Maths

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Threepairs.\\ Hence,thevalueofthefilleristhreepairs.\end{array}$

Let TV (r) denotes truth value of a statement $r$ .

$\text{Now, if}TV\left(p\right)=TV\left(q\right)=T$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_1\right)=\mathrm{F}$

Also, if $TV\left(p\right)=T$ and $TV\left(q\right)=F$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_2\right)=\mathrm{T}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}ThethreeaxesOX,OYandOZ\text{\hspace{0.17em}determine}threecoordinateplanes.\\ Hence,thefillervalueisthreecoordinateplanes.\end{array}$

$1+\left(1-2^2.1\right)+\left(1-4^2.3\right)+\dots \dots .+\left(1-{20}^2.19\right)$

$=\alpha -220\beta$

$=11-\left(2^2\cdot 1+4^2\cdot 3+\dots ..+{20}^2\cdot 19\right)$

$=11-2^2\cdot {\sum }_{r=1}^{10}?r^2(2r-1)=11-4\left(\frac{{110}^2}{2}-35*11\right)$

$=11-220\left(103\right)$

$\Rightarrow \alpha =11,\beta =103$

 $\mathrm{t}\mathrm{a}\mathrm{n}?\theta =\frac{10}{\mathrm{x}}=\frac{\mathrm{h}}{{\mathrm{x}}_2}\Rightarrow {\mathrm{x}}_2=\frac{\mathrm{h}\mathrm{x}}{10}$

$\mathrm{t}\mathrm{a}\mathrm{n}??=\frac{15}{\mathrm{x}}=\frac{\mathrm{h}}{\mathrm{x}}\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{h}\mathrm{x}}{15}$

Now,  $x_1+x_2=x=\frac{hx}{15}+\frac{hx}{10}$

$\Rightarrow 1=\frac{\mathrm{h}}{10}+\frac{\mathrm{h}}{15}\Rightarrow \mathrm{h}=6$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Weknowthatonx-axis,y=0andz=0.\\ So,therequiredcoordinatesare\left(3,0,0\right).\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

$(a+\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?x)(a-\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?y)=a^2-b^2$

$\Rightarrow a^2-\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?y+\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?x-2b^2\mathrm{c}\mathrm{o}\mathrm{s}?x\mathrm{c}\mathrm{o}\mathrm{s}?y=a^2-b^2$

Differentiating both sides:

$0-\sqrt[]{2}ab\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\sqrt[]{2}ab(-\mathrm{s}\mathrm{i}\mathrm{n}?x)-2b^2\left[\mathrm{c}\mathrm{o}\mathrm{s}?x\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?y(-\mathrm{s}\mathrm{i}\mathrm{n}?x)\right]=0$

At $\left(\frac{\pi }{4},\frac{\pi }{4}\right)$ :

$\begin{array}{rr}& ab\frac{dy}{dx}-ab-2b^2\left(-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}\right)=0\\ & \Rightarrow \frac{dx}{dy}=\frac{ab+b^2}{ab-b^2}=\frac{a+b}{a-b};a,b>0\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Weknowthatonxyplane,z=0\\ So,thecoordinatesofthe\text{\hspace{0.17em}point}Lare\left(3,4,0\right).\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Given\text{\hspace{0.17em}points}areA\left(5,8,10\right)andB\left(3,6,8\right)\\ \therefore AB=\sqrt[]{{\left(5-3\right)}^2+{\left(8-6\right)}^2+{\left(10-8\right)}^2}=\sqrt[]{4+4+4}=\sqrt[]{12}=2\sqrt[]{3}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Ontheyzplane,x=0\\ Hence,thelocusofthe\text{\hspace{0.17em}point}isyzplane.\\ Hence,thecorrectoptionis\left(b\right).\end{array}$