Multi-problem-solving "thinking Divergent development training counseling"

多题一解★01

\ (\fbox{example 1-1}\) The dynamic point on the line \ (y=x\) is \ (p\), the dynamic point on the function \ (y=lnx\ ) is \ (q\), beg \ (| pq|\) of the minimum value.

\ (\fbox{example 1-2}\) the point on the line \ (y=x\) is \ (P (x, y) \), the point on the function \ (y=lnx\) is \ (Q (m,n) \) , the minimum value of \ (\sqrt{(x-m) ^2+ (y-n) ^2}\) .

Analysis: Using parallel line method,

Set and line \ (y=x\) parallel with the function \ (y=lnx\) tangent line is \ (y=x+m\),

The tangent point is \ (p_0 (x_0,y_0) \), then there is

\ (\begin{cases} y_0=x_{0}+ m \ \ y_0=lnx_0 \ f ' (X_0) =\cfrac{1}{x_0}=1\end{cases}\);

thereby solving the (x_0=1,y_0=0,m=-1\)

So the minimum value of the point distance is converted to the point line spacing of the tangent point \ (p_0 (1,0) \) to the line \ (x-y=0\) ,

\ (d=\cfrac{|1-0|} {\sqrt{1^2+1^2}}=\cfrac{\sqrt{2}}{2}\).

or two lines \ (y=x,y=x-1\) of line spacing \ (d=\cfrac{|1-0|} {\sqrt{1^2+1^2}}=\cfrac{\sqrt{2}}{2}\). Courseware Address

多题一解★02

\ (\fbox{example 2-1}\) the known function \ (f (x) =x^2 +ax-2\ge 0\) can be set on the interval \ ([1,5]\) , and the range of parameters \ (a\) is obtained.

\ (\fbox{example 2-2}\) the known inequality \ (x^2 +ax-2\ge 0\) has a solution on the interval [1,5], seeking the range of parameters \ (a\) .

\ (\fbox{example 2-3}\) known inequality \ (x^2 +ax-2\ge 0\) in the interval [1,5] The solution set is not an empty, the parameter \ (a\) range of values.

\ (\fbox{example 2-4}\) the known inequality \ (x^2 +ax-2\ge 0\) has at least one solution on the interval [1,5] to find the range of parameters \ (a\) .

\ (\fbox{example 2-5}\) known proposition \ (p\): Inequality \ (x^2 +ax-2< 0\) in the interval [1,5] no real solution is a false proposition, parameter \ (a\) Range of values to be taken.

"FA 1": The same way to get \ (a≥\cfrac{2}{x}-x\) in the interval \ ([1,5]\) can be set up to transform into a novelty function \ (\cfrac{2}{x}-x\) in the \ ([1,5] \) on the minimum value.

Order \ (g (x) =\cfrac{2}{x}-x,g (x) =\cfrac{2}{x}-x\) is monotonically decreasing on the interval \ ([1,5]\) ,

So \ (g (x) _{min}=g (5) =-\cfrac{23}{5}\), so \ (a≥-\cfrac{23}{5}\),

The value range of \ (a\) is \ ([-\cfrac{23}{5},+\infty) \)

"FA 2": \ ( f (x) _{max}\ge 0\)on x\in [1,5]\] ,

The axis of symmetry is \ (x=-a\), for the x=-a\ and a given interval of the position of the relationship classification discussion, more cumbersome,

① when \ (-a\leq 1\) , that is \ (a\ge-1\) ,\ (f (x) \) is monotonically incrementing in the interval \ ([1,5]\] ),

So \ (f (x) _{max}=f (5) =5a+23\ge 0\), i.e. \ (a\ge-\cfrac{23}{5}\),

And because \ (a\ge-1\), to seek the intersection get \ (a\ge-1\);

② when \ (1<-a<5\) , i.e. \ ( -5<a<-1\) ,\ (f (x) \) in Interval \ ([1,5]\) There is no increase in monotony,

\ (f (x) _{max}=max{f (1), F (5)}\),

\ (f (1) =a-1\),\ (f (5) =5a+23\),

\ (f (5)-F (1) =4a+24\in [4,20]\], i.e. \ (f (5) >f (1) \),

So \ (f (x) _{max}=f (5) =5a+23\ge 0\), i.e. \ (a\ge-\cfrac{23}{5}\),

The intersection gets,\ (-\cfrac{23}{5}\leq a<-1\);

③ when \ (-a\ge 5\) , i.e. \ (a\leq-5\) ,\ (f (x) \) is monotonically decreasing in interval \ ([1,5]\] ),

So \ (f (x) _{max}=f (1) =a-1\ge 0\), i.e. \ (A\ge 1\),

The intersection gets \ (a\in \varnothing\);

In summary, get \ (a\in [-\cfrac{23}{5},+\infty) \).

The value range of \ (a\) is \ ([-\cfrac{23}{5},+\infty) \)

"FA 3": Converting to inequality \ (f (x) =x^2 +ax-2≥0\) has a solution on the interval \ ([1,5]\) ,

The solution is basically the same as the Method 2,

① when \ (-a\leq 1\) , must \ (f (5) \ge 0\), solution ( a\ge-1\);

② when \ (1<-a<5\) , must \ (f (5) \ge 0\), Solution (-\cfrac{23}{5}\leq a<-1\);

③ when \ (-a\ge 5\) , must \ (f (1) \ge 0\), Solution (a\in \varnothing\);

In summary, get \ (a\in [-\cfrac{23}{5},+\infty) \).

Multi-problem-solving "thinking Divergent development training counseling"