Maths NCERT Exemplar Solutions Class 12th Chapter One

$\left(p\Delta q\right)\Rightarrow \left(p\Delta \sim q\right)\vee \left(\sim p\Delta q\right)$

Case I

When  $\Delta$  is same as $\vee .$  

Then  $\left(p\Delta \sim q\right)\vee \left(-p\Delta q\right)$ becomes

$\left(p\vee \sim q\right)\vee \left(\sim p\vee q\right)$ which is always true, so x becomes tautology.

Case II

When  $\Delta$  is same as $\vee$  

Then  $\left(p\wedge q\right)\Rightarrow \left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$  becomes $p\wedge q$ is T, then $\left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$ is false, so x cannot be tautology.

Case III

When  $\Delta$  is same as $\Rightarrow$  

Then  $\left(p\Rightarrow \sim q\right)\vee \left(-p\Rightarrow q\right)$ is same as $\left(\sim p\vee \sim q\right)\vee \left(p\vee q\right)$ which is true, so x becomes tautology.

Case IV

When  $\Delta$ is same as $\iff$  

Then   $\left(p\iff q\right)\Rightarrow \left(p\iff \sim q\right)\vee \left(\sim p\iff q\right)$

$p\iff q$ is true when p and q have same truth values $p\iff \sim qand\sim p\iff q$  both are false. Hence x cannot be tautology.

  $\frac{sin\theta sin\theta }{cos\theta }+\frac{sin\theta }{cos\theta }=2sin\theta cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }\left[sin\theta +1\right]=2sin\theta cos\theta$            

$\therefore$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$  

$\theta =\frac{3\pi }{2},\theta =\frac{\pi }{6},\frac{5\pi }{6}$           

$\theta =\frac{3\pi }{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{\pi }{3}+cos\frac{5\pi }{3}=4$  

$\therefore$ T + n(s) = 4 + 5 = 9

$\frac{x^2-5x+6}{x^2-9}\ge -1\&\frac{x^2-5x+6}{x^2-9}\le 1$

$\frac{2x+1}{x+3}\ge 0,x\ne -3\&\frac{1}{x+3}\ge 0,x\ne -3\Rightarrow x>-3$

$x\in \left[-\frac{1}{2},\infty \right]...........(i)$

$x^2-3x+2>0and{x}^2-3x+1\ne 0$

(x – 2) (x – 1) > 0 and $x\ne \frac{3\pm \sqrt[]{5}}{2}$

$x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

From (i) and (ii)   $x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

n = 33, p = success, q = failure

3P (x = 0) = P (x = 1)

${\Rightarrow }^{33}{C}_0{p}^0{q}^{33}{=}^{33}{C}_{1}p{q}^{32}$            

$\Rightarrow p=\frac{1}{12},q=\frac{11}{12}\Rightarrow \frac{q}{p}=11$          

………. (i)

Subtracting, (ii) – (i), we get 1320

$\left|\widehat{a}\right|=1,\left|\widehat{b}\right|=1$

${\left|\widehat{b}\right|}^2={\left|\overrightarrow{c}+2\left(\overrightarrow{c}*\widehat{a}\right)\right|}^2$

$\Rightarrow 1={\left|\overrightarrow{c}\right|}^2+4{\left|\overrightarrow{c}\right|}^2{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)}^2$

${\left|\overrightarrow{c}\right|}^2\left[1+4*\frac{{\left(\sqrt[]{3}-1\right)}^2}{8}\right]={\left|\overrightarrow{c}\right|}^2\left(3-\sqrt[]{3}\right)$

$\therefore {\left|\overrightarrow{c}\right|}^2=\frac{1}{3-\sqrt[]{3}}=\frac{3+\sqrt[]{3}}{6}$

$\Rightarrow {\left|6\overrightarrow{c}\right|}^2=6\left(3+\sqrt[]{3}\right)$

$\lambda x-2y=\mu ........(i)and\frac{x^2}{\left(\frac{b^2}{a^2}\right)}-\frac{y^2}{b^2}=1.........(ii)$

Let (x₁, y₁) be a point on the curve equation of tangent of eq (ii)

  $y=mx\pm \sqrt[]{\frac{b^2}{a^2}{m}^2-b^2}..........(iii)$

From (i) and (ii) are identical

$\therefore m=\frac{\lambda }{2}and\frac{b^2}{a^2}{m}^2-b^2=\frac{{\mu }^2}{4}$               

$\Rightarrow \frac{b^2}{a^2}*\frac{{\lambda }^2}{4}-b^2=\frac{{\mu }^2}{4}$

$\Rightarrow {\left(\frac{\lambda }{a}\right)}^2-{\left(\frac{\mu }{b}\right)}^2=4$

3cos²2θ + 6cos2θ -   $10\frac{\left(1+cos2\theta \right)}{2}+5=0$

$cos2\theta \left(3cos2\theta +1\right)=0$

 Þ cos2θ = 0,     $\frac{-1}{3}$

Draw y = cos2θ, y = 0 and y =  $\frac{-1}{3},$  find the pt. of intersection.

Let A, A' be (, 2) AB and A'B subtends $\frac{\pi }{4}$ angle at (0, 0) slope of OA = $\frac{2}{\alpha }$

slope of OB = $\frac{3}{2}$

$tan\frac{\pi }{4}=\left|\frac{\frac{2}{\alpha }-\frac{3}{2}}{1+\frac{2}{\alpha }\cdot \frac{3}{2}}\right|$

$\Rightarrow 1+\frac{3}{\alpha }=\pm \left(\frac{4-3\alpha }{2\alpha }\right)$

$\Rightarrow \frac{\alpha +3}{\alpha }=\pm \left(\frac{4-3\alpha }{2\alpha }\right)\Rightarrow \alpha =10,-\frac{2}{5}$

now distance between A'A, (10, 2) & $\left(\frac{-2}{5},2\right)is\frac{52}{5}$

Use aRb = a is related to b, belongs to A iff a belongs to A.

In simple terms, aRb is true if both a & b belongs to the same set.

For reflexive

aRa, a $\in$ A, so it is true.

For symmetric

Let aRb be true

Þ a & b belongs to the same set.

Þ b & a also belongs to the same set

Þ bRa will be true

For transitive

Let aRb and bRc be true.

aRb Þ a, b belongs to the same set

bRc Þ b, c belongs to the same set

Þ (a, c) belongs to the same set

Þ so aRc will be true.

So R is an equivalence relation.

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$

Therefore square of $ar\left(\Delta PQR\right)$ = 153.