Class 12th

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

12 * 9 – 17 * 5 = 23

f (3n) – f (3n– 3) = n

n = 1

f (3) – f (0) = 1

f (0) = 0

n = 2

$\begin{array}{l}f\left(6\right)–f\left(3\right)=2\\ ?\text{?}??\end{array}$

f (3n) – f (3) = 2 + 3 + 4+…………….+ n

f (3n) = 1 + 2 + 3 +…………….+ n

= $n\frac{\left(n+1\right)}{2}$

$?$ f (1312) – f (3 * 311)

$\frac{3^{11}}{2}\left(3^{11}+1\right)=\frac{3^{22}+3^{11}}{2}$

Hence, remainder $\left[\frac{\frac{3^{22}+3^{11}}{2}}{10}\right]$ = 8

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

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