Class 12th

Assuming that

b – c = p b = c + p

c – a = q c = a + q

Hence b = c + p

= a + q + p

$?$ $y+\frac{1}{(b-c)a(c-a)}$ can be written as a + q + p + $\frac{1}{a.p.q}$

Apply AM GM in equally concept in above equation

$\frac{a+q+p\frac{1}{a.p.q}}{4}\ge 4\sqrt[]{a*p*q*\frac{1}{a.p.q}}$

$\frac{a+q+p\frac{1}{a.p.q}}{4}\ge 1$

$a+q+p\frac{1}{a.p.q}\ge 4$

Hence, minimum value is 4

Let the distance between Delhi (D) and Jaipur (J) be 8x and Akbar's initial speed be 4a and final speed be 7a.

Let the point where he increased his speed at P and let Q be the point on DJ such that $DP=\frac{5}{8}(DJ)$

$\frac{PQ}{PJ}=\frac{4}{7}$

Let PQ = 4y,

PJ = 7y

QJ = 7y – 4y = 3x

3y = 3x

y= x

PQ = 4x and PJ =7x

Required ratio = $\frac{DP}{DJ}$

= $\frac{x}{8x}$ = 1/8

f (a) = a² – 2ax + y

f (x) = x² – 2x² + y = 0

– x² + y =0

y = x²                           ………. (1)

f (y) = y² – 2xy + y = 0

y² – 2xy + x² = 0

  (y – x)² = 0

y – x = 0

y = x

Put value of y in equation 1

We get

x = y = 1

So, f (4) = 4² – 2 (4*1) + 1

= 16 – 8 +1

= 9

Electron emission is used for generating a beam of electrons in an electron microscope. An electron microscope requires a source of electrons for creating a beam of electrons. This source can be mostly a gun that uses electron emission for producing free electrons. Once the electrons are emitted, they are accelerated by anode and then, they are focused as a fine beam through a series of electromagnetic lenses. The beam is then directed onto the specimen that is being examined.

Let the airline has a free luggage allowance →'f' kg

Luggage by M →m kg

Luggage by S →s kg 

Rate charger beyond f kg →'e'/kg

Extra luggage of m = (m – f) kg

Extra luggage of s = (s – f) kg

ATQ

e (2m – f) = 2400 ______equation 1

e (2s – f) = 900 ______equation 2

add 1 and 2

e (m + s –f ) = 1650

Also, e (m – f) + e (s – f) = 1050

e (m + s –f ) – ef = 1050

1650 – ef= 1050

ef = 600

Extra luggage from m = e (m – f)

= $\frac{e(2m-2f)}{2}=\frac{e(2m-f)-ef}{2}$

$=\frac{2400-600}{2}$

= Rs. 900

$\frac{1}{2lo{g}_{a}10}+\frac{1}{2lo{g}_{b}10}+\frac{1}{2lo{g}_{c}10}+\frac{1}{2lo{g}_{d}10}$

$=\frac{lo{g}_{10}a}{2}+\frac{lo{g}_{10}b}{2}+\frac{lo{g}_{10}c}{2}+\frac{lo{g}_{10}d}{2}$

$=\frac{lo{g}_{10}a+lo{g}_{10}b+lo{g}_{10}c+lo{g}_{10}d}{2}$

$=\frac{lo{g}_{10}abcd}{2}$

$=\frac{lo{g}_{10}10^8}{2}$

=8 * ½ = 4

x (x – 6) > 2x – 12

x² – 6x > 2x – 12

x² – 8x > 12x > 0

(x – 2) (x – 6) > 0

x < 2 or x >6

The work function in electron emission is the minimum amount of energy needed for removing an electron from the surface of a material. This function is denoted as? , and it is measured in electron volts. An electron requires at least the work function as energy to escape the surface of a solid material into the vacuum. This represents the difference between energy of an electron at rest in vacuum and fermi level of the material. Work function varies from material to material.

Given equation is

x² – 25 <5

–5 < x² – 25 < 5

20 < x² < 30

$\sqrt[]{20}$ < x < $\sqrt[]{30}$

Hence x = 5

x = 5