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+21 votes

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. $3$
in Numerical Ability by Veteran (432k points)
edited by | 5.5k views
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].

Someone please explain this

4 Answers

+27 votes
Best answer

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 $[-5, 5]$.

C is correct answer.

by Boss (17.9k points)
edited by
good approach thanks
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.
+67 votes

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

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)
nice solution
nice approach.......
How to draw   −0.5x2+2 graph?
thumb up (y)
+15 votes

so ans should be C.

by Loyal (6.8k points)
this is exactly correct solution
+7 votes
@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)


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)
this is great, since the question is asked in aptitude section.
Yes. This is the statement of intermediate value theorem. In other words, the value will become zero once while changing signs.

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