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The number of roots of $e^{x}+0.5x^{2}-2=0$ in the range $[-5,5]$ is

1. $0$
2. $1$
3. $2$
4. $3$

edited | 5.5k views
0
Why here   f(5)f(-5) >0  ? ,we know that if f(a)f(b)<0  then there will exist at least one root in between [a,b].

Let $f(x)=e^x+0.5x^2-2$, and we want to see if $f(x)=0$ has any solution in $\left [ -5, 5 \right ]$

$f'(x)$ = $e^x+x$ and we definitely have solution of $f'(x)=0$ in $\left [ -5, 5 \right ]$ because $f'(x)$ is continuous and $f'(-5) \lt 0$,  $f'(5) \gt 0$

(Moreover $f'(x)$ has only one solution as it is a strictly increasing function.)

Let $k$ be one of the solution for $f'(x)$ hence $f'(k) = 0$  where $k \in [-5, 5]$, for "$\lt k$", $f'(x)$ is -ve and for "$\gt k$", $f'(x)$ is +ve. means for $\lt k$, $f(x)$ is decreasing and for $\gt k$, $f(x)$ is increasing.

Rough plot of $f(x)$ may go like this

Now it is clear that $f(x)$ has two solutions.

BUT, how can i sure that these solutions are in between $[-5, 5]$ ?

Yes I agree that $k \in [-5, 5]$, but there are many possibilities of solution to be in $[-5, 5]$

(in all three images $k \in [-5, 5]$)

You wanna know which image is correct ? :O

Just check sign of  $f(-5)$ and $f(5)$, u will get to know "Image-1" is correct one.

Hence, there are $2$ solutions in between of $[-5, 5]$.

by Boss (17.9k points)
edited by
0
good approach thanks
0
f(x) has two solutions or not depends on the value of f(k). If f(k) <0, only then f(x) has two solutions.

$e^x = -0.5x^2 + 2$

Plot $e^x$ and Downward Parabola $-0.5x^2 + 2$, And find intersection.

As, these two intersect at two points. So, (C) is correct answer.

by Boss (28.8k points)
+1
nice solution
+1
nice approach.......
+3
How to draw   −0.5x2+2 graph?
0
thumb up (y)

so ans should be C.

by Loyal (6.8k points)
0
this is exactly correct solution
@arjun sir

can we do like this

if we put x=0 in equation we get f(x) = -ve

if we put x=5 in eq we get f(x) = +ve

means there is one value of x between (0,5)  for which eq is 0 there one root is between (o,5)

similarly

when we put x=-5 we get f(x) = +ve

means other root is present between (-5,0)

therefor two roots
by Active (2.7k points)
0
this is great, since the question is asked in aptitude section.
0
Yes. This is the statement of intermediate value theorem. In other words, the value will become zero once while changing signs.