Maths NCERT Exemplar Solutions Class 11th Chapter Three

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

  $\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

  $\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

  $x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\left(a\right)sin\left(x+y\right)sin\left(x-y\right)=si{n}^{2}x-si{n}^{2}y\\ \therefore \left(a\right)\leftrightarrow \left(iv\right)\\ \left(b\right)cos\left(x+y\right)cos\left(x-y\right)=co{s}^{2}x-co{s}^{2}y\\ \therefore \left(b\right)\leftrightarrow \left(i\right)\\ \left(c\right)cot\left(\frac{\pi }{4}+\theta \right)=\frac{cot\frac{\pi }{4}cot\theta -1}{cot\theta +cot\frac{\pi }{4}}=\frac{cot\theta -1}{cot\theta +1}=\frac{1-tan\theta }{1+tan\theta }\\ \therefore \left(c\right)\leftrightarrow \left(ii\right)\\ \left(d\right)tan\left(\frac{\pi }{4}+\theta \right)=\frac{tan\frac{\pi }{4}+tan\theta }{1-tan\frac{\pi }{4}tan\theta }=\frac{1+tan\theta }{1-tan\theta }\\ \therefore \left(d\right)\leftrightarrow \left(iii\right)\\ Hence,\left(a\right)\leftrightarrow \left(iv\right),\left(b\right)\leftrightarrow \left(i\right),\left(c\right)\leftrightarrow \left(ii\right)and\left(d\right)\leftrightarrow \left(iii\right).\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:tan\left(\pi cos\theta \right)=cot\left(\pi sin\theta \right)\\ \Rightarrow tan\left(\pi cos\theta \right)=tan\left(\frac{\pi }{2}-\pi sin\theta \right)\\ \Rightarrow \pi cos\theta =\frac{\pi }{2}-\pi sin\theta \\ \Rightarrow \pi cos\theta +\pi sin\theta =\frac{\pi }{2}\\ \Rightarrow cos\theta +sin\theta =\frac{1}{2}\\ \Rightarrow \frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta =\frac{1}{2\sqrt[]{2}}\\ \Rightarrow cos\frac{\pi }{4}cos\theta +sin\frac{\pi }{4}sin\theta =\frac{1}{2\sqrt[]{2}}\\ \Rightarrow cos\left(\theta -\frac{\pi }{4}\right)=\pm \frac{1}{2\sqrt[]{2}}\left[?cos\left(\theta -\frac{\pi }{4}\right)orcos\left(\frac{\pi }{4}-\theta \right)\right]\\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:tan\theta +tan2\theta +\sqrt[]{3}tan\theta tan2\theta =\sqrt[]{3}\\ \Rightarrow tan\theta +tan2\theta =\sqrt[]{3}-\sqrt[]{3}tan\theta tan2\theta \\ \Rightarrow tan\theta +tan2\theta =\sqrt[]{3}\left(1-tan\theta tan2\theta \right)\\ \Rightarrow \frac{tan\theta +tan2\theta }{1-tan\theta tan2\theta }=\sqrt[]{3}\\ \Rightarrow tan\left(\theta +2\theta \right)=\sqrt[]{3}\\ \Rightarrow tan3\theta =tan\frac{\pi }{3}\therefore 3\theta =n\pi +\frac{\pi }{3}\\ So,\theta =\frac{n\pi }{3}+\frac{\pi }{9}\\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:cosecx=1+cotx\\ \Rightarrow \frac{1}{sinx}=1+\frac{cosx}{sinx}\\ \Rightarrow \frac{1}{sinx}=\frac{sinx+cosx}{sinx}\\ \Rightarrow sinx+cosx=1\\ \Rightarrow \frac{1}{\sqrt[]{2}}sinx+\frac{1}{\sqrt[]{2}}cosx=\frac{1}{\sqrt[]{2}}\\ \Rightarrow sin\frac{\pi }{4}sinx+cos\frac{\pi }{4}cosx=\frac{1}{\sqrt[]{2}}\\ \Rightarrow cos\left(x-\frac{\pi }{4}\right)=\frac{1}{\sqrt[]{2}}\\ \Rightarrow cos\left(x-\frac{\pi }{4}\right)=cos\frac{\pi }{4}\\ \Rightarrow x=2n\pi +\frac{\pi }{4}+\frac{\pi }{4}\Rightarrow x=2n\pi +\frac{\pi }{2}\\ orx=2n\pi +\frac{\pi }{4}-\frac{\pi }{4}\Rightarrow x=2n\pi \\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationis\\ si{n}^4\theta -2si{n}^2\theta -1=0\\ si{n}^2\theta =\frac{-\left(-2\right)\pm \sqrt[]{{\left(-2\right)}^2-4*1*-1}}{2*1}\\ =\frac{2\pm \sqrt[]{4+4}}{2}=\frac{2\pm \sqrt[]{8}}{2}=\frac{2\pm 2\sqrt[]{2}}{2}=1\pm \sqrt[]{2}\\ \therefore si{n}^2\theta =\left(1+\sqrt[]{2}\right)or\left(1-\sqrt[]{2}\right)\Rightarrow -1\le sin\theta \le 1\\ \Rightarrow si{n}^2\theta \le 1butsi{n}^2\theta =\left(1+\sqrt[]{2}\right)or\left(1-\sqrt[]{2}\right)\\ Whichisnotpossible.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}L.H.S.cos\frac{2\pi }{15}.cos\frac{4\pi }{15}.cos\frac{8\pi }{15}.cos\frac{16\pi }{15}\\ =cos24^0.cos48^0.cos96^0.cos192^0\\ =\frac{1}{16sin24^0}\left[\left(2sin24^{0}cos24^0\right)\left(2cos48^0\right)\left(2cos96^0\right)\left(2cos192^0\right)\right]\\ =\frac{1}{16sin24^0}\left[sin48^0.2cos48^0\left(2cos96^0\right)\left(2cos192^0\right)\right]\\ =\frac{1}{16sin24^0}\left[2sin48^{0}cos48^0\left(2cos96^0\right)\left(2cos192^0\right)\right]\\ =\frac{1}{16sin24^0}\left[sin96^0\left(2cos96^0\right)\left(2cos192^0\right)\right]\\ =\frac{1}{16sin24^0}\left[2sin96^{0}cos96^0\left(2cos192^0\right)\right]\\ =\frac{1}{16sin24^0}\left[sin192^0\left(2cos192^0\right)\right]\\ =\frac{1}{16sin24^0}\left[2sin192^{0}cos192^0\right]\\ =\frac{1}{16sin24^0}sin384^0=\frac{1}{16sin24^0}sin\left(360^0+24^0\right)\\ =\frac{1}{16sin24^0}*sin24^0\left[?sin\left(360^0+\theta \right)=sin\theta \right]\\ =\frac{1}{16}R.H.S.\\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:  

          $\begin{array}{l}Ifsin10^0>cos10^0\\ \Rightarrow sin10^0>cos\left(90^0-80^0\right)\\ \Rightarrow sin10^0>sin80^{0}whichisnotpossible.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$