Mathematics Test - 14

\(f(x)=\frac{\sin ^{2} x+4 \sin x+5}{2 \sin ^{2} x+8 \sin x+8}\)
\(\Rightarrow f(x)=\frac{\frac{1}{2}\left(2 \sin ^{2} x+8 \sin x+10\right)}{2 \sin ^{2} x+8 \sin x+8}\)
\(\Rightarrow f(x)=\frac{1}{2}\left[1+\frac{2}{2 \sin ^{2} x+8 \sin x+8}\right]\)
\(\Rightarrow f(x)=\frac{1}{2}\left[1+\frac{2}{2(\sin x+2)^{2}}\right]\)
\(\Rightarrow f(x)=\frac{1}{2}\left[1+\frac{1}{(\sin x+2)^{2}}\right]\)
Minimum value of \(\frac{1}{(\sin x+2)^{2}}=\frac{1}{9}\) when \(\sin x=1\)
And maximum value of \(\frac{1}{(\sin x+2)^{2}}=1\) when \(\sin x=-1\)

\(\Rightarrow \lambda \in(-\infty,-7] \cup[9,12]\)
but in question it os given that \(\lambda \geq 0\)
so \(\lambda \in[9,12]\)
So greatest value of \(\lambda\) is 12

Let \(\frac{1}{\cos ^{2} x}=\lambda \geq 1\)
(as \(\left.0 \leq \cos ^{2} x \leq 1\right)\)
\(\therefore \lim _{x \rightarrow \frac{\pi}{2}}\left[1^{1 / \cos ^{2} x}+2^{1 / \cos ^{2} x}+3^{1 \cos ^{2} x}+\ldots \ldots \ldots+10^{1 / \cos ^{2} x}\right]^{\cos ^{2} x}\)
\(=\lim _{\lambda \rightarrow \infty}\left(1^{\lambda}+2^{\lambda}+\ldots+10^{\lambda}\right)^{\frac{1}{\lambda}}\)
\(=\left(10^{\lambda}\right)^{\frac{1}{\lambda}} \lim _{\lambda \rightarrow \infty}\left[\left(\frac{1}{10}\right)^{\lambda}+\left(\frac{2}{10}\right)^{\lambda}+\ldots+1\right]^{\frac{1}{\lambda}}\)
\(=10[0+0+\ldots+1]^{0}=10\)

\(\vec{a} \perp(\vec{b}+\vec{c}), \vec{b} \perp(\vec{c}+\vec{a}), \vec{c} \perp(\vec{a}+\vec{b})\)
\(\vec{a} \cdot(\vec{b}+\vec{c})=0, \vec{b} \cdot(\vec{c}+\vec{a})=0, \vec{c} \cdot(\vec{a}+\vec{b})=0\)
\(2(a \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=2 \sum \vec{a} \cdot \vec{b}=0\)
\(\vec{a}+\vec{b}+\vec{c}^{2}=\sum a^{2}+2 \sum \vec{a} \cdot \vec{b}\)
\(|\vec{a}+\vec{b}+\vec{c}|^{2}=\sum a^{2}+0=4^{2}+4^{2}+5^{2}=57\)
\(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{57}\)

\(\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+\ldots+a_{q}}=\frac{p^{2}}{q^{2}}\)
\(\Rightarrow \frac{\frac{p}{2}\left[2 a_{1}+(p-1) d\right]}{\frac{q}{2}\left[2 a_{1}+(q-1) d\right]}=\frac{p^{2}}{q^{2}}\)
\(\Rightarrow \frac{\left(2 a_{1}-d\right)+p d}{\left(2 a_{1}-d\right)+q d}=\frac{p}{q}\)
\(\Rightarrow\left(2 a_{1}-d\right)(p-q)=0\)
As p is not equal to \(\Rightarrow \mathrm{a}_{1}=\frac{\mathrm{d}}{2}\)
\(\mathrm{Now} \frac{\mathrm{a}_{6}}{\mathrm{a}_{21}}=\frac{\mathrm{a}_{1}+5 \mathrm{d}}{\mathrm{a}_{1}+20 \mathrm{d}}\)
\(=\frac{\frac{d}{2}+5 d}{\frac{d}{2}+20 d}=\frac{11 d}{41 d}\)
\(=\frac{11}{41}\)

I. Ellipse. \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1 a^{2}=25 \cdot b^{2}=9\)
Now \(b^{2}=a^{2}\left(1-e^{2}\right) 9=25\left(1-e^{2}\right) e=\frac{4}{5}\)
II. Hyperbola ae \(=5\left(\frac{4}{5}\right)=4 .\) Focus \(S\) is at \(S(a e, 0)\) or \(S(4,0)\) Eccentricity \(=e_{1}=2\) Focus is at \(\mathrm{S}\left(\mathrm{ae}_{1}, 0\right)\) or \(\mathrm{S}(2 \mathrm{a}, 0) \ldots(\mathrm{ii})\)
By assumption (i) \(\&\) (ii) represent the same position. \(\therefore 2 a=4 a=2\)
Now \(a=2, e_{1}=2 b^{2}=a^{2}\left(e_{1}^{2}-1\right)\)
\(b^{2}=4(4-1) b^{2}=12\)
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) becomes \(\frac{x^{2}}{4}-\frac{y^{2}}{12}=1\)

\(|\bar{z} w|=|\bar{z}||w|=|z||w|=1\) and

 

\(\begin{aligned} \arg (\bar{z} w) &=\arg (z)+\arg (w) \\ &=-\arg (z)+\arg (w) \\ &=-[\arg (z)-\arg (w)] \\ &=-\frac{\pi}{2} \end{aligned}\)
so, modulus of this complex number is 1 and argument is \(-\frac{\pi}{2} \quad\) so complex no. is "- \(-1^{n}\)

Let the point P be ( \(h\), 0). Hence equation of the chord passing through \(P(h, 0)\) in parametric form is given by \(y\).
\(\frac{x-h}{\cos \theta}=\frac{y}{\sin \theta}=r x=h+r \cos \theta, y=r \sin \theta\)
Solving the chord \& the parabola:
\(2 r^{2} \sin ^{2} \theta-3 r \cos \theta-3(h-2)=0\)
\(\therefore \mathrm{r}_{1}+\mathrm{r}_{2}=\frac{3 \mathrm{cos} \theta}{2 \sin ^{2} \theta}, \mathrm{r}_{1} \mathrm{r}_{2}=\frac{-3(\mathrm{h}-2)}{2 \sin ^{2} \theta}\)
(note that \(\left.r_{1}=P A, r_{2}=P B\right)\)
\(\therefore \frac{1}{\mathrm{r}_{1}^{2}}+\frac{1}{\mathrm{r}_{2}^{2}}=\frac{\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)^{2}-2 \mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}^{2} \mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}+\mathrm{r}_{2}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)^{2}-\frac{2}{\mathrm{r}_{1} \mathrm{r}_{2}}\)
\(=\frac{\cos ^{2} \theta}{(h-2)^{2}}+\frac{4 \sin ^{2} \theta}{3(h-2)}\)
Let \(f(\theta)=\frac{\cos ^{2} \theta}{(h-2)^{2}}+\frac{4 \sin ^{2} \theta}{3(h-2)}=\) constant
\(f^{\prime}(\theta)=0 \forall \theta \in R\)
\(\left(\frac{-1}{(h-2)^{2}}+\frac{4}{3(h-2)}\right) 2 \sin \theta \cos \theta=0 \forall \theta \in R\)
\(\left(\frac{-1}{(h-2)^{2}}+\frac{4}{3(h-2)}\right)=0 h=11 / 4\)
\(P\) is \(\left(\frac{11}{4}, 0\right)\)