Class 11th

$\begin{array}{l}\left(i\right)Giventhat:35\in \left\{x|xhasexactlyfourpositivefactor\right\}\\ \therefore Factorsof35are1,5,7,35\\ Hence,thestatement\left(i\right)is\text{'}True\text{'}.\\ \left(ii\right)Giventhat:128\in \left\{y|thesumofallpositivefactorsofyis2y\right\}\\ \therefore Factorsof128are1,2,4,8,16,32,64and128.\\ Sumofallfactors=1+2+4+8+16+32+64+128\\ =255\ne 2*128\\ Hence,thestatement\left(ii\right)is\text{'}False\text{'}.\\ \left(iii\right)Giventhat:3\in \left\{x|x^4-5x^3+2x^2-112x+6=0\right\}\\ \therefore x^4-5x^3+2x^2-112x+6=0\\ Nowforx=3,wehave\\ {\left(3\right)}^4-5{\left(3\right)}^3+2{\left(3\right)}^2-112\left(3\right)+6\\ \Rightarrow 81-135+18-336+6\Rightarrow -366\ne 0\\ Hence,thestatement\left(iii\right)is\text{'}True\text{'}.\\ \left(iv\right)Giventhat:496\notin \left\{y|thesumofallpositivefactorsofyis2y\right\}\\ \therefore Thepositivefactorsof496are1,2,4,8,16,31,62,124,248and496.\\ \therefore Thesumofallpositivefactors=1+2+4+8+16+31+62+124+248+496\\ =992=2*496\\ Hence,thestatement\left(iv\right)is\text{'}False\text{'}.\end{array}$

$\begin{array}{l}Given,Y=\left\{x|xisapositivefactorofthenumber{2}^{p-1}\left(2^p-1\right),where{2}^{p-1}isaprimenumber\right\}.\\ Since,thefactorsof{2}^{p-1}are1,2,2^2,2^3,\dots ,2^{p-1}andfactorsof{2}^p-1are1and{2}^{p-1}\\ \therefore Y=\left\{1,2,2^2,2^3,\dots ,2^{p-1},2^p-1\right\}\\ =2\left(2^p-1\right),2^2\left(2^p-1\right),\dots ,2^{p-1}\left(2^p-1\right)\end{array}$

$\begin{array}{l}\left(i\right)Wehave,D=\left\{t|t^3=t,t\in R\right\}\\ \therefore t^3=t\\ \Rightarrow t^3-t=0\Rightarrow t\left(t^2-1\right)=0\\ \Rightarrow t\left(t-1\right)\left(t+1\right)=0\Rightarrow t=0,1,-1\\ \therefore D=\left\{-1,0,1\right\}\\ \left(ii\right)Wehave,E=\left\{w|\frac{w-2}{w+3}=3,w\in R\right\}\\ \therefore \frac{w-2}{w+3}=3\\ \Rightarrow w-2=3w+9\Rightarrow w-3w=9+2\\ \Rightarrow -2w=11\Rightarrow w=\frac{-11}{2}\\ \therefore E=\left\{\frac{-11}{2}\right\}\\ \left(iii\right)Wehave,F=\left\{x|x^4-5x^2+6=0,x\in R\right\}\\ \therefore x^4-5x^2+6=0\\ \Rightarrow x^4-3x^2-2x^2+6=0\Rightarrow x^2\left(x^2-3\right)-2\left(x^2-3\right)=0\\ \Rightarrow \left(x^2-3\right)\left(x^2-2\right)=0\Rightarrow x=\pm \sqrt[]{3},\pm \sqrt[]{2}\\ \therefore F=\left\{-\sqrt[]{3},-\sqrt[]{2},\sqrt[]{2},\sqrt[]{3}\right\}\end{array}$

$\begin{array}{l}\left(i\right)Wehave,A=\left\{x:x\in R,2x+11=15\right\}\\ \therefore 2x+11=15\\ \Rightarrow 2x=15-11\\ \Rightarrow 2x=4\Rightarrow x=2\\ \therefore A=\left\{2\right\}\\ \left(ii\right)Wehave,B=\left\{x|x^2=x,x\in R\right\}\\ \therefore x^2=x\\ \Rightarrow x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow x=0,1\\ \therefore B=\left\{0,1\right\}\\ \left(iii\right)Wehave,C=\left\{x|xisapositivefactorofprimenumberp.\right\}\\ Since,positivefactorsofaprimenumberare1andthenumberitself.\\ \therefore C=\left\{1,p\right\}\end{array}$

Answer: (a) A catalyst favours the formation of products

Answer: (d) decreases

Answer: (a) increases

(i) f(x)=sin x cos x

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{sin(x+h)cos(x+h)?sinxcosx}{h}$

$=\underset{h?0}{lim}\frac{1}{2h}*[2sin(x+h)cos(x+h)?2sinxcosx]$

$=\underset{h?0}{lim}\frac{1}{2h}[sin2(x+h)?sin2x]$

$=\underset{h?0}{lim}\frac{1}{2h}\left[2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)?2x}{2}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[cos\left(2x+h\right)sinh\right]$

$=\underset{h?0}{lim}cos(2x+h)*\underset{h?0}{lim}\frac{sinh}{h}$

$=cos(2x+0)$

$=cos2x$

$\mathrm{(ii)\; f}(x)=secx$

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[sec(x+h)?secx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{?2sin\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{(?1sin\left(\frac{2x+h}{2}\right)}{cos(x+h)cosx}*\underset{h?0}{lim}(?1)\frac{sinh/2}{h/2}$

$=\frac{sinx}{cosx?cosx}*1$

$=tanx?secx.$

(iii) Given f(x)=5 sec x+4 cosx.

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$\begin{array}{l}\\ =\underset{h?0}{lim}\frac{5sec(x+h)+4cos(x+h)?[5secx+4cosx]}{h}.\end{array}$

$=\underset{h?0}{lim}\frac{5}{h}[sec(x+h)?secx]+\underset{h?0}{lim}\frac{4}{h}[cos(x+h)?cosx]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h?x}{2}\right)\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)(cosx)}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{2x+h}{2}\right)sin\frac{h}{2}\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]?4\underset{h?0}{lim}sin\left(\frac{2x+h}{2}\right)\underset{h?0}{lim}\frac{sinh/2}{h/2}$