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Calculus Level 5

Two functions f ( x ) , g ( x ) f(x),g(x) are such that f ( x ) = ( x − 2 ) e x − 1 − 1 2 x 2 + x + 1 2 f(x)=(x-2)e^{x-1}-\dfrac{1}{2}x^2+x+\dfrac{1}{2} , g ( x ) = a x 2 − x + 4 a cos ⁡ x + ln ⁡ ( x + 1 ) g(x)=ax^2-x+4a\cos x+\ln(x+1) , where a a is a parameter and a ∈ R a \in \mathbb R .

Function max ⁡ { m , n } \max\{m,n\} denotes the maximum value of m , n m,n . Let F ( x ) = max ⁡ { f ( x ) , g ( x ) } ( x > − 1 ) F(x)=\max \{f(x),g(x)\}(x>-1) .

Casework the number of real roots for F ( x ) = 0 F(x)=0 .

The result is as follows:

  • When a > A a>A or a < B a<B , there are N 1 N_1 roots.

  • When a = A a=A or a = B a=B , there are N 2 N_2 roots.

  • When B < a < A B<a<A , there are N 3 N_3 roots.

Submit ⌊ 10000 ( A + B + N 1 + N 2 + N 3 ) ⌋ \lfloor 10000(A+B+N_1+N_2+N_3) \rfloor .

Note: Thank for Jon Haussmann for revising the answer.


The answer is 60970.

Alice Smith by Alice Smith , 20, China

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1 solution

Making the graph of f ( x ) , g ( x ) and F ( x ) on graph plotter we see that : \text{ Making the graph of }f(x)\,,\, g(x)\text{ and }F(x) \text{ on graph plotter we see that :}

1. x = 1 is the only solution of f ( x ) = 0. 2. Curve y = g ( x ) moves downward as a decreases. 3. From the above figure we see that for a = 0.2 there is only 1 solution of F ( x ) = 0. \text{1. }x = 1\text{ is the only solution of }f(x) = 0.\newline\text{2. Curve }y = g(x) \text{ moves downward as }a\text{ decreases. }\newline\text{3. From the above figure we see that for }a = 0.2\text{ there is only }1\text{ solution of }F(x) = 0.

As a decreases curve y = g ( x ) tends to intersect the curve y = f ( x ) at x = 1. At this value of a , x = 1 is a solution of g ( x ) = 0 i.e. g ( 1 ) = 0 ⇒ a − 1 + 4 a cos ( 1 ) + ln ( 2 ) = 0 ⇒ a = 1 − ln ( 2 ) 1 + 4 cos ( 1 ) = 0.09706 \text{As }a\text{ decreases curve }y = g(x) \text{ tends to intersect the curve }y = f(x)\text{ at }x = 1.\text{ At this value of }a\,,\,x = 1\text{ is a solution of }g(x) = 0\; \text{i.e.}\newline\hspace{15pt} g(1) = 0\newline\Rightarrow a - 1 +4a\,\text{cos}(1) + \text{ln}(2) = 0\newline\large\Rightarrow a = \frac{1 - \text{ln}(2)}{1 + 4\text{cos}(1)} = 0.09706

⇒ A = 0.09706 and N 2 = 2 , N 1 = 1 \newline\Rightarrow A = 0.09706\hspace{30pt} \text{and}\; N_2 = 2\,,\,N_1 = 1

There are three solution for a < A upto certain number B ⇒ N 3 = 3. On decreasing the value of a from a = A we see that curve y = g ( x ) touches the x-axis when a = 0. At this value of a number of solutions of F ( x ) = 0 reduces to 2. \text{There are three solution for }a < A \text{ upto certain number }B \Rightarrow N_3 = 3.\text{ On decreasing the value of }a \text{ from }a = A \text{ we see that curve }y = g(x) \text{ touches the x-axis when }a = 0.\text{ At this value of }a\text{ number of solutions of }F(x) = 0 \text{ reduces to 2. }

On further decreasing a number of solution of F ( x ) = 0 reduces to 1 . So B = 0 \text{On further decreasing }a\text{ number of solution of }F(x) = 0\text{ reduces to }1\text{. So }B = 0

A = 0.09706 , B = 0 , N 1 = 1 , N 2 = 2 , N 3 = 3 ⇒ A + B + N 1 + N 2 + N 3 = 0.09706 + 0 + 1 + 2 + 3 = 6.09706 ⇒ 10000 ( A + B + N 1 + N 2 + N 3 ) = 60970.6 ⇒ ⌊ 10000 ( A + B + N 1 + N 2 + N 3 ) ⌋ = 60970 A = 0.09706\,,\,B = 0\,,\, N_1 = 1\,,\,N_2 = 2\,,\,N_3 = 3\newline\Rightarrow A+B+N_1+N_2+N_3 = 0.09706+0+1+2+3=6.09706\newline\Rightarrow 10000(A+B+N_1+N_2+N_3) = 60970.6\newline\Rightarrow\lfloor10000(A+B+N_1+N_2+N_3)\rfloor = \boxed{60970}

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