0

enter image description here

q 23

my solutionenter image description here

I attempted like above.Then to check i plotted graph on desmos. so it showed me three roots.One before zero also. So i tried using first dervivative to get monotonocity of the function, yet am not able to reach a proper solution.

maveric
  • 2,168
  • question is itself saying to equate to 2x^2-2 – maveric Oct 02 '18 at 19:48
  • 1
    sir, passage 7, question 23. – maveric Oct 02 '18 at 19:51
  • As you have worked it, the vertex of the parabola is at $(-\frac 12, \frac 32)$ But as you have drawn it, you have up it at $(0,2)$ The parabola should be cutting more steeply at $x = 0$ dropping below the exponential on the left of 0, before rising again. – Doug M Oct 02 '18 at 21:41

2 Answers2

1

Let us define the continuous function

$$ U(x)=f(x)-h(x)-2x^2+2=2^x+3^x-2(x^2+x+1) $$

We know that as $x$ increases the exponential portion will eventually overpower the polynomial portion and the function will be positive from that point on. So begin by plotting a few points.

  1. $U(0)=0$ so we have at least one solution
  2. $U(1)=-1$ so there is at least one positive solution since it must eventually be positive
  3. $U(3)=9$ confirmed. So because the graph is continuous there is at least one zero between $1$ and $3$.

Taking the first derivative gives

$$ U^\prime(x)=2^x\ln 2+3^x\ln 3-4x-2 $$

Then $U^\prime(0)=\ln 6-2<0$. So the function is decreasing at $0$ which means it is positive in value just to the left of $0$. So for some small number $a<0$ we have $U(a)>0$. But $U(-1)=-\frac{7}{6}<0$. So there is a change of sign on the interval $(-1,a)$ from negative to positive. Thus there is at least one zero between $-1$ and $a$. So there are at least three zeros.

ADDENDUM: See the comment of @String below.

John Wayland Bales
  • 21,814
0

Number of roots=3. enter image description here

enter image description here

I am usingGeogebra for drawing this curve.

Avinash N
  • 1,163