+1 vote
51 views

Let $0.01^x+0.25^x=0.7$ . Then

1. $x\geq1$
2. $0\lt x\lt1$
3. $x\leq0$
4. no such real number $x$ is possible.

recategorized | 51 views
0
$\mathbf B$ should be the answer by simple substitution.

+1 vote
$\underline{\mathbf{Answer:B}}\;$ $\bbox[orange,5px,border: 2px dotted red]{\color{blue}{\mathbf{0<x<1}}}$

$\underline{\mathbf{Explanation:}}$

Substitute $\mathbf {x = 0} \Rightarrow (0.001)^0 + (0.25)^0 = 1 + 1 = 2 \;\text{which is }\; \color{red}{> 0.7}$

Similarly, subsitute $\mathbf {x = 1} \Rightarrow (0.01)^1 + (0.25)^1 = 0.26\;\text{which is }\; \color{red}{< 0.7}$

$\therefore \mathbf x$ lies between $\mathbf 0$ and $\mathbf 1$.

$\therefore \mathbf B$ is the right option.
by Boss (18.9k points)
edited by
0

Can you please verify this answer, whether this method is correct or not?

@ankitgupta.1729

+1

@`JEET

yeah, it is correct and the reason is, both functions $0.01^x$ and $0.25^x$ are strictly decreasing for interval $(-\infty,\infty)$.

So, if we consider $f(x)=0.01^x + 0.25^x$

then $f(1) = 0.26$  and since both functions are decreasing.So, value of sum of both functions for $x>1$ will also be decreasing. So, $f(>1)$  will always be less than $0.26$ and this way, we can't get $0.7$. So, option (A) is eliminated.

Now, $f(0) = 2$. Now, when $x<0$, both functions will be strictly increasing and so their sum will also be increasing. So, $f(<0)$ will always be $>2$ and so we will not get $0.7$ for $x\leq 0$. So, option (C) eliminated.

Now, since, $f(0) = 2$ and $f(1) = 0.26$ and $f(x)$ is decreasing in $(-\infty,\infty)$and continuous function, so there must be some value of $x$ which is between $0$ and $1$ for which $f(x)$ will be $0.7$.

+1
Thanks.
+1
How to find the functions are decreasing, monotonically decreasing?

By differentiation, right?
+1

Yeah, right.

Or we can simply remember that if $f(x) = a^x$

Then in case of $a > 1$ , $f(x)$ will be strictly increasing for all $x$

And in case of $0<a<1$, $f(x)$ will be strictly decreasing for all x.

It is quite intuitive also.