Let f (x) = e ^ ax (a > 0). The intersection of a straight line passing through point P (a, 0) and parallel to Y-axis and curve C: y = f (x) is Q, and the tangent of curve C passing through point Q intersects X-axis at point R, then The minimum area of triangle PQR is () A.1 B.√(2e)/2 C.e/2 D.e^2/4

Let f (x) = e ^ ax (a > 0). The intersection of a straight line passing through point P (a, 0) and parallel to Y-axis and curve C: y = f (x) is Q, and the tangent of curve C passing through point Q intersects X-axis at point R, then The minimum area of triangle PQR is () A.1 B.√(2e)/2 C.e/2 D.e^2/4

B
f(x) = e^(ax)
Q(a,e^(a²))
f'(x) = ae^(ax)
f'(a) = ae^(a²)
Tangent passing through point Q: Y - e ^ (A & # 178;) = AE ^ (A & # 178;) (x - a)
y = 0,x = a - 1/a,R(a - 1/a,0)
Area of triangle PQR s = (1 / 2) RP * PQ
= (1/2)(a - a + 1/a)*e^(a²)
= [e^(a²)]/(2a)
Derivation of a: s' = 2 [e ^ (A & # 178;)] (2a & # 178; - 1) / A & # 178; = 0
a² = 1/2
a = 1/√2
S = √(2e)/2

Given that the point P is on the curve f (x) = x4-x and the tangent of the curve at point P is parallel to the straight line 3x-y = 0, then the coordinate of point P is ()
A. （0，0）B. （1，1）C. （0，1）D. （1，0）

The derivative of the function is y ′ = f ′ (x) = 4x3-1, the tangent of the curve at point P is parallel to the straight line 3x-y = 0, and the tangent slope of the curve at point P is k = 3. Let P (a, b), that is, k = f ′ (a) = 4a3-1 = 3, then A3 = 1, the solution is a = 1, and then B = f (1) = 0, that is, the tangent point P (1, 0), so choose D

Find limsn =? From the proportional sequence an a1 + A2 + a3 = 18 A2 + a3 + A4 = - 9?

（a2+a3+a4）/（a1+a2+a3）=q=-（1/2）
a1+a2+a3=a1+（-1/2）a1+（1/4）a1=（3\4）a1=18
a1=24,q=-（1/2）
Sn=（a1（1-（-1/2）^n））\（1+1/2）=16-16(-1/2)^n

Given that the common ratio Q ∈ R, a1 + A2 + a3 = 9, A4 + A5 + A6 = - 3, Sn is the sum of the first n terms of the sequence {an}, then LiMn →∞ Sn equals ()
A. 36175B. 48175C. 6D. 274

∵ a1 + a1q + a1q2 = 9, Q3 (a1 + a1q + a1q2) = - 3, ∵ Q3 = - 13, then a11 − q = 274 ∵ Sn = A1 (1 − QN) 1 − q = 274 (1-qn) ∵ LiMn →∞ Sn = 274 (1-qn) = 274, so D is selected

A 1 + a 2 + a 3 = 18, a 2 + a 3 + a 4 = - 9, limsn =?

∵a1+a2+a3=18 ∴18q=q(a1+a2+a3)=a2+a3+a4=-9
∴ q=-1/2
∵18=a1+a2+a3=a1(1+q+q^2)=a1×(1-1/2+1/4)=3/4×a1
∴ a1=24
∴Sn=a1(q^n-1)/(q-1)
limSn(n->∞)=a1/(1-q)=24/(1+1/2)=16

An is an equal ratio sequence, A1 = 2A3 + 1A4 is an equal difference sequence, the general term formula of an and the sum SN of the first n terms are obtained

The general term formula is the nth power of 2, q is equal to 2. You can find the sum of the first n terms by yourself

If Sn = a1 + A2 +... + an, A3 = 2 * S2 + 1, A4 = 2 * S3 + 1, then the common ratio q is

a4=2*S3+1
a3=2*S2+1
Subtraction of two formulas
a4-a3=2a3
a4=3a3
a4/a3=3
q=3

In the equal ratio sequence {an}, Sn represents the first n terms and A3 = 2 × S2 + 1, A4 = 2 × S3 + 1, then the common ratio q is?

Because A3 = A1 × Q ^ 2 = 2 × S2 + 1; A4 = A1 × Q ^ 3 = 2 × (S2 + A1 × Q ^ 2) + 1;
So a4-a3 = A1 × (Q ^ 3-q ^ 2) = 2A1 × Q ^ 2
Q ^ 3-q ^ 2 = 2q ^ 2, q = 3

In the known arithmetic sequence an, a1 + a3 = 10, A4 + A6 = 10, find the sum of the fourth term and the first five terms

a4+a6
=a1q^3+a3q^3
=q^3(a1+a3)
10=q^3*10
q^3=10
q=1
a1+a3=10
a1+a1q^2=10
a1+a1=10
2a1=10
a1=5
an=a1q^(n-1)
=5
a4=5
S5=a1+a2+a3+a4+a5
=5+5+5+5+5
=25

In the known sequence {an}, a1 + a3 = 10, A4 + A6 = 54, find the sum of the fourth term and the first five terms

Let the common ratio be q According to the known results, a1 + a1q2 = 10a1q3 + a1q5 = 54 (3 points) ② A1 (1 + Q2) = 10a1q3 (1 + Q2) = 54 (5 points) ② △ ① get & nbsp; Q3 = 18, that is, q = 12 (7 points) substituting q = 12 into (1) to get A1 = 8 (8) A4 = a1q3 = 8 × (12