Mathematics Test - 22

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

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

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

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

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

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

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

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

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

If a set containing n elements then number of elements in their subset = 2ⁿ

For a given set \(\mathrm{A}\), a set \(\mathrm{B}\) is a subset of set \(\mathrm{A}\) if all elements of set \(\mathrm{B}\) are also elements of set \(A\). Set \(A\) is called the super-set of set \(B\). Null set "{ }" or "\(\phi\)" is a subset of all sets.

As we know,

The equation of the line that passes through the two points \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is given by the formula,

\(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)

The line passing through the points \((1,2,-1)\) and \((3,-1,2)\) is,

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

Let, \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=\lambda \)

\(\therefore x=2 \lambda+1, y=-3 \lambda+2 \text { and } z=3 \lambda-1\)

Line meets the yz-plane,

So, \(x =0 \)

\(\Rightarrow 2 \lambda+1=0\)

\(\Rightarrow 2 \lambda=-1\)

\(\Rightarrow \lambda=\frac{-1 }{ 2}\)

Now,

\(y=-3 \lambda+2\)

\(\Rightarrow y=-3 \times \frac{-1}{2}+2\)

\(\Rightarrow y =\frac{3}{2}+2\)

\(\Rightarrow y=\frac{7 }{ 2}\)

Now,

\(z=3 \lambda-1\)

\(\Rightarrow z=3 \times \frac{-1}{2}-1\)

\(\Rightarrow z =\frac{-3}{2}-1\)

\(\Rightarrow z=\frac{-5 }{ 2}\)

\(\therefore\) Point \((x, y, z)\) is \(\left(0, \frac{7}{2},-\frac{5}{2}\right)\).

Given,

\(\mathrm{S}_{\mathrm{n}}=\mathrm{n}+12\)

By substituting n = 2 in the given equation, we get

\(\mathrm{S}_{2}=2+12=14\)

Similarly by substituting n = 3 in the given equation, we get

\( S_{3}=3+12=15\)

\(\mathrm{S}_{\mathrm{n}}\) denotes the sum of the first \(\mathrm{n}\) terms of the sequence.

\(\mathrm{T}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}\)

For n = 3 and n= 2,

\(\Rightarrow a_{3}=S_{3}-S_{2}\)

\(\Rightarrow a_{3}=15-14\)

\(\Rightarrow a_{3}=1\)

Let P(h,k) be the orthocentre of △ABC

Then

\(\mathrm{h}=\frac{\sin \mathrm{t}+\cos \mathrm{t}+\mathrm{a}}{3}, \mathrm{k}=\frac{-\cos \mathrm{t}+\sin \mathrm{t}+\mathrm{b}}{3}\)

(orthocentre coincide with centroid)

\(\begin{aligned}& \therefore(3 h-a)^2+(3 k-b)^2=2 \\& \therefore\left(h-\frac{a}{3}\right)^2+\left(k-\frac{b}{3}\right)^2=\frac{2}{9}\end{aligned}\)

\(\because\) orthocentre lies on circle with centre \(\left(1, \frac{1}{3}\right)\)

\(\begin{aligned}& \therefore \mathrm{a}=3, \mathrm{~b}=1 \\& \therefore \mathrm{a}^2-\mathrm{b}^2=8\end{aligned}\)

\(\begin{aligned}& 3+2 \sqrt{-54}=3+6 \sqrt{6} i \\& \text { Let } \sqrt{3+6 \sqrt{6} i}=a+i b \\& \Rightarrow a^2-b^2=3 \text { and } a b=3 \sqrt{6} \\& \Rightarrow a^2+b^2=\sqrt{\left(a^2-b^2\right)^2+4 a^2 b^2}=15\end{aligned}\)

So, \(a= \pm 3\) and \(b= \pm \sqrt{6}\)

\(\sqrt{3+6 \sqrt{6}} i= \pm(3+\sqrt{6} i)\)

Similarly, \(\sqrt{3-6 \sqrt{6}} i= \pm(3-\sqrt{6} i)\)

\(\operatorname{lm}(\sqrt{3+6 \sqrt{6}} i-\sqrt{3-6 \sqrt{6}} i)= \pm 2 \sqrt{6}\)

Given,

The line passing through the points \((3,2)\) and \((1,-1)\)

Equation of a line passing through \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Therefore,

\(\frac{y-2}{x-3}=\frac{-1-2}{1-3} \)\(\quad\quad(\because \frac{y-2}{x-3}=\frac{-1-2}{1-3} )\)

\(3 x-2 y-5=0\)

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

\(\Rightarrow\) Slope \(\left(m_{1}\right)=\frac{3}{2}\) and \(c_{1}=\frac{-5}{2}\)

Now for the slope of the perpendicular line \(\left(\mathrm{m}_{2}\right)\)

\(\mathrm{m}_{1} \times \mathrm{m}_{2}=-1 \)

\(\Rightarrow \frac{3}{2} \times \mathrm{m}_{2}=-1 \)

\(\Rightarrow \mathrm{m}_{2}=\frac{-2}{3}\)

Given:

\(\lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

As we know,

\(\lim _{x \rightarrow a}[f(x) \cdot g(x)]=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)\)

\(\therefore \lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

\(=\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}\)

\(=1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{x} \quad\) \( (\because\lim _{x \rightarrow 0} \frac{\sin x}{x}=1)\)

\(=\lim _{x \rightarrow 0} \frac{\log (1+(-x))}{-(-x)}\)

\(=-1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{(-x)}\)

\(=-1 \times 1 \quad\) \((\because\lim _{x \rightarrow 0} \frac{\log x}{x}=1)\)

\(=-1\)

\(\begin{aligned}& \mathrm{S}=\{1,2,3,4,5,6,9\} \\& 3 \mathrm{n} \text { Type numbers } 3,6,9 \\& 3 \mathrm{n}-1 \text { Type numbers } 2,5 \\& 3 \mathrm{n}-2 \text { Type numbers } 1,4 \\& \text { Let } \mathrm{N}_{\mathrm{P}}=\text { Number of Subset of } \mathrm{S} \\& \text { containing } \mathrm{p} \text { element which are not divisible by } 3 \text {. } \\& \text { For } \mathrm{P}=1 \\& { }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1=4\end{aligned}\)

\(\begin{aligned}& \text { For } \mathrm{P}=2 \\& { }^3 \mathrm{C}_1{ }^2 \mathrm{C}_1+{ }^3 \mathrm{C}_1^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2+{ }^2 \mathrm{C}_2=14 \\& \text { For } \mathrm{P}=3 \\& \left.{ }^3 \mathrm{C}_1{ }^2 \mathrm{C}_2+{ }^2 \mathrm{C}_2\right)+{ }^3 \mathrm{C}_2\left({ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1\right)+{ }^2 \mathrm{C}_2{ }^2 \mathrm{C}_1 \\& +{ }^2 \mathrm{C}_1{ }^2 \mathrm{C}_2=22\end{aligned}\)

\(\begin{aligned}& \text { For } \mathrm{P}=4 \\& { }^3 \mathrm{C}_1\left[{ }^2 \mathrm{C}_2{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1{ }^2 \mathrm{C}_2\right]+{ }^3 \mathrm{C}_2\left({ }^2 \mathrm{C}_2+{ }^2 \mathrm{C}_2\right) \\& +{ }^3 \mathrm{C}_3\left({ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1\right)=22 \\& \text { For } \mathrm{P}=5 \\& { }^3 \mathrm{C}_2\left({ }^2 \mathrm{C}_2{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1{ }^2 \mathrm{C}_2\right)+{ }^3 \mathrm{C}_3\left({ }^2 \mathrm{C}_2+{ }^2 \mathrm{C}_2\right)=14 \\& \text { For } \mathrm{P}=6 \\& { }^3 \mathrm{C}_3\left({ }^2 \mathrm{C}_2{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1{ }^2 \mathrm{C}_2\right)=4\end{aligned}\)

Total Subsets

\(=4+14+22+22+14+4=80\)