Class 11th

S₁ = $\frac{1}{2}$ a (p – 1)²

S₂ = $\frac{1}{2}$ ap²

S₁ + S₂ = $\frac{1}{2}$ a [ (p – 1)² + p²] = $\frac{1}{2}$ at²

$t=\sqrt[]{2p^2+1-2p}$

$h=\frac{2Scos\theta }{\rho gR}$

as T­, S¯

The correct answer is Option (2).

  $\frac{GMm}{{\left(R+x\right)}^2}=\frac{GMm\left(R-x\right)}{R^3}$

->R³ = (R + x)² (R – x)

->R³ = (R²– x²) (R + x)

->x² + Rx – R² = 0

$x=\frac{-R\pm \sqrt[]{R+4R^2}}{2}$

$x=\frac{\left(\sqrt[]{5R}-1\right)}{2}R$

  $\frac{v^2}{2a}=s$ (a = mg)

v² = 2 * mgs

v² = 2 * (0.4) * 10 * 10

v² = 80

W_f = Dk

= $-\frac{1}{2}*100*80$  

W_f = – 4000

n P (A) = 2⁷ = 128

f : A → B

Number of function = 128 * 128….128 = 128⁷

$={\left(2^7\right)}^7=2^{49}$

->mⁿ = 2⁴⁹

m + n = 49 + 2 = 51

$N={\displaystyle \sum _{}^{}f}=27$

$\left(\frac{N}{2}\right)=\frac{27}{2}=13.5$

So, we have median lies in the class 12 – 16

I₁ = 12, f = 8, h = 4, c.f. = 13

So, here we apply formula

$M=I_1+\frac{\frac{N}{2}-c.f.}{f}*h=12+\frac{13.5-13}{8}*4$

$=12+\frac{5}{2}$

$M=\frac{24.5}{2}=12.25$

20 M = 20 * 12.25

= 245

n (C) = 16, n (P) = 20, n (M) = 25

n (MÇP) = n (MÇC) = 15, n (PÇC) = 10,

n (MÇCÇP) = x.

n (CÈPÈM) £ n (U) = 40

n (CÈPÈM) = n (C) + n (P) + n (M) – n (C? M) – n (PÈM) – n (CÇP) + n (CÇPÇM)

40 ³ 16 + 20 + 25 – 15 – 15 – 10 + x

40 ³ 61 – 40 + x

19 ³ x

So maximum number of students that passed all the exams is 19.

$T_{n+1}={}^n{C}_{r}11^{\frac{1}{6}}\cdot 7^{\frac{824-4}{2}}$

For integral term

6 should divide r

and  $\frac{824-r}{2}$  must be integer

->2 most divide r

->r divisible by 6

->possible values of r Î {0, 1, 2, …824}

->For integer terms

r Î {0, 6, 12, …822} (822 = 0 + (n – 1)6 Þ n = 138)

= 138 terms

$z^2=-\overline{iz}$

$\left|z^2\right|=\left|-\overline{iz}\right|$

$\left|z^2\right|=\left|z\right|$

$\left|z^2\right|-\left|z\right|=0$

$\left|z\right|\left(\left|z\right|-1\right)=0$

|z| = 0 (not acceptable)

|z| = 1

|z|² = 1

Eqⁿ : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct