Class 12th

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}IfAandBaretwoinvertiblematricesthen\\ \left(a\right)adjA=\left|A\right|.A^{-1}iscorrect\\ \left(b\right)det{\left(A\right)}^{-1}={\left[det\left(A\right)\right]}^{-1}=\frac{1}{det\left(A\right)}iscorrect\\ \left(c\right)Also,{\left(AB\right)}^{-1}=B^{-1}{A}^{-1}iscorrect\\ \left(d\right){\left(A+B\right)}^{-1}=\frac{1}{\left|A+B\right|}.adj\left(A+B\right)\\ \therefore {\left(A+B\right)}^{-1}\ne B^{-1}+A^{-1}\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Wehave,\\ A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill \lambda \hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\Rightarrow \left|A\right|=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill \lambda \hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|\\ Expandingalong{R}_1=2\left|\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right|-\lambda \left|\begin{array}{cc}\hfill 0\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right|-3\left|\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right|\\ =2\left(6-5\right)-\lambda \left(0-5\right)-3\left(0-2\right)\\ =2+5\lambda +6=8+5\lambda \\ If{A}^{-1}existsthen\left|A\right|\ne 0\\ \therefore 8+5\lambda \ne 0So\lambda \ne \frac{-8}{5}\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(d) The speed of the image of the car would appear to increase as the distance between the cars decreases.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill x-a\hfill & \hfill x-b\hfill \\ \hfill x+a\hfill & \hfill 0\hfill & \hfill x-c\hfill \\ \hfill x+b\hfill & \hfill x+c\hfill & \hfill 0\hfill \end{array}\right|\\ f\left(a\right)=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill a-b\hfill \\ \hfill 2a\hfill & \hfill 0\hfill & \hfill a-c\hfill \\ \hfill a+b\hfill & \hfill a+c\hfill & \hfill 0\hfill \end{array}\right|\\ Expandingalong{R}_1=\left(a-b\right)\left|\begin{array}{cc}\hfill 2a\hfill & \hfill 0\hfill \\ \hfill a+b\hfill & \hfill a+c\hfill \end{array}\right|\\ =\left(a-b\right)\left[2a\left(a+c\right)\right]=\left(a-b\right).2a.\left(a+c\right)\ne 0\\ f\left(b\right)=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill b-a\hfill & \hfill 0\hfill \\ \hfill b+a\hfill & \hfill 0\hfill & \hfill b-c\hfill \\ \hfill 2b\hfill & \hfill b+c\hfill & \hfill 0\hfill \end{array}\right|\\ Expandingalong{R}_1=-\left(b-a\right)\left|\begin{array}{cc}\hfill b+a\hfill & \hfill b-c\hfill \\ \hfill 2b\hfill & \hfill 0\hfill \end{array}\right|\\ =-\left(b-a\right)\left[\left(-2b\right)\left(b-c\right)\right]=2b\left(b-a\right)\left(b-c\right)\ne 0\\ f\left(0\right)=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill -b\hfill \\ \hfill a\hfill & \hfill 0\hfill & \hfill -c\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill 0\hfill \end{array}\right|\\ Expandingalong{R}_1=a\left|\begin{array}{cc}\hfill a\hfill & \hfill -c\hfill \\ \hfill b\hfill & \hfill 0\hfill \end{array}\right|-b\left|\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill b\hfill & \hfill c\hfill \end{array}\right|\\ =a\left(bc\right)-b\left(ac\right)=abc-abc=0\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

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

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1-cos2x}{x^2}=\underset{h\to 0}{lim}\frac{1-cos2\left(0-h\right)}{{\left(0-h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{1-cos\left(-2h\right)}{h^2}=\underset{h\to 0}{lim}\frac{1-cos\left(2h\right)}{h^2}\\ =\underset{h\to 0}{lim}\frac{2si{n}^{2}h}{h^2}\left[?1-cos\theta =2si{n}^2\frac{\theta }{2}\right]\\ =\underset{h\to 0}{lim}\frac{2sinh}{h}.\frac{sinh}{h}=2.1.1=2\left[\underset{x\to 0}{lim}\frac{sinx}{x}=1\right]\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{1-cos2x}{x^2}\\ =\underset{h\to 0}{lim}\frac{1-cos2\left(0+h\right)}{{\left(0+h\right)}^2}=\underset{h\to 0}{lim}\frac{1-cos\left(2h\right)}{h^2}\\ =\underset{h\to 0}{lim}\frac{2si{n}^{2}h}{h^2}=\underset{h\to 0}{lim}\frac{2sinh}{h}.\frac{sinh}{h}=2.1.1=2\\ \underset{x\to 0}{lim}f\left(x\right)=5\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Here, light ray goes from (optically) rarer medium air to optically denser terpentine, then it bends towards the normal i.e., i>r whereas when it goes from to optically denser medium terpentine to rarer medium water. then it bends away the normal i.e., i

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1+sin\theta \hfill & \hfill 1\hfill \\ \hfill 1+cos\theta \hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|\\ C_1\to C_1-C_2,C_2\to C_2-C_3\\ \Rightarrow =\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -sin\theta \hfill & \hfill sin\theta \hfill & \hfill 1\hfill \\ \hfill cos\theta \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\\ Expandingalong{R}_1\\ \Rightarrow =1\left|\begin{array}{cc}\hfill -sin\theta \hfill & \hfill sin\theta \hfill \\ \hfill cos\theta \hfill & \hfill 0\hfill \end{array}\right|=-sin\theta cos\theta \\ \Rightarrow =-\frac{1}{2}.2sin\theta cos\theta =-\frac{1}{2}sin2\theta \\ butmaximumvalueofsin2\theta =1\Rightarrow \left|-\frac{1}{2}.1\right|=\frac{1}{2}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

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

Sol:

$\begin{array}{l}\underset{x\to 2^{+}}{lim}f\left(x\right)=3x+5\\ =\underset{h\to 0}{lim}3\left(2+h\right)+5=11\\ \underset{x\to 2}{lim}f\left(x\right)=3x+5=3\left(2\right)+5=11\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=x^2=\underset{h\to 0}{lim}{\left(2-h\right)}^2\\ =\underset{h\to 0}{lim}{\left(2\right)}^2+h^2-4h={\left(2\right)}^2=4\\ Since\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)\ne \underset{x\to 2^{-}}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=2.\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Wehavef\left(t\right)=\left[\begin{array}{ccc}\hfill cost\hfill & \hfill t\hfill & \hfill 1\hfill \\ \hfill 2sint\hfill & \hfill t\hfill & \hfill 2t\hfill \\ \hfill sint\hfill & \hfill t\hfill & \hfill t\hfill \end{array}\right]\\ Expandingalong{R}_1\\ =cost\left|\begin{array}{cc}\hfill t\hfill & \hfill 2t\hfill \\ \hfill t\hfill & \hfill t\hfill \end{array}\right|-t\left|\begin{array}{cc}\hfill 2sint\hfill & \hfill 2t\hfill \\ \hfill sint\hfill & \hfill t\hfill \end{array}\right|+1\left|\begin{array}{cc}\hfill 2sint\hfill & \hfill t\hfill \\ \hfill sint\hfill & \hfill t\hfill \end{array}\right|\\ =cost\left(t^2-2t^2\right)-t\left(2tsint-2tsint\right)+\left(2tsint-tsint\right)\\ =t^{2}cost+tsint\\ \therefore \frac{f\left(t\right)}{t^2}=\frac{-t^{2}cost+tsint}{t^2}\\ \Rightarrow \frac{f\left(t\right)}{t^2}=-cost+\frac{sint}{t^2}\\ \Rightarrow \underset{t\to 0}{lim}\frac{f\left(t\right)}{t^2}=\underset{t\to 0}{lim}\left(-cost\right)+\underset{t\to 0}{lim}\frac{sint}{t^2}=-1+1=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$