{k}\)

⇒ \(a=k, b=k, c=k\)

Putting \(a=k, b=k\) and \(c=k\) in (i), we get

The required equation of the plane as \(\mathrm{k}(x-1)+\mathrm{k}(y-7)+\mathrm{k}(z-7)=0\)

⇒ \(x+(y-7)+(z+7)=0\)

⇒ \(x+y+z=0\)

Total number of teams that play the world cup = \(12\) teams

One important formula used in solving is \({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

All \(12\) teams are divided into two groups. In one group we have \(6\) teams and \(2\) teams will play one match. So, for two groups having \(6\) teams. Putting \(n = 6\) and \(r = 2\), we get:

Matches in the first round:

\(={ }^{6} C_{2}+{ }^{6} C_{2}\)

\(=\frac{6 !}{4 ! 2 !}+\frac{6 !}{4 ! 2 !}\)

\(=\frac{6 \times 5 \times 4 !}{4 ! 2 !}+\frac{6 \times 5 \times 4 !}{4 ! 2 !}\)

\(=15+15\)

\(=30\)

From each group three top teams will qualify for the next round and each team will play against another one. So, we have only \(6\) teams left after the first round and \(2\) teams will play a match. By using combination formula and putting \(n = 6\) and \(r = 2\), we get:

Match in the second round:

\(={ }^{6} C_{2}\)

\(=\frac{6 !}{4 ! 2 !}\)

\(=\frac{6 \times 5 \times 4 !}{4 ! 2 !}\)

\(=15\)

We get top \(4\) from this round where each team will play against the other three. Now, putting \(n = 4\) and \(r = 2\) in combination formula, we get:

Matches in the semi finals:

\(={ }^{4} C_{2}\)

\(=\frac{4 !}{2 ! 2 !}\)

\(=\frac{4 \times 3 \times 2 !}{2 ! 2 !}\)

\(=6\)

Final is best of three so if one team wins the first match and second match then there will not be a third match. Hence, the minimum number of matches are:

The total number of matches = Matches in the first round + Matches in the second round + Matches in the semi finals round + \(2\) matches in finals

Total number of matches \(=30+15+6+2=53\)

Given:

\(\left(\frac{d y}{d x}\right)=\left(\frac{y}{x}\right)-\cos ^{2}\left(\frac{y}{x}\right)\)

\(\operatorname{Let}\left(\frac{y}{x}\right)=t\)

\(\Rightarrow\left(\frac{d y}{d x}\right)=x\left(\frac{d t}{d x}\right)+t\)

\(\Rightarrow x\left(\frac{d t}{d x}\right)+t=t-\cos ^{2} t\)

\(\Rightarrow x\left(\frac{d t}{d x}\right)+\cos ^{2} t=0\)

\(\Rightarrow\left[\frac{(x \cdot d t)}{\left(d x \cdot \cos ^{2} t\right)}\right]+1=0\)

\(\Rightarrow\left[\frac{(d t)}{\left(\cos ^{2} t\right)}\right]+\left(\frac{d x}{x}\right)=0\)

On integrating,

\(\Rightarrow \int \sec ^{2} t d t+\int\left(\frac{d x}{x}\right)=0\)

\(\Rightarrow \tan \left(\frac{y}{x}\right)+\log x=c\)

\(\Rightarrow \tan \left(\frac{y}{x}\right)+\log x=c\)

when \(x=1, y=\left(\frac{\pi}{4}\right)\) hence

\(\Rightarrow \tan \left(\frac{\pi}{4}\right)=c \text { i.e. } c=1\)

\(\Rightarrow \tan \left(\frac{y}{x}\right)+\log x=\log e\)

\(\Rightarrow \tan \left(\frac{y}{x}\right)+\log \left(\frac{e}{x}\right)\)

\(\therefore\) The equation of the curve is \(y=x \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]\)

The given equation of the ellipse is \(2 x^{2}+3 y^{2}=6\).

Divide throughout by 6 to get the equation in the standard form.

\(\frac{x^{2}}{3}+\frac{y^{2}}{2}=1\)

Therefore, \(a^{2}=3\) and \(b^{2}=2\)

Now, for any ellipse using the relation we can write:

\(b^{2}=a^{2}\left(1-{e}^{2}\right)\)

\(\Rightarrow2=3\left(1-{e}^{2}\right)\)

\(\Rightarrow e^{2}=\frac{1}{3}\)

\(\Rightarrow{e}=\frac{1}{\sqrt{3}}\)

Therefore, the eccentricity of the given ellipse is \(\frac{1}{\sqrt{3}}\).

Hence the correct option is (A).

Given,

\(\frac{\cos x-\sin x+1}{\cos x+\sin x-1}\)

\(=\frac{\cos x-(\sin x-1)}{\cos x+(\sin x-1)} \times \frac{\cos x-(\sin x-1)}{\cos x-(\sin x-1)}\)

\(=\frac{[\cos x-(\sin x-1)]^{2}}{\cos ^{2} x-(\sin x-1)^{2}}\)

\(=\frac{\cos ^{2} x+(\sin x-1)^{2}-2 \cos x(\sin x-1)}{\cos ^{2} x-\left(\sin ^{2} x-2 \sin x+1\right)}\)

\(=\frac{\cos ^{2} x+\sin ^{2} x-2 \sin x+1-2 \cos x(\sin x-1)}{1-\sin ^{2} x-(\sin 2 x-2 \sin x+1)}\)

\(=\frac{2-2 \sin x-2 \cos x(\sin x-1)}{1-\sin 2 x-\sin ^{2} x+2 \sin x-1}\)

\(=\frac{2(1-\sin x)+2 \cos x(1-\sin x)}{2 \sin x-2 \sin 2 x}\)

\(=\frac{(1-\sin x)(1+\cos x)}{\sin x\left(1-\sin ^{2} x\right)}\)

\(=\frac{1+\cos x}{\sin x}\)\(=\operatorname{cosec} x+\cot x\)

Given,

\(y=f(x)=x^{3}\)

\(\Rightarrow y^{\prime}=f^{\prime}(x)=3 x^{2}\)

\({~m}={f}^{\prime}(1)=3 \times 1^{2}=3\),

Equation of the tangent at \((1,1)\) will be:

\((y-b)=m(x-a)\)

\(\Rightarrow(y-1)=3(x-1)\)

\(\Rightarrow y-1=3 x-3\)

\(\Rightarrow 3 x-y-2=0\)

Objective function \(: \mathrm{Z}=-3 \mathrm{x}+4 \mathrm{y}\)

We have to minimize \(\mathrm{Z}\) on constraints

\(\mathrm{x}+2 \mathrm{y} \leq 8 \)

\(3 \mathrm{x}+2 \mathrm{y} \leq 12 \)

\(\mathrm{x} \geq 0, \mathrm{y} \geq 0\)

After plotting inequalities, we got feasible region as shown in the image.

Now, there are 3 corner points \((0,4),(2,3)\) and \((4,0)\).

At \((0,4)\), value of \(Z=-3 \times 0+4 \times 4=16\)

At \((2,3)\), value of \(\mathrm{Z}=-3 \times 2+4 \times 3=6\)

\(\operatorname{At}(4,0)\), value of \(\mathrm{Z}=-3 \times 4+4 \times 0=-12\)

So, minimum value of \(\mathrm{Z}=-12\).

Given:

Group of 6 people having weights 45, 50, 65, 72, 63, 35 (in kgs)

Formula Used:

Mean \(=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}\)

Standard deviation \(=\sqrt{\left[\frac{(\sum \mathrm{X}^{2})}{\mathrm{N}}\right]}\) where \(\sum x^{2}\) is the summation of the square of deviation of all observations from Mean.

Sum of all observations = (45 + 50 + 65 + 72 + 63 + 35) = 330

Mean \(= \frac{330}{6}\)

⇒ Mean = 55

\(\Sigma x^{2}=(100+25+100+289+64+400)\)

\(\Rightarrow \Sigma x^{2}=978\)

Standard Deviation \((\sigma)=\sqrt{\left(\frac{978}{6}\right)}\)

\(\Rightarrow \sigma=\sqrt{163}\)

∴ The standard deviation of given data set of weights is \(\sqrt{163}\).