Class 11th

(y – 2)² = (x – 1)

2 (y – 2)    $\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}\left(2,3\right)=\frac{1}{2\left(3-2\right)}=\frac{1}{2}$              

Equation of tangent at P (2, 3):

$y-3=\frac{1}{2}\left(x-2\right)$  

2y – 6 = x – 2

x – 2y + 4 = 0

Q (-4, 0)

Required area = ${\displaystyle \underset{0}{\overset{3}{\int }}\left({\left(y-2\right)}^2+1-\left(2y-4\right)\right)dy=9}$

For divisibility by 5 last digit must be 0 or 5 but 0 is not possible in palindrome

so it will be 5.

So, required no. = 10 * 10 = 100

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

Equation of tangent to given ellipse at

$P:\frac{x.cos\theta }{b}+\frac{y.sin\theta }{2a}=1$

A (b sec θ. 0) 7 B (0, 2a cosec θ)

area of    $\Delta OAB$

$=\frac{2ab}{2sin\theta cos\theta }=\frac{2ab}{sin2\theta }$

For minimum area sin 2θ = 1

So minimum area = 2ab

=>k = 2

  $\underset{x\to \infty }{lim}\left(\sqrt[]{x^2-x+1}-ax\right)=b$

$\underset{x\to \infty }{lim}\left|\frac{x^2-x+1-a^2{x}^2}{\sqrt[]{x^2-x+1}+ax}\right|=b$

For existence of limit 1 – a² = 0 i.e. a = 1 only

$\underset{x\to \infty }{lim}\frac{1-x}{\sqrt[]{x^2-x+1}+x}=b$

$\Rightarrow b=\frac{-1}{2}$

So, (a, b) =   $\left(1,-\frac{1}{2}\right)$

$r=\sqrt[]{{\left(\frac{p}{2}\right)}^2+{\left(\frac{1-p}{2}\right)}^2-5}=\sqrt[]{\frac{p^2+1+p^2-2p-20}{4}}=\sqrt[]{\frac{2p^2-2p-19}{2}}$

Since $r\in \left(0,5\right)so0<\sqrt[]{2p^2-2p-19}<10$

$\Rightarrow 2p^2-2p-19>0\&2p^2-2p-19<100$     

$P\in \left[\frac{1-\sqrt[]{239}}{2},\frac{1-\sqrt[]{39}}{2}\right)\cup \left(\frac{1+\sqrt[]{39}}{2},\frac{1+\sqrt[]{239}}{2}\right]$

So number of integral values of P² is 61.

  $\left|x-2\right|>1givesx-2<-1orx-2>1$ $$

i.e. x < 1 or x > 3 . (i) represent set A

$\sqrt[]{x^2-3}>1gives{x}^2-3>1or{x}^2-4>0$

x < 2 or x > 2 . (ii) represent set B

$\left|x-4\right|\ge 2givesx-4\le -2orx-4\ge 2$

$$  $x\le 2orx\ge 6$ . (iii)respect set C

so number of subset = 28 = 256

Molarity (M) = $\frac{w*1000}{molecularmass*volumeofsolution\left(ml\right)}$

= $\frac{6.3*1000}{126*250}=\frac{4}{20}=0.2M$

Molecular mass of oxalic acid $\left(H_2{C}_2{O}_4.2H_{2}O\right)$

= 1 * 2 + 12 * 2 + 16 * 4 + 2 * 18

              = 26 + 64 + 36 = 126

              M = 2 * 10^-1 M

              = 20 * 10^-2M

$\therefore x*10^{-2}=20*10^{-2}$

$\therefore x=20$

Ans. = 20

$A_3{B}_{2\left(s\right)}?3A_{\left(aq\right)}^{+2}+2B_{\left(aq\right)}^{3-}$

Solubility $\frac{x}{M}$ $\frac{3x}{M}$ $\frac{x}{M}$

$\therefore Ksp={\left[A^{+2}\right]}^3{\left[B^{-3}\right]}^2$

$={\left(\frac{3x}{M}\right)}^3{\left(\frac{2x}{M}\right)}^2$

$=108{\left(\frac{x}{M}\right)}^5$

$Ksp=a{\left(\frac{x}{M}\right)}^5=108{\left(\frac{x}{M}\right)}^5$

Ans. $\therefore$ a = 108

(a) Torque ® ML²T^-2 ↑ (iii)

(b) Impulse ® MlT^-1 ↑ (i)

(c) Tension ↑ MLT^-2 ↑ (iv)

(d) Surface Tension ↑  $\frac{ML{T}^{-2}}{L}=M{T}^{-2}$ (ii)