Class 12th

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

I? = ∫? ¹ (1 − x? )? · 1 dx
= (1 - x? )? x|? ¹ - ∫? ¹ 7 (1 - x? )? (-4x³)xdx = -28∫? ¹ (1 − x? )? (1 − x? − 1)dx
I? = −28∫? ¹ (1 − x? )? dx + 28∫? ¹ (1 − x? )? dx
I? = -28I? + 28I?
29I? = 28I?
I? /I? = 28/29 ⇒ (29/4) * (I? /I? ) = (29/4) * (28/29) = 7

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

  $\left(\lambda ,2\lambda ,3\lambda \right)$

$cos\theta =\frac{6}{\sqrt[]{42}}\Rightarrow sin\theta =\frac{\sqrt[]{6}}{\sqrt[]{42}}$ (where q is the angle between L₁ & L₂)

$Area\Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)sin\theta$

$=\frac{1}{2}\left(\sqrt[]{3}\right)\left|\lambda \right|\left(\sqrt[]{14}\right)\frac{\sqrt[]{6}}{\sqrt[]{42}}=\sqrt[]{6}\Rightarrow \lambda =\pm 2$

${\mathrm{M}\mathrm{a}}_2{\mathrm{b}}_2{\mathrm{c}}_2=5$ Geometrical Isomers

M (AA)b2c2:3  $\mathrm{M}\left(\mathrm{A}\mathrm{A}\right){\mathrm{b}}_2{\mathrm{c}}_2:3$  Geometrical Isomers

dy/dx = (e? – x)? ¹
dx/dy = e? - x
dx/dy + x = e? ⇒ I.F. = e^ (∫dy) = e?
∴ xe? = ∫e? dy + c
xe? = e²? /2 + c
y (0) = 0 ⇒ c = -1/2
∴ x = e? /2 - (1/2)e? ⇒ e? – e? = 2x
e²? – 2xe? - 1 = 0
e? = x ± √ (x² + 1) ⇒ e? = x + √ (x² + 1)
y = ln (x + √ (x² + 1)

Kindly consider the following Image 

${\mathrm{C}\mathrm{r}\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}+2{\mathrm{H}}_2{\mathrm{O}}_2?\underset{\text{Chromium Pentoxide}}{{\mathrm{C}\mathrm{r}\mathrm{O}}_5+3{\mathrm{H}}_2\mathrm{O}}$

Molarity $=\frac{\text{moles of}\text{solute}}{\text{volume of solution in}L}=\frac{1*1.15}{1342}*1000=0.86\mathrm{M}$

f' (x) = cosx + sinx − k ≤ 0∀x ∈ R
k ≥ √2