Class 12th

$1-log10_{10}5=\frac{1}{3}\left[log10_{10}\frac{x*5^{1/3}}{2}\right]$

3 [log101010 – log10105] = $log10_{10}\frac{x*5^{1/3}}{2}$

log101023 = $log10_{10}\frac{x*5^{1/3}}{2}$

23 = $\frac{x*5^{1/3}}{2}$

x = $\frac{2^4}{5^{1/3}}$

Hence, m + 3n = $4+3*\frac{1}{3}$ = 5

Let the interest rate be r percent interest earned from 1.2 Lacs at the end of year

= 1,200,000 * r /100

= 1200r

From 1.8 lakh, = 8/12 * 1, 80,000 * 2r/100

= 2400r

Total interest = 3600r = 54000

r = 15 percent

Let the quotient obtained when the numbers is divided by 3, 5 and 6 be x₁, x₂ and x₃

Respectively

N = 3 x₁ + 1

x₁ = 5 x₂ + 3

x₂ = 6 x₃ + 2

N = 3 [5 (6 x₃ + 2) + 3] + 1

=90 x₃ + 40 < 1000

x₃ can take 11 values

In solar cells, electron emission converts the light energy into electrical energy through the photoelectric effect. Solar cells work due to photoelectric effect, in which photons from sunlight strike solar cell surface. When a photon that has sufficient energy hits the surface, it transfers its energy to an electron in the material of solar cell surface. This causes the electron emission from its bound state. Solar cells can use photoelectric effect to generate electron-hole pairs that are separated and collected for producing electrical current. This provides a renewable source of power.

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