Mathematics Test - 40

\(a, b, c\) are in \(A P \Rightarrow 2 b=a+c\)
Also, \(a+b+c=3 / 2 \Rightarrow 3 b=3 / 2\)
\(\Rightarrow b=1 / 2\)
\(\Rightarrow a+c=1\)
Next, \(\mathrm{a}^{2}, \mathrm{b}^{2}, \mathrm{c}^{2}\) are in GP.
\(\Rightarrow b^{2}=\sqrt{a^{2} c^{2}} \Rightarrow b^{2}=\pm a c\)
\(\Rightarrow a c=1 / 4\)
or
\(\mathrm{ac}=-1 / 4\)
From (1) \& (2) a \& c are roots of a quadratic \(x^{2}-(a+c) x+a c=0\)
\(\Rightarrow x^{2}-x+(1 / 4)=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^{2}=0\)

\(S_{n}=12+16+24+40+\ldots \ldots \ldots \ldots+t_{n}\)
\(S_{n}=0+12+16+24+\ldots \ldots \ldots . t_{n-1}+t_{n}\)
On Subtracting. \(0=12+4+8+16+\ldots+\left(t_{n}-t_{n-1}\right)-t_{n}\)
or, \(t_{n}=12+[4+8+16+\ldots(n-1)\) terms \(]\)
\(=12+\frac{4\left(2^{n-1}-1\right)}{2-1}\)
\(=12+2^{n-1+2}-4\) or, \(t_{n}=8+2^{n+1}=8+2.2^{n}\)
\(\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{t}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(8+2.2^{\mathrm{n}}\right)=8 \mathrm{n}+2\left[\frac{2 \cdot\left(2^{\mathrm{n}}-1\right)}{2-1}\right]\)
\(=8 n+4\left(2^{n}-1\right)\) as \(\sum 2.2^{n}=2 \sum 2^{n}=2\left(2+2^{2}+2^{3}+\ldots\right)\)

\(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{1}{2} \overrightarrow{\mathrm{b}}\)
\(\Rightarrow(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}=\frac{1}{2} \overrightarrow{\mathrm{b}}\)
Comparing on both sides of the equation \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\frac{1}{2}\)
\(\Rightarrow \mathrm{ac} \cos \theta=\frac{1}{2}(\theta \quad\) is the angle between a and \(\mathrm{c})\)
\(\Rightarrow \cos \theta=\frac{1}{2} \quad\) since a, b,c have unit magnitude
\(\Rightarrow \theta=\frac{\pi}{3}\)

\(\dot{\alpha}=\dot{p}+\dot{q}, p-\dot{q}=\dot{\beta}\)
where \(\dot{\mathrm{p}}=5 \mathrm{a}+2 \mathrm{b}, \dot{\mathrm{q}}=\mathrm{a}-3 \dot{\mathrm{b}},\) then \(\dot{\alpha}=\dot{\mathrm{p}}+\dot{\mathrm{q}}=6 \mathrm{a}-\dot{\mathrm{b}}\)
\(\alpha^{2}=(6 \dot{a}-\dot{b})^{2}=36 a^{2}+b^{2}-12 a^{} b\)
\(=36(8)+9-12(2 \sqrt{2})(3) \cos \left(\frac{\pi}{4}\right)=(15)^{2}\)
\(\beta^{2}=(\vec{p}-\vec{q})^{2}=(4 \vec{a}+5 \vec{b})^{2}=16 a^{2}+25 b^{2}+40 \vec{a} \cdot \vec{b}\)
\(=16(8)+25(9)+40(2 \sqrt{2})(3) \cos \left(\frac{\pi}{4}\right)=593\)
\(\alpha=15, \beta=\sqrt{593}, \beta>\alpha\)
\(\beta=\sqrt{593}\) is required length. \((c)\)

The direction ratios of the line joining \(\mathrm{P}(-1,2,4)\) and \(\mathrm{Q}(1,0,5)\) are proportional to 2,-2,1
\(\therefore\) Its direction cosines are \(\frac{2}{3},-\frac{2}{3}, \frac{1}{3}\)
Thus, the projection of the line joining \(\mathrm{A}(3,4,5)\) and \(\mathrm{B}(4,6,3)\) on \(\mathrm{PQ}\) is given by
\(=\left|(4-3) \times \frac{2}{3}+(6-4) \times-\frac{2}{3}+(3-5) \times \frac{1}{3}\right|\)
\(=\left|\frac{2}{3}-\frac{4}{3}-\frac{2}{3}\right|=\frac{4}{3}\)

\({ }^{9} C_{2}(1 / 3)^{2}+{ }^{9} C_{3}\left(\frac{1}{3}\right)^{3}+\ldots \ldots . .^{9} C_{9}\left(\frac{1}{3}\right)^{9}\)
\(=\left[{ }^{9} C_{0}+{ }^{9} C_{1}\left(\frac{1}{3}\right)+{ }^{9} C_{2}\left(\frac{1}{3}\right)^{2}+\ldots . . .^{9} C_{9}\left(\frac{1}{3}\right)^{9}\right]-\left[{ }^{9} C_{0}+{ }^{9} C_{1}\left(\frac{1}{3}\right)\right]\)
\(=\left(1+\frac{1}{3}\right)^{9}-\left(1+\frac{9}{3}\right)\)
\(=\left(\frac{4}{3}\right)^{9}-4\)

\(\log _{2}\left(4^{x+1}+4\right) \log _{2}\left(4^{x}+1\right)=\log _{2} 8\)
\(\Rightarrow \log _{2} 4\left(4^{x}+1\right) \log _{2}\left(4^{x}+1\right)=\log _{2} 8\)
\(\Rightarrow\left(\log _{2} 4+\log _{2}\left(4^{x}+1\right) \log _{2}\left(4^{x}+1\right)=\log _{2} 8\right.\)
\(\operatorname{Let} \log _{2}\left(4^{x}+1\right)=y\)
\(\Rightarrow\left(\log _{2} 2^{2}+y\right)(y)=\log _{2} 2^{3} \Rightarrow(2+y)(y)=3\)
\(\Rightarrow y^{2}+2 y-3=0 \Rightarrow y=1,-3 \Rightarrow \log _{2}\left(4^{x}+1\right)=1,-3\)
\(\Rightarrow 4^{x}+1=2^{1}\) or \(4^{x}+1=2^{-3}\)
\(\Rightarrow 4^{\mathrm{x}}=1\) or \(4^{\mathrm{x}}=\frac{1}{8}-1=-\frac{7}{8}\)
\(\Rightarrow \mathrm{x}=0\) or \(\mathrm{x} \in \phi,\left\{\right.\) as \(_{4} \mathrm{x}\) is always positive \(\}\)
\(\Rightarrow x=0\)

\(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{a(0+\sin \theta)}{a(1-\cos \theta)}=\frac{\sin \theta}{1-\cos \theta}\)
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{\sin \theta}{1-\cos \theta}\right)\)
\(=\frac{\mathrm{d}}{\mathrm{d} \theta}\left(\frac{\sin \theta}{1-\cos \theta}\right) \frac{\mathrm{d} \theta}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{d} \theta}\left(\frac{\sin \theta}{1-\cos \theta}\right)\left(\frac{1}{\mathrm{dx} / \mathrm{d} \theta}\right)\)
\(=\frac{(1-\cos \theta)(\cos \theta)-\sin \theta(\sin \theta)}{(1-\cos \theta)^{2}} \frac{1}{a(1-\cos \theta)}=\frac{\cos \theta-\cos ^{2} \theta-\sin ^{2} \theta}{a(1-\cos \theta)^{3}}\)
\(=\frac{\cos \theta-1}{a(1-\cos \theta)^{3}}=-\frac{1}{a(1-\cos \theta)^{2}}\)
\(\left.\frac{d^{2} y}{d x^{2}}\right]_{\text {at } \theta=\frac{\pi}{2}}=-\frac{1}{a\left(1-\cos \frac{\pi}{2}\right)^{2}}=-\frac{1}{a}\)
\(\left.\frac{d^{2} y}{d x^{2}}\right|_{\text {at } \theta=\frac{\pi}{3}}=-\frac{1}{a\left(1-\cos \frac{\pi}{3}\right)^{2}}=-\frac{4}{a}\)