Question

prove...............!!

Answer 1

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}} </p><p>$

$Let \: f(x) = \: {e}^{x} - 2$

f(x) is continuous for all x

f(0) = - 1 < 0

and

f(1) = e - 2 < 0

So By the Intermediate Value Theorem

f(c) = 0 for some number c between x = 0 and x = 1

So, f has at least one real root

Assume that f(x) has another real root say, b Such that f(b) = f(c) = 0

Hence

f is continuous on [b , c]

f is differentiable on (b, c)

So, by Rolle's Theorem, there exists a real number say d in (b, c) such that f'(d) = 0

But

$f'(x) = {e}^{x} > 0 \: for \: all \: x$

Therefore f'(d) ≠ 0

------- This leads to a contradiction

So, our assumption that f(x) has more than 1 real root is false.

Therefore, f(x) has only one real root

Hence the proof follows

Answer 2

Answer:

Letf(x)=e

x

−2

f(x) is continuous for all x

f(0) = - 1 < 0

and

f(1) = e - 2 < 0

So By the Intermediate Value Theorem

f(c) = 0 for some number c between x = 0 and x = 1

So, f has at least one real root

Assume that f(x) has another real root say, b Such that f(b) = f(c) = 0

Hence

f is continuous on [b , c]

f is differentiable on (b, c)

So, by Rolle's Theorem, there exists a real number say d in (b, c) such that f'(d) = 0

But

f'(x) = {e}^{x} > 0 \: for \: all \: xf

′

(x)=e

x

>0forallx

Therefore f'(d) ≠ 0

------- This leads to a contradiction

So, our assumption that f(x) has more than 1 real root is false.

Therefore, f(x) has only one real root

Hence the proof follows