Mathematics Test - 30

since 3 does not occur in \(1000,\) so we have to count the number of times 3 occurs when we list the integers from 1 to 999 . Any number between 1 and 999 is of the form abc, where \(0 \leq a\) and \(b, c \leq 9\). Clearly, number of times 3 occurs \(=\) (number of numbers in which 3 occurs exactly at one place) \(+2 \times(\) number of numbers in which 3 occurs exactly at two places) \(+3 \times\) (number of numbers in which 3 occurs exactly at three places) Now, Number of numbers in which 3 occurs exactly at one place:
since 3 can occur at one place in \({ }^{3} \mathrm{C}_{1}\) ways and each of the remaining two places can
be filled in 9 ways;
So, number of numbers in which 3 occurs exactly at one place \(={ }^{3} \mathrm{C}_{1} \times 9 \times 9\)
Number of numbers in which 3 occurs exactly at two places:
since 3 can occur exactly at two places in \({ }^{3} \mathrm{C}_{2}\) ways and the remaining place can be
filled in 9 ways; So, number of numbers in which 3 occurs exactly at two places \(={ }^{3} \mathrm{C}_{2} \times 9\)
Number of numbers in which 3 occurs at all the three places:
since 3 can occur in all the three digits in one way only, so the number of numbers in which 3 occurs at all the three places is one.
Hence, number of times 3 occurs \(={ }^{3} \mathrm{C}_{1} \times 9 \times 9+2\left({ }^{3} \mathrm{C}_{2} \times 9\right)+3 \times 1=300\)

We have \(2(2 b)=a+3 c\) and \(b^{2}=a c\)
\(\therefore \quad \frac{1}{16}(a+3 c)^{2}=a c\)
\(\Rightarrow\)
\(a^{2}+9 c^{2}+6 a c=16 a c\)
\(\Rightarrow \quad a^{2}-10 a c+9 c^{2}=0\)
\(\Rightarrow \quad(a-c)(a-9 c)=0 \Rightarrow a=9 c[\because a \neq c]\)
\(\Rightarrow\)
\(\frac{c}{a}=\frac{1}{9}\) or \(r^{2}=\frac{1}{9} \Rightarrow r=\pm \frac{1}{3}\)

\(I_{n}=\int_{0}^{\pi / 4}\left(\sec ^{2} \theta-1\right) \tan ^{n-2} \theta d \theta\)
\(=\int_{0}^{\pi / 4} \sec ^{2} \theta \tan ^{n-2} \theta d \theta-\int_{0}^{\pi / 4} \tan ^{n-2} \theta d \theta\)
\(=\left[\frac{\tan ^{n-1} \theta}{n-1}\right]_{0}^{\pi / 4}-\ln -2 \Rightarrow L_{n}+\ln -2=\frac{1}{n-1}\)
Hence, \(\lg +16=\frac{1}{8-1}=\frac{1}{7}\)

As the centre of \(C_{3}\). lies above the \(x\) -axis, we take \(\lambda=\) \(-\sqrt{3}\) and thus an equation of \(C_{3}\) is \(x^{2}+y^{2}-x-\sqrt{3} y=0\) Since \(C_{1}\) and \(C_{3}\) intersect and are of unit radius, their common tangents are parallel to the line joining their centres (0,0) and \((1 / 2, \sqrt{3} / 2)\)
So, let the equation of a common tangent be \(\sqrt{3} x-y+k=0\)
It will touch \(C_{1},\) if \(\left|\frac{k}{\sqrt{3+1}}\right|=1 \Rightarrow k=\pm 2\)
From the figure we observe that the required tangent makes positive intercept on the \(y\) -axis and negative on the \(x\) -axis and hence its equation is \(\sqrt{3} x-y+2=0\)

\(\mathrm{As}^{n} \mathrm{C}_{4} \cdot{ }^{n} \mathrm{C}_{5}\) and \({ }^{n} \mathrm{C}_{6}\) are in A.P.
\(\begin{aligned} & 2\left({ }^{n} C_{5}\right)={ }^{n} C_{4}+{ }^{n} C_{6} \\ \Rightarrow \quad 2 &=\frac{{ }^{n} C_{4}}{{ }^{n} C_{5}}+\frac{{ }^{n} C_{6}}{{ }^{n} C_{5}} \\ &=\frac{n !}{4 !(n-4) !} \frac{5 !(n-5) !}{n !}+\frac{n !}{6 !(n-6) !} \frac{5 !(n-5) !}{n !} \\ \Rightarrow \quad 2 &=\frac{5}{n-4}+\frac{n-5}{6} \quad \Rightarrow \quad n^{2}-21 n+98=0 \\ \Rightarrow \quad & n=7,14 \end{aligned}\)

\(\cos ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=\log a\)
\(\Rightarrow \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\cos \log a=A(\operatorname{say})\)
Putting \(u=y / x\) and applying componendo and dividendo, we have
\(-1 / u^{2}=(A+1) /(A-1)\)
\(u^{2}=(1-A) /(1+A)\)
\(\Rightarrow y x=\sqrt{(1-A) /(1+A)}\)
\(\Rightarrow \quad x \mathrm{d} y / \mathrm{d} x-y=0\)
\(\Rightarrow \mathrm{d} y / \mathrm{d} x=y / x\)

\(S=(x-1)^{10}+(x-1)^{9}(x+1)+(x-1)^{8}(x+1)^{2}+\ldots \ldots+(x+1)^{10}\)
\(=(x-1)^{10}\left\{\frac{\left(\frac{x+1}{x-1}\right)^{11}-1}{\frac{x+1}{x-1}-1}\right\}\)
\(=\frac{1}{2}\left\{(x+1)^{11}-(x-1)^{11}\right.\)
\(={ }^{11} C_{1} x^{10}+{ }^{11} C_{3} x^{8}+{ }^{11} C_{5} x^{6}+{ }^{11} C_{7} x^{4}+{ }^{1} C_{9} x^{2}+1\)
\(\Rightarrow\) The constant term in expansion of \(\mathrm{S}=1\)

\(T_{r+1}=\sqrt[3 n]{C}_{r} x^{3 n-r}\left[\sin \left(\frac{1}{n}\right)\right]^{r} x^{-2 r}\)
For this term to be independent of \(x,\) set
\(3 n-r-2 r=0 \Rightarrow r=n\)
\(\therefore \quad a_{n}={ }^{3 n} C_{n} \sin ^{n}\left(\frac{1}{n}\right)\)
Now, \(\frac{n ! a_{n}}{3 n}=\sin ^{n}\left(\frac{1}{n}\right) \rightarrow 0\) as \(n \rightarrow \infty\)

Centre of the circle is \((g,-g)\) and radius is \(1 g\) ।. If it touches the line \(x \cos \alpha+y \sin \alpha=2\),
then perepndicular distance of centre from line = radius of circle
\(g \cos \alpha-g \sin \alpha-2=\pm g\)
\(\Rightarrow \quad g(\cos \alpha-\sin \alpha \pm 1)=2\)
\(\Rightarrow \quad g=\frac{2}{\cos \alpha-\sin \alpha \pm 1}\)
which is satisfied by ( 2 ).