Quadratic equations from a function perspective Use the graph of a quadratic function to judge the existence and number of real roots, and understand the relation between a function's zeros and an equation's roots.
Practice questions 1 下列命题为真命题的是( ) (\quad) ( ) ∃ x 0 ∈ R \text{∃} x_{0}\in R ∃ x 0 ∈ R x 0 2 + 4 x 0 + 6 ⩽ 0 x_{0}^{2}\text{+}4x_{0}+6\leqslant 0 x 0 2 + 4 x 0 + 6 ⩽ 0 y = tan x y=\tan x y = tan x R R R 2 下列四个命题是假命题的是( ) (\quad) ( ) { 0 } \{0\} { 0 } a ∈ N a\in N a ∈ N − a ∉ N -a\notin N − a ∈ / N { x ∈ R ∣ x 2 − 2 x + 1 = 0 } \{x\in R|x^{2}-2x+1=0\} { x ∈ R ∣ x 2 − 2 x + 1 = 0 } 2 2 2 { x ∈ Q ∣ 6 x ∈ N } \{x\in Q|\dfrac{6}{x}\in N\} { x ∈ Q ∣ x 6 ∈ N } 3 已知函数f ( x ) = e x − 1 f(x)={{e}^{x}}-1 f ( x ) = e x − 1 g ( x ) = sin x ( 1 ) g(x)=\sin x(1) g ( x ) = sin x ( 1 ) F ( x ) = f ( x ) − g ( x ) F(x)=f(x)-g(x) F ( x ) = f ( x ) − g ( x ) x ∈ [ 0 , + ∞ ) x\in[0,+\infty) x ∈ [ 0 , + ∞ ) x ∈ [ 0 , π ] x\in[0,\pi] x ∈ [ 0 , π ] f ( x ) ⩾ a g ( x ) ( a ∈ R ) f(x)\geqslant ag(x)(a\in R) f ( x ) ⩾ a g ( x ) ( a ∈ R ) 4 若抛物线y = x 2 − 2 x + k y=x^{2}-2x+k y = x 2 − 2 x + k x x x k k k 5 在平面直角坐标系中，如果点P P P P P P P ( 1 , 1 ) P(1,1) P ( 1 , 1 ) ( − 2 , − 2 ) (-2,-2) ( − 2 , − 2 ) ( 0.5 , 0.5 ) (0.5,0.5) ( 0.5 , 0.5 ) y = a x 2 + 7 x + c ( a ≠ 0 ) y=ax^{2}+7x+c(a≠0) y = a x 2 + 7 x + c ( a = 0 ) 6 若方程2 a x 2 − x − 1 = 0 2a{{x}^{2}}-x-1=0 2 a x 2 − x − 1 = 0 ( 0 , 1 ) (0,1) ( 0 , 1 ) a a a ( ) () ( ) a < − 1 a < -1 a < − 1 a > 1 a > 1 a > 1 − 1 < a < 1 -1 < a < 1 − 1 < a < 1 0 ⩽ a < 1 0\leqslant a < 1 0 ⩽ a < 1 7 Complete the table of values for y = x 2 + 2 x − 1 y = x^2 + 2x - 1 y = x 2 + 2 x − 1 y = x 2 + 2 x − 1 y = x^2 + 2x - 1 y = x 2 + 2 x − 1 − 4 ≤ x ≤ 2 -4 \le x \le 2 − 4 ≤ x ≤ 2 8 函数f ( x ) = x a x + b ( a , b f(x)=\dfrac{x}{ax+b}(a,b f ( x ) = a x + b x ( a , b ) ) ) ( 2 , 1 ) (2,1) ( 2 , 1 ) f ( x ) f(x) f ( x ) f ( x ) = x f(x)=x f ( x ) = x f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x ) ( − 2 , + ∞ ) (-2,+∞) ( − 2 , + ∞ ) 9 设椭圆 C C C x 2 a 2 + y 2 b 2 = 1 ( a > b > 0 ) \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a>b>0) a 2 x 2 + b 2 y 2 = 1 ( a > b > 0 ) F 1 F_{1} F 1 F 2 F_{2} F 2 2 2 \dfrac{\sqrt{2}}{2} 2 2 10 若函数y = ( a − 1 ) x 2 − 4 x + 2 a y=(a-1)x^{2}-4x+2a y = ( a − 1 ) x 2 − 4 x + 2 a x x x a a a 11 已知函数y = a x 2 + 2 x + c y=ax^{2}+2x+c y = a x 2 + 2 x + c x x x ( ) (\quad) ( ) a c > 1 ac>1 a c > 1 y > 0 y>0 y > 0 a a a c c c y y y 0 0 0 12 已知函数f ( x ) = a x 2 − 3 x + b f\left({x}\right)=ax^{2}-3x+b f ( x ) = a x 2 − 3 x + b a a a b ∈ R b\in R b ∈ R x x x f ( x ) > 0 f\left({x}\right)>0 f ( x ) > 0 ( − 4 , 1 ) \left({-4,1}\right) ( − 4 , 1 ) 13 有下列命题：①m x 2 + 2 x − 1 = 0 mx^{2}+2x-1=0 m x 2 + 2 x − 1 = 0 y = a x 2 + 2 x − 1 y=ax^{2}+2x-1 y = a x 2 + 2 x − 1 x x x ( ) (\quad) ( ) 1 1 1 14 已知圆C 1 C_{1} C 1 x 2 + y 2 = r 2 ( r > 0 ) x^{2}+y^{2}=r^{2}(r > 0) x 2 + y 2 = r 2 ( r > 0 ) C 2 C_{2} C 2 x 2 = 2 p y ( p > 0 ) x^{2}=2py(p > 0) x 2 = 2 p y ( p > 0 ) ( 2 , − 2 ) ( \sqrt {2},-2) ( 2 , − 2 ) C 1 C_{1} C 1 C 2 C_{2} C 2 15 在直角坐标系x O y xOy x O y C 1 {{C}_{1}} C 1 y = k ∣ x ∣ + 2. y=k\left| x \right|+2. y = k ∣ x ∣ + 2. x x x C 2 {{C}_{2}} C 2 ρ 2 + 2 ρ cos θ − 3 = 0 {{\rho }^{2}}+2\rho \cos \theta -3=0 ρ 2 + 2 ρ cos θ − 3 = 0 16 已知函数f ( x ) = x 2 + a x + b − a ( a , b ∈ R ) f(x)=x^{2}+ax+b-a(a,b∈R) f ( x ) = x 2 + a x + b − a ( a , b ∈ R ) x x x f ( x ) > 0 f(x) > 0 f ( x ) > 0 ( − ∞ , − 1 ) ∪ ( 3 , + ∞ ) (-∞,-1)∪(3,+∞) ( − ∞ , − 1 ) ∪ ( 3 , + ∞ ) a a a b b b a = 2 a=2 a = 2 f ( x ) > b 2 − 3 b f(x) > b^{2}-3b f ( x ) > b 2 − 3 b 17 若关于x x x m ( x − 1 ) > x 2 − x m(x-1) > {{x}^{2}}-x m ( x − 1 ) > x 2 − x { x ∣ 1 < x < 2 } \left\{ x|1 < x < 2 \right\} { x ∣1 < x < 2 } m m m 18 一次函数y 1 = x + 1 {{y}_{1}}=x+1 y 1 = x + 1 y 2 = 2 x {{y}_{2}}=\dfrac{2}{x} y 2 = x 2 ( 1 , 2 ) , ( − 2 , − 1 ) (1,2),(-2,-1) ( 1 , 2 ) , ( − 2 , − 1 ) x + 1 = 2 x x+1=\dfrac{2}{x} x + 1 = x 2 x 1 = 1 , x 2 = 2 {{x}_{1}}=1,{{x}_{2}}=2 x 1 = 1 , x 2 = 2 19 已知命题p:关于 x x x m x − 1 ≥ 0 mx-1≥0 m x − 1 ≥ 0 2 ∈ A 2∈A 2 ∈ A x 2 − 2 x + m = 0 x ^ { 2 } -2x+m=0 x 2 − 2 x + m = 0 20 已知双曲线x 2 − y 2 3 = 1 x^{2}- \dfrac {y^{2}}{3}=1 x 2 − 3 y 2 = 1 M M M N N N y = x + m y=x+m y = x + m M N MN MN y 2 = 18 x y^{2}=18x y 2 = 18 x m m m 21 下列几个命题： ①函数y = x 2 - 1 + 1 - x 2 y\text{=}\sqrt{x^{2}\text{-}1}\text{+}\sqrt{1\text{-}x^{2}} y = x 2 - 1 + 1 - x 2 { a > 0 Δ = b 2 - 4 a c ⩽ 0 \left\{\begin{matrix}a\text{ > }0\\ {\Delta=}b^{2}\text{-}4ac{\leqslant}0\\\end{matrix}\right. { a > 0 Δ = b 2 - 4 a c ⩽ 0 22 若不等式a x 2 + b x + 2 > 0 ax^{2}+bx+2 > 0 a x 2 + b x + 2 > 0 { x ∣ − 1 2 < x < 1 3 } \{x|- \dfrac {1}{2} < x < \dfrac {1}{3}\} { x ∣ − 2 1 < x < 3 1 } a − b a-b a − b 23 关于函数f ( x ) = { 2 x − a , 0 ⩽ x < 2 b − x , x ⩾ 2 f(x)=\begin{cases}{2^{x}-a,0\leqslant x< 2}\\ {b-x,x\geqslant 2}\end{cases} f ( x ) = { 2 x − a , 0 ⩽ x < 2 b − x , x ⩾ 2 a a a b ∈ R b\in R b ∈ R 6 6 6 24 在数学中，布劳威尔不动点定理是拓扑学里一个非常重要的不动点定理，它可以应用到有限维空间并构成了一般不动点定理的基石．布劳威尔不动点定理得名于荷兰数学家鲁伊兹・布劳威尔( L . E . J . B r o u w e r ) (L.E.J.Brouwer) ( L . E . J . B ro u w er ) 25 在直角坐标系x o y xoy x oy l l l M ( 3 , 4 ) M(3,4) M ( 3 , 4 ) 45 ∘ 45^{\circ} 4 5 ∘ x x x x o y xoy x oy C C C ρ = 4 sin θ ρ=4\sin θ ρ = 4 sin θ 26 函数y = ( x + 1 ) 2 − 9 y=(x+1) ^{2} -9 y = ( x + 1 ) 2 − 9 x x x 27 已知函数f ( x ) = e x − a x f(x)= \dfrac {e^{x}-a}{x} f ( x ) = x e x − a g ( x ) = a ln x + a g(x)=a\ln x+a g ( x ) = a ln x + a ( 1 ) a = 1 (1)a=1 ( 1 ) a = 1 F ( x ) = f ( x ) − g ( x ) F(x)=f(x)-g(x) F ( x ) = f ( x ) − g ( x ) x > 1 x > 1 x > 1 y = f ( x ) y=f(x) y = f ( x ) 28 在平面直角坐标系中，抛物线y 1 = x 2 − 2 x + a {y}_{1}={x}^{2}-2x+a y 1 = x 2 − 2 x + a x x x A A A y 2 = x 2 + 2 a + 1 + 2 a {y}_{2}={x}^{2}+2a+1+2a y 2 = x 2 + 2 a + 1 + 2 a x x x B B B A A A B B B 29 设a ∈ R a\in R a ∈ R x ⩽ 0 x\leqslant 0 x ⩽ 0 ( a x + 1 ) ( x 2 − x − 2 a ) ⩽ 0 (ax+1)(x^{2}-x-2a)\leqslant 0 ( a x + 1 ) ( x 2 − x − 2 a ) ⩽ 0 a = a= a = 30 函数f ( x ) = x 2 + m x − 6 f(x)=x^{2}+mx-6 f ( x ) = x 2 + m x − 6 − 6 -6 − 6 31 已知函数f ( x ) = 1 − 2 2 a x − 1 + 1 ( a > 0 f(x)=1- \dfrac {2}{2a^{x-1}+1}(a > 0 f ( x ) = 1 − 2 a x − 1 + 1 2 ( a > 0 a ≠ 1 ) a\neq 1) a = 1 ) R R R ( ( ( ) ) ) a a a ( ( ( ) ) ) f ( x ) f(x) f ( x ) 32 对∀ x ∈ R \forall x\in R ∀ x ∈ R M ( x ) M(x) M ( x ) f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) M ( x ) = max { f ( x ) , g ( x ) } M(x)=\max\{f(x),g(x)\} M ( x ) = max { f ( x ) , g ( x )} M ( x ) = { − x + 3 , ( x − 1 ) 2 } M(x)=\{-x+3,{{(x-1)}^{2}}\} M ( x ) = { − x + 3 , ( x − 1 ) 2 } M ( x ) M(x) M ( x ) 33 已知关于x x x a x 2 + 4 a x − 3 < 0. ( 1 ) ax^{2}+4ax-3< 0.(1) a x 2 + 4 a x − 3 < 0. ( 1 ) { x ∣ x < − 3 \left\{\begin{array}{l}x\left|{x< -3}\right.\end{array}\right. { x ∣ x < − 3 x > − 1 } x>-1\} x > − 1 } 34 已知函数f ( x ) = ln x x − a ( a ∈ R ) . ( 1 ) f\left(x\right)=\dfrac{{\rm\ln}x}{x}-a\left(a\in R\right).(1) f ( x ) = x ln x − a ( a ∈ R ) . ( 1 ) f ( x ) f(x) f ( x ) a a a f ( x ) f(x) f ( x ) m m m 35 若α ∈ ( 0 , π 2 ) \alpha \in \left( 0,\dfrac{\pi }{2} \right) α ∈ ( 0 , 2 π ) cos 2 α = 2 5 5 sin ( α + π 4 ) \cos 2\alpha =\dfrac{2\sqrt{5}}{5}\sin \left( \alpha +\dfrac{\pi }{4} \right) cos 2 α = 5 2 5 sin ( α + 4 π ) 36 已知α，β是关于x的方程（x-a）（x-b）-1=0的两实根，实数a、b、α、β的大小关系可能是（ ） A. α＜a＜b＜β B. a＜α＜β＜b C. a＜α＜b＜β D. α＜a＜β＜b 37 已知等差数列{ a n } \{a_{n}\} { a n } d > 0 d>0 d > 0 n n n S n S_{n} S n a 2 a 3 = 45 a_{2}a_{3}=45 a 2 a 3 = 45 a 1 + a 4 = 14. ( 1 ) a_{1}+a_{4}=14.(1) a 1 + a 4 = 14. ( 1 ) { a n } \{a_{n}\} { a n } n n n 38 已知函数f ( x ) = e x f(x)=e^{x} f ( x ) = e x g ( x ) = a 2 x + b ( a , b ∈ R ) g(x)= \dfrac {a}{2}x+b(a,b∈R) g ( x ) = 2 a x + b ( a , b ∈ R ) h ( x ) = f ( x ) g ( x ) h(x)=f(x)g(x) h ( x ) = f ( x ) g ( x ) b = 1 − a 2 . b=1- \dfrac {a}{2}. b = 1 − 2 a . h ( x ) h(x) h ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] 39 已知函数f ( x ) = e x ( x 2 − 2 x + a ) ( f(x)=e^{x}(x^{2}-2x+a)( f ( x ) = e x ( x 2 − 2 x + a ) ( a ∈ R a∈R a ∈ R a a a e e e ) ) ) ( ( ( ) ) ) f ( x ) f(x) f ( x ) ( ( ( ) ) ) y = f ( x ) y=f(x) y = f ( x ) 40 已知点P ( x 0 , y 0 ) P(x_{0},y_{0}) P ( x 0 , y 0 ) C 1 C_{1} C 1 C 2 C_{2} C 2 y 2 = 4 x y^{2}=4x y 2 = 4 x A A A B B B P A PA P A P B PB PB C 2 C_{2} C 2 A A A 41 已知函数f ( x ) = x x 2 + 4 , x ∈ ( − 3 , 3 ) f(x)=\dfrac{x}{x^{2}+4},x\in(-3,3) f ( x ) = x 2 + 4 x , x ∈ ( − 3 , 3 ) x ∈ ( − 3 , 3 ) x\in(-3,3) x ∈ ( − 3 , 3 ) a ∈ [ − 1 , 2 ] a\in[-1,2] a ∈ [ − 1 , 2 ] f ( x ) + a t + t 2 − 1 4 < 0 f(x)+at+t^{2}-\dfrac{1}{4}< 0 f ( x ) + a t + t 2 − 4 1 < 0 42 已知函数f ( x ) = { ln x , x > 0 x 2 + 2 x , x ⩽ 0 f(x)=\left\{\begin{array}{l}\ln x,x>0\\ {x}^{2}+2x,x⩽0\end{array}\right. f ( x ) = { ln x , x > 0 x 2 + 2 x , x ⩽ 0 g ( x ) = f ( x ) + 1 g(x)=f(x)+1 g ( x ) = f ( x ) + 1 43 若集合A＝{x∈R|ax2＋ax＋1＝0}中只有一个元素，则a＝( ) A. 4 B. 2 C. 0 D. 0或4 44 已知函数f ( x ) = 3 x f(x)=3^{x} f ( x ) = 3 x g ( x ) = ∣ x + a ∣ − 2 ( a ∈ R ) . g(x)=|x+a|-\:2(a\in R). g ( x ) = ∣ x + a ∣ − 2 ( a ∈ R ) . y = f ( g ( x ) ) y=f(g(x)) y = f ( g ( x )) a = a= a = ; ; ; y = g ( f ( x ) ) y=g(f(x)) y = g ( f ( x )) a a a 45 已知函数f ( x ) = ( log 2 x ) 2 − log 2 x 2 + 3 f\left(x\right)={\left({\log }_{2}x\right)}^{2}-{\log }_{2}{x}^{2}+3 f ( x ) = ( log 2 x ) 2 − log 2 x 2 + 3 x ∈ [ 1 , 4 ] x∈\left[1,4\right] x ∈ [ 1 , 4 ] f ( x ) f\left(x\right) f ( x ) 46 已知函数f ( x ) = − 2 x ln x + x 2 − 2 a x + a 2 . f(x)=-2x\ln x+x^{2}-2ax+a^{2}. f ( x ) = − 2 x ln x + x 2 − 2 a x + a 2 . g ( x ) g(x) g ( x ) f ( x ) f(x) f ( x ) y = f ( x ) y=f(x) y = f ( x ) ( 1 , f ( 1 ) ) (1,f(1)) ( 1 , f ( 1 )) x + y + 3 = 0 x+y+3=0 x + y + 3 = 0 a a a 47 已知映射f f f A → B A→B A → B A = B = R A=B=R A = B = R f f f x → y = x 2 − 2 x + 2 x→y=x^{2}-2x+2 x → y = x 2 − 2 x + 2 k ∈ B k∈B k ∈ B A A A k k k ( ) () ( ) k ⩽ 1 k\leqslant 1 k ⩽ 1 48 已知函数f ( x ) = log a ( a x + 1 ) + b x ( a > 0 f(x)=\log _{a}(a^{x}+1)+bx(a > 0 f ( x ) = log a ( a x + 1 ) + b x ( a > 0 a ≠ 1 a\neq 1 a = 1 b ∈ R ) b∈R) b ∈ R ) y y y f ( 0 ) = 1 f(0)=1 f ( 0 ) = 1 ( ( ( ) ) ) a a a b b b ( ( ( 49 已知过抛物线y2＝4x的焦点F的弦与抛物线交于A，B两点，过A，B分别作y轴的垂线，垂足分别为C，D，则|AC|＋|BD|的最小值是___． 50 已知抛物线M M M y 2 = 8 x y^{2}=8x y 2 = 8 x F F F F F F M M M A A A B B B S △ A O B = 4 5 ( O S_{\triangle AOB}=4\sqrt{5}(O S △ A OB = 4 5 ( O ) ) ) ( ) (\quad) ( ) 51 对于二次函数y = ( x − 1 ) 2 + 2 y=(x-1)^{2}+2 y = ( x − 1 ) 2 + 2 ( ) () ( ) ( 1 , 2 ) (1,2) ( 1 , 2 ) x = − 1 x=-1 x = − 1 x x x 52 下列叙述中不正确的是( ) (\quad) ( ) a , b , c ∈ R a,b,c\in R a , b , c ∈ R a x 2 + b x + c ⩾ 0 a{{x}^{2}}+bx+c\geqslant 0 a x 2 + b x + c ⩾ 0 b 2 − 4 a c ⩽ 0 {{b}^{2}}-4ac\leqslant 0 b 2 − 4 a c ⩽ 0 53 已知函数f ( x ) = 1 3 x 3 − ( 2 k + 1 ) x 2 + 3 k ( k + 2 ) x + 1 f(x)= \dfrac {1}{3}x^{3}-(2k+1)x^{2}+3k(k+2)x+1 f ( x ) = 3 1 x 3 − ( 2 k + 1 ) x 2 + 3 k ( k + 2 ) x + 1 k k k k = − 1 k=-1 k = − 1 f ( x ) f(x) f ( x ) [ 0 , 6 ] [0,6] [ 0 , 6 ] 54 抛物线y = 2 ( x − 3 ) ( x + 2 ) y=2(x-3)(x+2) y = 2 ( x − 3 ) ( x + 2 ) x x x ( ) (\quad) ( ) 1 1 1 2 2 2 3 3 3 5 5 5 55 函数f ( x ) = { x − 1 , x > 0 x 2 − 4 , x < 0 f(x)=\begin{cases}{x-1,x>0}\\ {x^{2}-4,x< 0}\end{cases} f ( x ) = { x − 1 , x > 0 x 2 − 4 , x < 0 56 若函数f ( x ) = x 2 − 2 x + a ln x f(x)=x^{2}-2x+a\ln x f ( x ) = x 2 − 2 x + a ln x a a a ( ) (\quad) ( ) a > 1 2 a>\dfrac{1}{2} a > 2 1 − 1 2 < a < 0 -\dfrac{1}{2}< a< 0 − 2 1 < a < 0 a < 1 2 a< \dfrac {1}{2} a < 2 1 57 已知集合A = { a ∣ A=\{a| A = { a ∣ x x x x + a x 2 − 1 = 1 \dfrac{x+a}{x^{2}-1}=1 x 2 − 1 x + a = 1 a ∈ R } a∈R\} a ∈ R } A = A= A = 58 已知抛物线y = 2 x 2 + 4 x + k − 1 y=2x^{2}+4x+k-1 y = 2 x 2 + 4 x + k − 1 x x x k k k 59 抛物线y = a x 2 + b x + c y=ax^{2}+bx+c y = a x 2 + b x + c x x x ( − 1 , 0 ) (-1,0) ( − 1 , 0 ) ( 3 , 0 ) (3,0) ( 3 , 0 ) x x x a x 2 + b x + c = 0 ax^{2}+bx+c=0 a x 2 + b x + c = 0 60 下列各小题中，p p p q q q ( ) ① p ()①p ( ) ① p m < − 2 m < -2 m < − 2 m > 6 m > 6 m > 6 q q q y = x 2 + m x + m + 3 y=x^{2}+mx+m+3 y = x 2 + m x + m + 3 ② p ： f ( − x ) f ( x ) = 1 ②p： \dfrac {f(-x)}{f(x)}=1 ② p ： f ( x ) f ( − x ) = 1