Question

The correct answer will be marked as the brainliest, and the silly answers will be reported and removed. ..........XD

Answer 1

Let,

• the strawberry = $\displaystyle\sf {a}$

• the orange = $\displaystyle\sf {b\neq a}$

• the pear = $\displaystyle\sf {c}$

• the grapes = $\displaystyle\sf {d}$

• the lemon = $\displaystyle\sf {e}$

• the brinjal = $\displaystyle\sf {f}$

• the watermelon = $\displaystyle\sf {g}$

• the cherry = $\displaystyle\sf {h}$

From second equation,

$\displaystyle\longrightarrow\sf{5a=b}$

Then first equation becomes,

$\displaystyle\longrightarrow\sf{ab=b}$

$\displaystyle\longrightarrow\sf{5a^2-5a=0}$

$\displaystyle\longrightarrow\sf{5a(a-1)=0}$

$\displaystyle\longrightarrow\sf{a=0\quad;\quad a=1}$

If $\displaystyle\sf {a=0,}$

$\displaystyle\longrightarrow\sf{b=0}$

But given that $\displaystyle\sf {a\neq b.}$ Hence,

$\displaystyle\longrightarrow\sf{a=1}$

And so,

$\displaystyle\longrightarrow\sf{b=5}$

But how the fourth equation is possible?

$\displaystyle\longrightarrow\sf{\dfrac {1}{d}=1-1}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{d}=0}$

$\displaystyle\longrightarrow\sf{d=\infty}$

From the third equation,

$\displaystyle\longrightarrow\sf{c=(1+1)\times 5^{1+1+1}-5-5-1}$

$\displaystyle\longrightarrow\sf{c=2\times5^3-11}$

$\displaystyle\longrightarrow\sf{c=239}$

Consider the fifth one.

$\displaystyle\longrightarrow\sf{e=\lim_{n\to\infty}\left (1+\dfrac {1}{n}\right)^n}$

Let,

$\displaystyle\longrightarrow\sf{y=\left(1+\dfrac{1}{n}\right)^n}$

$\displaystyle\longrightarrow\sf{\ln y=n\cdot\ln\left(1+\dfrac{1}{n}\right)}$

$\displaystyle\longrightarrow\sf{\ln y=\dfrac{\ln\left(1+\dfrac{1}{n}\right)}{\left(\dfrac{1}{n}\right)}}$

$\displaystyle\longrightarrow\sf{\lim_{n\to\infty}\ln y=\lim_{n\to\infty}\dfrac{\ln\left(1+\dfrac{1}{n}\right)}{\left(\dfrac{1}{n}\right)}}$

As $\displaystyle\sf{n\to\infty,\ \dfrac{1}{n}\to0.}$ then,

$\displaystyle\longrightarrow\sf{\lim_{n\to\infty}\ln y=\lim_{\frac{1}{n}\to0}\dfrac{\ln\left(1+\dfrac{1}{n}\right)}{\left(\dfrac{1}{n}\right)}}$

But we know that $\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\ln(1+x)}{x}=1.}$ Then,

$\displaystyle\longrightarrow\sf{\lim_{n\to\infty}\ln y=1}$

$\displaystyle\longrightarrow\sf{\lim_{n\to\infty}y=e}$

Thus,

$\displaystyle\longrightarrow\sf{\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e}$

And hence, the lemon,

$\displaystyle\longrightarrow\sf{e=e}$

Consider the sixth equation.

$\displaystyle\longrightarrow\sf{f=(5-1)\times((5-1)\cot^{-1}(5)-\cot^{-1}(239))}$

$\displaystyle\longrightarrow\sf{f=4\left(4\tan^{-1}\left(\dfrac{1}{5}\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right)}$

But,

$\displaystyle\longrightarrow\sf{\tan(4x)=\dfrac{4\tan x-4\tan^3x}{1-6\tan^2x+\tan^4x}}$

implies,

$\displaystyle\longrightarrow\sf{4\tan^{-1}(a)=\tan^{-1}\left(\dfrac{4a-4a^3}{1-6a^2+a^4}\right)}$

only if $\displaystyle\sf{a=\tan x\quad\iff\quad x=\tan^{-1}(a).}$

Then,

$\displaystyle\longrightarrow\sf{f=4\left(\tan^{-1}\left(\dfrac{\dfrac{4}{5}-\dfrac{4}{125}}{1-\dfrac{6}{25}+\dfrac{1}{625}}\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right)}$

$\displaystyle\longrightarrow\sf{f=4\left(\tan^{-1}\left(\dfrac{120}{119}\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right)}$

Since $\displaystyle\sf{\tan^{-1}(a)-\tan^{-1}(b)=\tan^{-1}\left(\dfrac{a-b}{1+ab}\right),}$

$\displaystyle\longrightarrow\sf{f=4\tan^{-1}\left(\dfrac{\dfrac{120}{119}-\dfrac{1}{239}}{1+\dfrac{120}{119\times239}}\right)}$

$\displaystyle\longrightarrow\sf{f=4\tan^{-1}\left(\dfrac{120\times239-119}{119\times239+120}\right)}$

$\displaystyle\longrightarrow\sf{f=4\tan^{-1}\left(1\right)}$

$\displaystyle\longrightarrow\sf{f=4\cdot\dfrac{\pi}{4}}$

$\displaystyle\longrightarrow\sf{f=\pi}$

Consider the seventh equation,

$\displaystyle\longrightarrow\sf{\dfrac {g}{h}=e+f}$

$\displaystyle\longrightarrow\sf{\dfrac {g}{h}=e+\pi}$

But it's given that $\displaystyle\sf{g,\ h\in\mathbb{Z}.}$ So the LHS $\displaystyle\sf{\dfrac{g}{h}}$ is rational but the RHS $\displaystyle\sf{e+\pi}$ is irrational.

Hence there are no possible integer values for the watermelon and the cherry.

Answer 2

|| ✰✰ ANSWER ✰✰ ||

First :- Strawberry * Orange = Orange

Second :- 5 (Strawbery) = Orange

Conclude :-

→ Strawberry = 1

→ Orange = 5

_______________

Now,

Third :- (1 + 1) * 5^(1+1+1) - 5 - 5 - 1 = Pear

→ Pear = 2 * 5³ - 11 = 250 - 11 = 239.

_______________

So, Now,

Fourth :- (1/grapes) = 1 - 1 = 0

→ (1/grapes) = 0

→ grapes = (1/0)

→ Grapes = ∞

_______________

Fifth :- lim(n→∞) [ 1 + 1/n]^n = Mango

→ Mango = e

_______________

Sixth :- (Putting all values)

→ (5-1)*[ (5-1) * cot^(-1)5 - cot^(-1)239 ] = Eggplant

→ 4[4cot^(-1)5 - cot^(-1)239 ] = Eggplant

⟪ Now For value of [ 4cot^(-1)5 - cot^(-1)239 ] Refer To My Image .. ⟫

we get :-

→ 4 * (π/4) = Eggplant

→ Eggplant = π.

_______________

Seventh :-

→ (watermelon /Lichi ) = e + π

But given , watermelon & Lichi € Z.

But value of (e + π) cant be Rational Number .

Therfore, We conclude That, Both our Values are un-defined.