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This is a multiple choice answer as classified in NCERT Exemplar

(b) The PQ ray of light passes through focus F and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray-2 in the figure.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill -1\hfill & \hfill cosC\hfill & \hfill cosB\hfill \\ \hfill cosC\hfill & \hfill -1\hfill & \hfill cosA\hfill \\ \hfill cosB\hfill & \hfill cosA\hfill & \hfill -1\hfill \end{array}\right|\\ C_1\to a{C}_1+b{C}_2+c{C}_3\\ \Rightarrow \left|\begin{array}{ccc}\hfill -a+bcosC+ccosB\hfill & \hfill cosC\hfill & \hfill cosB\hfill \\ \hfill acosC-b+ccosA\hfill & \hfill -1\hfill & \hfill cosA\hfill \\ \hfill acosB+bcosA-C\hfill & \hfill cosA\hfill & \hfill -1\hfill \end{array}\right|\\ \Rightarrow \left|\begin{array}{ccc}\hfill -a+a\hfill & \hfill cosC\hfill & \hfill cosB\hfill \\ \hfill -b+b\hfill & \hfill -1\hfill & \hfill cosA\hfill \\ \hfill -c+c\hfill & \hfill cosA\hfill & \hfill -1\hfill \end{array}\right|\left[\begin{array}{l}?Fromprotectionformula\\ a=bcosC+ccosB\\ b=acosC+ccosA\\ c=bcosA+acosB\end{array}\right]\\ \Rightarrow \left|\begin{array}{ccc}\hfill 0\hfill & \hfill cosC\hfill & \hfill cosB\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill cosA\hfill \\ \hfill 0\hfill & \hfill cosA\hfill & \hfill -1\hfill \end{array}\right|=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\left|\begin{array}{ccc}\hfill sinx\hfill & \hfill cosx\hfill & \hfill cosx\hfill \\ \hfill cosx\hfill & \hfill sinx\hfill & \hfill cosx\hfill \\ \hfill cosx\hfill & \hfill cosx\hfill & \hfill sinx\hfill \end{array}\right|=0\\ C_1\to C_1+C_2+C_3\\ \Rightarrow \left|\begin{array}{ccc}\hfill 2cosx+sinx\hfill & \hfill cosx\hfill & \hfill cosx\hfill \\ \hfill 2cosx+sinx\hfill & \hfill sinx\hfill & \hfill cosx\hfill \\ \hfill 2cosx+sinx\hfill & \hfill cosx\hfill & \hfill sinx\hfill \end{array}\right|=0\left(Taking2cosx+sinxcommonfrom{C}_1\right)\\ \Rightarrow \left(2cosx+sinx\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill cosx\hfill & \hfill cosx\hfill \\ \hfill 1\hfill & \hfill sinx\hfill & \hfill cosx\hfill \\ \hfill 1\hfill & \hfill cosx\hfill & \hfill sinx\hfill \end{array}\right|=0\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ \Rightarrow \left(2cosx+sinx\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill cosx-sinx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill sinx-cosx\hfill & \hfill cosx-sinx\hfill \\ \hfill 1\hfill & \hfill cosx\hfill & \hfill sinx\hfill \end{array}\right|=0\\ \Rightarrow \left(2cosx+sinx\right)\left[1\left|\begin{array}{cc}\hfill cosx-sinx\hfill & \hfill 0\hfill \\ \hfill sinx-cosx\hfill & \hfill cosx-sinx\hfill \end{array}\right|\right]\\ \Rightarrow \left(2cosx+sinx\right){\left(cosx-sinx\right)}^2=0\\ 2cosx+sinx=0and{\left(cosx-sinx\right)}^2=0\\ 2+tanx=0and\left(cosx-sinx\right)=0\\ \therefore tanx=-2and\Rightarrow tanx=1\\ But-\frac{\pi }{4}\le x\le \frac{\pi }{4}and\Rightarrow tanx=tan\frac{\pi }{4}\\ \therefore x=\frac{\pi }{4}\in \left[-\frac{\pi }{4},\frac{\pi }{4}\right]\\ So,x\text{\&thins}\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthaty=f\left(x\right)willbecontinuousatx=aif\\ \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)\\ Givenf\left(x\right)=x^3+2x^2-1\\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left(1+h\right)}^3+2{\left(1+h\right)}^2-1=1+2-1=2\\ \underset{x\to 1}{lim}f\left(x\right)={\left(1\right)}^3+2{\left(1\right)}^2-1=1+2-1=2\\ \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left(1+h\right)}^3+2{\left(1+h\right)}^2-1=1+2-1=2\\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=2.\\ Hence,f\left(x\right)iscontinuousatx=1.\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill b^2-ab\hfill & \hfill b-c\hfill & \hfill bc-ac\hfill \\ \hfill ab-a^2\hfill & \hfill a-b\hfill & \hfill b^2-ab\hfill \\ \hfill bc-ac\hfill & \hfill c-a\hfill & \hfill ab-a^2\hfill \end{array}\right|\\ \Rightarrow =\left|\begin{array}{ccc}\hfill b\left(b-a\right)\hfill & \hfill b-c\hfill & \hfill c\left(b-a\right)\hfill \\ \hfill a\left(b-a\right)\hfill & \hfill a-b\hfill & \hfill b\left(b-a\right)\hfill \\ \hfill c\left(b-a\right)\hfill & \hfill c-a\hfill & \hfill a\left(b-a\right)\hfill \end{array}\right|\left(Taking\left(b-a\right)commonfrom{C}_{1}and{C}_3\right)\\ \Rightarrow ={\left(b-a\right)}^2\left|\begin{array}{ccc}\hfill b\hfill & \hfill b-c\hfill & \hfill c\hfill \\ \hfill a\hfill & \hfill a-b\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill c-a\hfill & \hfill a\hfill \end{array}\right|\\ C_1\to C_1-C_3\\ \Rightarrow ={\left(a-b\right)}^2\left|\begin{array}{ccc}\hfill b-c\hfill & \hfill b-c\hfill & \hfill c\hfill \\ \hfill a-b\hfill & \hfill a-b\hfill & \hfill b\hfill \\ \hfill c-a\hfill & \hfill c-a\hfill & \hfill a\hfill \end{array}\right|\left(C_{1}and{C}_{2}areidenticalcolumns.\right)\\ \Rightarrow ={\left(a-b\right)}^2.0=0\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) The phenomenon involved in the reflection of radio waves by ionosphere is similar to total internal reflection of light in air during a mirage i.e., angle of incidence is greater than critical angle.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Areaoftrianglewithvertices\left(x_1,y_1\right),\left(x_2,y_2\right)and\left(x_3,y_3\right)willbe:\\ \Delta =\frac{1}{2}\left|\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill 1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill 1\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill 1\hfill \end{array}\right|\Rightarrow \Delta =\frac{1}{2}\left|\begin{array}{ccc}\hfill -3\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill k\hfill & \hfill 1\hfill \end{array}\right|\\ \Rightarrow =\frac{1}{2}\left[-3\left|\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill k\hfill & \hfill 1\hfill \end{array}\right|-0\left|\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|+1\left|\begin{array}{cc}\hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill k\hfill \end{array}\right|\right]\\ \Rightarrow =\frac{1}{2}\left[-3\left(-k\right)-0+1\left(3k\right)\right]\\ \Rightarrow =\frac{1}{2}\left(3k+3k\right)\Rightarrow =\frac{1}{2}\left(6k\right)=3k\\ 3k=9\Rightarrow k=3\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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