Mathematics Test - 27

\(\begin{aligned} & x+1-2 \log _2\left(3+2^x\right)+2 \log _4\left(10-2^{-x}\right)=0 \\ & \Rightarrow \\ & \mathrm{x}+1-2 \log _2\left(3+2^{\mathrm{x}}\right)+\log _2\left(\frac{10 \cdot 2^{\mathrm{x}}-1}{2^{\mathrm{x}}}\right)=0 \\ & \Rightarrow \mathrm{x}+1-2 \log _2\left(3+2^{\mathrm{x}}\right)+\log _2\left(10 \cdot 2^{\mathrm{x}}-1\right) \\ & \Rightarrow 1+\log _2\left(\frac{10 \cdot 2^{\mathrm{x}}-1}{\left(3+2^{\mathrm{x}}\right)^2}\right)=0 \\ & \Rightarrow \frac{10 \cdot 2^{\mathrm{x}}-1}{9+\left(2^{\mathrm{x}}\right)^2+6 \cdot 2^{\mathrm{x}}}=\frac{1}{2} \\ & \Rightarrow\left(2^x\right)^2-14 \cdot 2^x+11=0 \\ & \text { Let } 2^x=y \\ & \Rightarrow \mathrm{y}^2-14 \mathrm{y}+11=0 \\ & \text { Let } 2^{\mathrm{x}}=\mathrm{y} \\ & \Rightarrow \mathrm{y}^2-14 \mathrm{y}+11=0 \\ & \mathrm{y}=\frac{14 \pm \sqrt{152}}{2}=7 \pm \frac{\sqrt{152}}{2} \\ & \mathrm{y}_1=7+\frac{\sqrt{152}}{2}, \\ & \mathrm{y}_2=7-\frac{\sqrt{152}}{2} \\ & \Rightarrow 2^{\mathrm{x}_1}=7+\frac{\sqrt{152}}{2} \text {, } \\ & 2^{\mathrm{x}_2}=7-\frac{\sqrt{152}}{2} \\ & \Rightarrow \mathrm{x}_1=\log _2\left(7+\frac{\sqrt{152}}{2}\right) \\ & \mathrm{x}_2=\log _2\left(7-\frac{\sqrt{152}}{2}\right) \\ & \therefore \text { Sum of roots }=\mathrm{x}_1+\mathrm{x}_2 \\ & =\log _2\left(49-\frac{152}{4}\right)=\log _2 11 \\ & \end{aligned}\)

\(\begin{aligned} & A=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right],|A|=6 \\ & A^{-1}=\frac{\operatorname{ad} j A}{|A|}=\frac{1}{6}\left[\begin{array}{cc}4 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right] \\ & {\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]=\left[\begin{array}{ll}\alpha & 0 \\ 0 & \alpha\end{array}\right]+\left[\begin{array}{cc}\beta & 2 \beta \\ -\beta & 4 \beta\end{array}\right]} \\ & \left.\begin{array}{l}\alpha+\beta=\frac{2}{3} \\ \beta=-\frac{1}{6}\end{array}\right\} \Rightarrow \alpha=\frac{2}{3}+\frac{1}{6}=\frac{5}{6} \\ & 4(\alpha-\beta)=4(1)=4\end{aligned}\)

Given, equation of curve is \(y^2=4 a(x+a)\)

\(\Rightarrow y^2=4 a x+4 a^2 \ldots \text {..(i) }\)

Differentiating Eq. (i) w.r.t. \(x\), we get

\(2 y \frac{d y}{d x}=4 a \\\)

\(\Rightarrow a=\left(\frac{y}{2}\right) \cdot \frac{d y}{d x} \ldots \text { (ii) }\)

\(\therefore\) Required differential equation is

\(y^2=4 \times \frac{y}{2} \times \frac{d y}{d x}+4\left(\frac{y}{2} \cdot \frac{d y}{d x}\right)^2 \quad \text { [From Eqs. (i) and (ii)] } \\\)

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

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

\(\text { As, } y \neq 0 \\\)

\(\therefore y\left(\frac{d y}{d x}\right)^2+2 x\left(\frac{d y}{d x}\right)-y=0\)

Given that:

\(\lim _{{x} \rightarrow \infty} \frac{{x}^{4}+3 {x}^{2}+5}{{x}^{4}+{x}^{2}-6}\)

On putting the limits in above equation, we get:

\(\lim _{{x} \rightarrow \infty} \frac{x^{4}+3 {x}^{2}+5}{{x}^{4}+{x}^{2}-6}=\frac{\infty}{\infty}\)

We know that:

L-Hospital Rule :

\(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Applying L-Hospital Rule

\({L}(\) say \() =\lim _{{x} \rightarrow \infty} \frac{\frac{d}{dx}({x}^{4}+3 {x}^{2}+5)}{\frac{d}{dx}({x}^{4}+{x}^{2}-6)}\quad\) \((\because \frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.)\)

\(=\lim _{{x} \rightarrow \infty} \frac{4 {x}^{3}+6 {x}}{4 {x}^{3}+2 {x}}\)

\(=\lim _{{x} \rightarrow \infty} \frac{4 {x}^{2}+6}{4 {x}^{2}+2}\) ....(1)

\(=\frac{\infty}{\infty}\)

Again applying L-Hospital Rule on equation (1), we get:

\(L=\lim _{{x} \rightarrow \infty} \frac{\frac{d}{dx}(4 {x}^{2}+6)}{\frac{d}{dx}(4 {x}^{2}+2)}\)

\(L=\lim _{x \rightarrow \infty} \frac{8 x}{8 x}\)

\(L=\lim _{{x} \rightarrow \infty} 1\)

\(L=1\)

We know that,

\((i) ^{2}=-1\)

\((-i)^{4}=1\)

\(\left(\frac{1-i}{1+i}\right)^{n^{2}}=1\)

On rationalizing,

\(\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^{n^{2}}=1 \)

\(\Rightarrow \left(\frac{(1-i)^{2}}{1-i^{2}}\right)^{n^{2}}=1 \)

\(\Rightarrow\left(\frac{1+i^{2}-2 i}{1-(-1)}\right)^{n^{2}}=1 \)

\(\Rightarrow\left(\frac{1-1-2 i}{1+1}\right)^{n^{2}}=1 \)

\(\Rightarrow\left(\frac{-2 i}{2}\right)^{n^{2}}=1 \)

\(\Rightarrow(-i)^{n^{2}}=1\)

If we put \(n =2\) then,

\( (-i)^{2^{2}}=1\)

\( (-i)^{4}=1\)

Satisfy the equation.

Given:

\(\overrightarrow{\mathrm{a}}=-4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and\(\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{c}}=\mathrm{ma}+\mathrm{n} \overrightarrow{\mathrm{b}}\)

\(2 \hat{\imath}+3 \hat{\jmath}=m(-4 \hat{\imath}+2 \hat{j})+n(2 \hat{\imath}+\hat{\jmath})\)

\(2 \hat{\imath}+3 \hat{\jmath}=(-4 m+2 n) \hat{\imath}+(2 m+n) \hat{\jmath}\)

Equating the scalar coefficients, we get:

\(-4 m+2 n=2\) ..(1)

\(2 m+n=3\) ...(2)

Multiplying equation (2) by 2 and adding to equation (1), we get:

\(4 n=8\)

\(n=2\)

Using either of the equations above, we also get:

\(\mathrm{m}=\frac{1}{2}\)

\(\therefore m+n=2+\frac{1}{2}=\frac{5}{2}\)

Given,

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

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

\(\Rightarrow 2\left[\tan ^{-1} 1-\tan ^{-1} x\right]=\tan ^{-1} x\)

\(\Rightarrow 2\left(\tan ^{-1} 1\right)=3 \tan ^{-1} x\)

\(\Rightarrow 2 \times \frac{\pi}{4}=3 \tan ^{-1} x\)

\(\Rightarrow \frac{\pi}{6}=\tan ^{-1} x\)

\(\Rightarrow x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\)

\( P^2=I-P \\\)

\( P^\alpha+P^\beta=\gamma I-29 P \\\)

\( P^\alpha-P^\beta=\delta I-13 P \\\)

\( P^4=(I-P)^2=I+P^2-2 P \\\)

\( P^4=I+I-P-2 P=2 I-3 P \\\)

\( P^8=\left(P^4\right)^2=(2 I-3 P)^2=4 I+9 P^2-12 P \\\)

\( =4 I+9(I-P)-12 P \\\)

\( P^8=13 I-21 P \ldots(1) \\\)

\( P^6=P^4 \cdot P^2=(2 I-3 P)(I-P) \\\)

\( =2 I-5 P+3 P^2 \\\)

\( =2 I-5 P+3(I-P) \\\)

\( =5 I-8 P \ldots(2)\)

\( (1)+(2) \\\)

\( \mathrm{P}^8+\mathrm{P}^6=18 \mathrm{I}-29 \mathrm{P}\)

(1) - (2)

\(\mathrm{P}^8-\mathrm{P}^6=8 \mathrm{I}-13 \mathrm{P}\)

From (A) \(\alpha=8, \quad \beta=6\)

\(\gamma=18 \\\)

\(\delta=8 \\\)

\( \alpha+\beta+\gamma-\delta=32-8=24\)

Given:

\(P(n): 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{2 n}{n+1}\)

For \(\boldsymbol{n}=\mathbf{1}\)

L.H.S \(=1\) and R. H. S. \(=\frac{2 \times 1}{1+1}=\frac{2}{2}=1\)

Thus, \(P(1)\) is true.

Let \(P(n)\) be true for some \(n=k\)

\(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}\) \(=\frac{2 k}{k+1}\)

Now, we have to prove that \(P(n)\) is true \(n=k+1\)

i.e., \(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+(k+1)}\)

\(\frac{=2(k+1)}{k+2}\)

Adding \(\frac{1}{1+2+3+\ldots+(k+1)}\) on both sides of \((1)\), we get

\(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots \frac{1}{1+2+3+\ldots+k}+\frac{1}{1+2+3+\ldots+(k+1)} \)

\(=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots+(k+1)} \)

\(=\frac{2 k}{k+1)}+\frac{1}{\frac{(k+1)(k+2)}{2}} \)

\({\left[\text{Using}~ 1+2+3+\ldots+n=\frac{n(n+1)}{2}\right]} \)

\(=\frac{2 k}{(k+1)}+\frac{2}{(k+1)(k+2)} \)

\(=\frac{2}{k+1}\left(\frac{k(k+2)+1}{k+2}\right) \)

\(=\frac{2(k+1)^{2}}{(k+1)(k+2)} \)

\(=\frac{2(k+1)}{(k+2)}\)

Thus, \(P(k+1)\) is true whenever \(P(k)\) is true.

By the principle of mathematical induction, statement \(\mathrm{P}(\mathrm{n})\) is true for all natural numbers.

Given \(-2<2 x-1<2\)

\(\Rightarrow-2+1<2 x<2+1\)

\(\Rightarrow-1<2 x<3\)

\(\Rightarrow\frac{-1}2

\(\Rightarrow x \in(\frac{-1}2,\frac32)\)