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+1 vote

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.
in Numerical Ability by Boss (17.5k points)
recategorized by | 53 views
$\mathbf B$ should be the answer by simple substitution.

1 Answer

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


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 (19.2k points)
edited by

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





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 $0.7$.

How to find the functions are decreasing, monotonically decreasing?

By differentiation, right?

@`JEET 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.

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