Maths

${\left|2\overrightarrow{a}+3\overrightarrow{b}\right|}^2={\left|3\overrightarrow{a}+\overrightarrow{b}\right|}^2\Rightarrow 4{\left|\overrightarrow{a}\right|}^2+9{\left|\overrightarrow{b}\right|}^2+12\overrightarrow{a}.\overrightarrow{b}=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}$

$\Rightarrow -5{\left|\overrightarrow{a}\right|}^2+8{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}=0.........\left(i\right)$

$cos60°=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{2}\therefore \overrightarrow{a}.\overrightarrow{b}=4\left|\overrightarrow{b}\right|and\left|\overrightarrow{a}\right|=8$

from (i) we get $\left|\overrightarrow{b}\right|=5$

Applying Leibniz theorem,  

$\sqrt[]{1-{\left(f\text{'}\left(x\right)\right)}^2}=f\left(x\right)\Rightarrow \frac{f\text{'}\left(x\right)}{\sqrt[]{1-{\left(f\left(x\right)\right)}^2}}=1$ on integrating both sides, we get

$f\left(x\right)=sinx+C$ put x = 0 and f (0) = 0 we get C = 0

$Now\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}}{x^2},\left(\frac{0}{0}\right)$ by L' Hospital rule $\underset{x\to 0}{lim}\frac{f\left(x\right)}{2x}=\underset{x\to 0}{lim}\frac{sinx}{2x}=\frac{1}{2}$

$\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y}\Rightarrow {\displaystyle \int \frac{2^{y}dy}{2^y-1}={\displaystyle \int 2^{x}dx\Rightarrow put{2}^y-1=t\Rightarrow 2^{y}ln2dy=dt}}$

$\frac{1}{ln2}{\displaystyle \int \frac{dt}{t}={\displaystyle \int 2^{x}dx\Rightarrow \frac{1}{ln2}lnt=\frac{2^x}{ln2}+\frac{C}{ln2}}}$    put x = 0 and y = 1 we get C = -1

$\therefore ln\left(2^y-1\right)=2^x-1$    put x = 1 and we get y = log₂ (1 + e)

$?\frac{3}{1^2*2^2}+\frac{5}{2^2*3^2}+\frac{7}{3^2*4^2}+........$

$\therefore t_r=\frac{2r+1}{r^2{\left(r+1\right)}^2}=\frac{1}{r^2}-\frac{1}{{\left(r+1\right)}^2}$

$S_{10}={\displaystyle \sum _{r=1}^{10}{t}_r=}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{r^2}-\frac{1}{{\left(r+1\right)}^2}\right)=1-\frac{1}{11^2}=\frac{120}{121}}$

${\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^{3/4}{\left(x+2\right)}^{5/4}}={\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^2}put\frac{x-1}{x+2}}=t\Rightarrow \frac{3}{{\left(x+2\right)}^2}dx=dt}$

$\therefore \frac{1}{3}{\displaystyle \int \frac{dt}{t^{3/4}}=\frac{1}{3}.\frac{t^{1/4}}{\frac{1}{4}}+C=\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{1/4}+C}$

$LHS=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to \infty }{lim}\frac{-1}{h}ln\left(\frac{1-\frac{h}{a}}{1+\frac{h}{b}}\right)=\underset{h\to 0}{lim}\frac{ln\left(1-\frac{h}{a}\right)}{a\left(\frac{-h}{a}\right)}+\underset{h\to 0}{lim}\frac{ln\left(1+\frac{h}{b}\right)}{b\left(\frac{h}{b}\right)}=\frac{1}{a}+\frac{1}{b}.............\left(i\right)$

$RHS=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{cos2x-1}{x^2+1-1}\left(\sqrt[]{x^2+1}+1\right)=\underset{x\to 0}{lim}\frac{-2si{n}^{x}x}{x^2}*2=-4...............\left(ii\right)$

$and\underset{x\to 0}{lim}f\left(x\right)=k.............\left(iii\right)$

$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=-4=k$

From (i), (ii) and (iii) we get $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$

Since the line $\frac{x}{5}cos\theta +\frac{y}{12}sin\theta =1$ is the equation of tangent to the ellipse $\frac{x^2}{5^2}+\frac{y^2}{12^2}=1$ at (5 cos q, 12 sin q). Hence option (A) is correct option.

Shiksha's NCERT notes are extremely useful for efficient preparation. We offer structured chapter-wise notes for the latest CBSE  syllabus and provide concise summaries and key formulas. These notes are designed for quick and last-minute revision. These short revision notes offer step-by-step explanations for conceptual clarity and include important questions from previous years' papers and NCERT Textbooks. Our notes are equally beneficial for competitive exams like JEE Mains, NEET and other exams as well.