1974 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If

x\ne0

or

4

and

y\ne0

or

6,

then

\frac2x+\frac3y=\frac12

is equivalent to

4x+3y=xy

y=\frac{4x}{6-y}

\frac{x}{2}+\frac{y}{3}=2

\frac{4y}{y-6}=x

none of these

Concepts:rational equation factoring

Difficulty rating: 1330

Small hint:

Clear all three denominators at once

Big hint:

After multiplying by

2xy,

isolate the terms containing

x

and factor

Solution:

Multiplying by

2xy

gives

4y+6x=xy,

so

4y=x(y-6).

Since

y\ne6,

x=\frac{4y}{y-6}.

Therefore, the correct answer is D.

2.

Let

x_1

and

x_2

be such that

x_1\ne x_2

and

3x_i^2-hx_i=b,

i=1,

2.

Then

x_1+x_2

equals

-\frac h3

\frac h3

\frac b3

2b

-\frac b3

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1410

Small hint:

Both given numbers satisfy the same quadratic equation

Big hint:

Move

b

to the left and use the sum-of-roots formula

Solution:

The distinct numbers

x_1,x_2

are the two roots of

3x^2-hx-b=0.

By Vieta’s formulas, their sum is

-\frac{(-h)}{3}=\frac{h}{3}.

Therefore, the correct answer is B.

3.

The coefficient of

x^7

in the polynomial expansion of

(1+2x-x^2)^4

is

-8

12

6

-12

none of these

Concepts:binomial theorem polynomial

Difficulty rating: 1830

Small hint:

A degree-

7

term must select degrees

2,

2,

2,

1

from the four factors

Big hint:

Choose which factor contributes

2x

; the other three contribute

-x^2

Solution:

The only way four selected terms can have total degree

7

is to choose

2x

once and

-x^2

three times. There are

4

choices for the first factor, so the coefficient is

4(2)(-1)^3=-8.

Therefore, the correct answer is A.

4.

What is the remainder when

x^{51}+51

is divided by

x+1?

0

1

49

50

51

Concepts:polynomial substitution exponent

Difficulty rating: 1410

Small hint:

For division by

x+1,

evaluate the polynomial at

-1

Big hint:

The exponent

51

is odd

Solution:

By the polynomial remainder theorem, the remainder is

(-1)^{51}+51=-1+51=50.

Therefore, the correct answer is D.

5.

Given a quadrilateral

ABCD

inscribed in a circle with side

AB

extended beyond

B

to point

E,

if

\angle BAD=92^\circ

and

\angle ADC=68^\circ,

find

\angle EBC.

66^\circ

68^\circ

70^\circ

88^\circ

92^\circ

Concepts:cyclic quadrilateral angle chasing

Difficulty rating: 1460

Small hint:

Opposite angles of a cyclic quadrilateral are supplementary

Big hint:

The desired exterior angle is also supplementary to

\angle ABC

Solution:

Because

ABCD

is cyclic,

\angle ABC+\angle ADC=180^\circ.

Since

BE

extends

BA,

\angle EBC+\angle ABC=180^\circ

as well. Hence

\angle EBC=\angle ADC=68^\circ.

The given

92^\circ

angle is unnecessary.

Therefore, the correct answer is B.

6.

For positive real numbers

x

and

y

define

x*y=\frac{x\cdot y}{x+y};

then

“

*

” is commutative but not associative

“

*

” is associative but not commutative

“

*

” is neither commutative nor associative

“

*

” is commutative and associative

none of these

Concepts:custom operation fraction

Difficulty rating: 1740

Small hint:

The formula is visibly unchanged when

x

and

y

are interchanged

Big hint:

Rewrite the operation through its reciprocal:

\frac1{x*y}=\frac1x+\frac1y

Solution:

Symmetry in

x,

y

makes the operation commutative. Also,

\frac1{x*y}=\frac1x+\frac1y.

Thus both

(x*y)*z

and

x*(y*z)

have reciprocal

\frac1x+\frac1y+\frac1z,

so they are equal. The operation is associative too.

Therefore, the correct answer is D.

7.

A town’s population increased by

1{,}200

people, and then this new population decreased by

11\%.

The town now had

32

less people than it did before the

1{,}200

increase. What is the original population?

1{,}200

11{,}200

9{,}968

10{,}000

none of these

Concepts:percentage linear equation

Difficulty rating: 1310

Small hint:

Let

P

be the original population and apply the decrease to

P+1200

Big hint:

The final population is both

0.89(P+1200)

and

P-32

Solution:

Let the original population be

P.

Then

0.89(P+1200)=P-32.

Therefore

0.11P=1100,

giving

P=10{,}000.

Therefore, the correct answer is D.

8.

What is the smallest prime number dividing the sum

3^{11}+5^{13}?

2

3

5

3^{11}+5^{13}

none of these

Concepts:parity prime

Difficulty rating: 920

Small hint:

Both bases are odd

Big hint:

The sum of two odd integers is divisible by the smallest prime

Solution:

Both

3^{11}

and

5^{13}

are odd, so their sum is even. Hence it is divisible by

2,

the smallest prime.

Therefore, the correct answer is A.

9.

The integers greater than one are arranged in five columns as follows:

	$2$	$3$	$4$	$5$
$9$	$8$	$7$	$6$
	$10$	$11$	$12$	$13$
$17$	$16$	$15$	$14$
	$\vdots$	$\vdots$	$\vdots$	$\vdots$

(Four consecutive integers appear in each row; in the first, third and other odd numbered rows, the integers appear in the last four columns and increase from left to right; in the second, fourth and other even numbered rows, the integers appear in the first four columns and increase from right to left.)

In which column will the number

1{,}000

fall?

first

second

third

fourth

fifth

Concepts:pattern recognition modular arithmetic

Difficulty rating: 1430

Small hint:

Look at where the displayed multiples of

8

occur

Big hint:

Each two-row cycle contains eight consecutive integers

Solution:

Each pair of rows is a repeating block of eight integers. In every such block, its multiple of

8

lies in the second column, as do

8

and

16.

Since

1000=125\cdot8,

it also lies in the second column.

Therefore, the correct answer is B.

10.

What is the smallest integral value of

k

such that

2x(kx-4)-x^2+6=0

has no real roots?

-1

2

3

4

5

Concepts:quadratic inequality

Difficulty rating: 1650

Small hint:

First rewrite the equation in standard quadratic form

Big hint:

No real roots means the discriminant is negative

Solution:

The equation is

(2k-1)x^2-8x+6=0.

Its discriminant is

64-24(2k-1)=88-48k.

This is negative exactly when

k\gt\frac{11}{6},

so the least integer possible is

2.

Therefore, the correct answer is B.

11.

If

(a,b)

and

(c,d)

are two points on the line whose equation is

y=mx+k,

then the distance between

(a,b)

and

(c,d),

in terms of

a,

c,

and

m,

is

|a-c|\sqrt{1+m^2}

|a+c|\sqrt{1+m^2}

\frac{|a-c|}{\sqrt{1+m^2}}

|a-c|(1+m^2)

|a-c||m|

Concepts:coordinate geometry distance formula

Difficulty rating: 1390

Small hint:

Use the line equation to express

b-d

in terms of

a-c

Big hint:

Substitute

b-d=m(a-c)

into the distance formula

Solution:

Since both points are on the line,

b-d=m(a-c).

Their distance is therefore

\begin{aligned} \ell&=\sqrt{(a-c)^2+(b-d)^2}\\ &=\sqrt{(a-c)^2(1+m^2)}\\ &=|a-c|\sqrt{1+m^2}. \end{aligned}

Therefore, the correct answer is A.

12.

If

g(x)=1-x^2

and

f(g(x))=\frac{1-x^2}{x^2}

when

x\ne0,

then

f(\frac{1}{2})

equals

\frac{3}{4}

1

3

\frac{\sqrt2}{2}

\sqrt2

Concepts:function substitution

Difficulty rating: 1620

Small hint:

Choose a nonzero

x

for which

g(x)=\frac{1}{2}

Big hint:

The equation

1-x^2=\frac{1}{2}

determines

x^2

directly

Solution:

If

g(x)=\frac{1}{2},

then

x^2=\frac{1}{2},

and either corresponding

x

is nonzero. Hence

f(\frac{1}{2})=\frac{1-x^2}{x^2} =\frac{\frac{1}{2}}{\frac{1}{2}}=1.

Therefore, the correct answer is B.

13.

Which of the following is equivalent to “If

P

is true then

Q

is false”?

“

P

is true or

Q

is false.”

“If

Q

is false then

P

is true.”

“If

P

is false then

Q

is true.”

“If

Q

is true then

P

is false.”

“If

Q

is true then

P

is true.”

Concepts:logical deduction

Difficulty rating: 1450

Small hint:

An implication is equivalent to its contrapositive

Big hint:

Reverse the implication and negate both “

P

true” and “

Q

false”

Solution:

The contrapositive of “

P

true implies

Q

false” is “

Q

not false implies

P

not true.” This is exactly “If

Q

is true then

P

is false.”

Therefore, the correct answer is D.

14.

Which statement is correct?

If

x\lt0,

then

x^2\gt x.

If

x^2\gt0,

then

x\gt0.

If

x^2\gt x,

then

x\gt0.

If

x^2\gt x,

then

x\lt0.

If

x\lt1,

then

x^2\lt x.

Concepts:inequality counterexample

Difficulty rating: 1160

Small hint:

For a negative

x,

compare

x^2

first with zero

Big hint:

Simple test values such as

-1,

\frac12,

and

2

reject the other implications

Solution:

If

x\lt0,

then

x^2\gt0\gt x,

so choice A always holds. The values

x=-1,

x=-1,

x=2,

and

x=-1

respectively disprove B, C, D, and E.

Therefore, the correct answer is A.

15.

If

x\lt-2

then

\bigl|1-|1+x|\bigr|

equals

2+x

-2-x

x

-x

-2

Concepts:absolute value inequality

Difficulty rating: 1430

Small hint:

Resolve the inner absolute value first using

1+x\lt-1

Big hint:

After replacing

|1+x|

by

-1-x,

determine the sign of

2+x

Solution:

Since

x\lt-2,

1+x\lt0,

so

|1+x|=-1-x.

Thus

\bigl|1-|1+x|\bigr|

becomes

|1-(-1-x)|,

which is

|2+x|=-2-x

because

2+x\lt0.

Therefore, the correct answer is B.

16.

A circle of radius

r

is inscribed in a right isosceles triangle, and a circle of radius

R

is circumscribed about the triangle. Then

\frac{R}{r}

equals

1+\sqrt2

\frac{2+\sqrt2}{2}

\frac{\sqrt2-1}{2}

\frac{1+\sqrt2}{2}

2(2-\sqrt2)

Concepts:special right triangle incircle, incenter, and inradius circumcircle, circumcenter, and circumradius

Difficulty rating: 1670

Small hint:

Let each leg have length

a

and express both radii in terms of

a

Big hint:

For a right triangle, the circumradius is half the hypotenuse; the inradius is

\frac{a+a-a\sqrt2}{2}

Solution:

With both legs of length

a

and hypotenuse

a\sqrt2,

\begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2}. \end{aligned}

Therefore

\begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2. \end{aligned}

Therefore, the correct answer is A.

17.

If

i^2=-1,

then

(1+i)^{20}-(1-i)^{20}

equals

-1024

-1024i

0

1024

1024i

Concepts:complex number exponent

Difficulty rating: 1540

Small hint:

Compute the fourth power of each conjugate base

Big hint:

Both

(1+i)^4

and

(1-i)^4

equal

-4

Solution:

Since

(1+i)^2=2i,

we have

(1+i)^4=-4.

Likewise

(1-i)^2=-2i

and

(1-i)^4=-4.

Hence each twentieth power equals

(-4)^5=-1024,

and their difference is

0.

Therefore, the correct answer is C.

18.

If

\log_8 3=p

and

\log_3 5=q,

then, in terms of

p

and

q,

\log_{10}5

equals

pq

\frac{3p+q}{5}

\frac{1+3pq}{p+q}

\frac{3pq}{1+3pq}

p^2+q^2

Concepts:logarithm

Difficulty rating: 1780

Small hint:

Convert every logarithm to base

2

Big hint:

The givens imply

\log_2 3=3p

and

\log_2 5=3pq

Solution:

From

p=\log_8 3,

\log_2 3=3p.

Therefore

\log_2 5=(\log_2 3)(\log_3 5)=3pq.

Since

\log_2 10=1+\log_2 5=1+3pq,

\log_{10}5=\frac{\log_2 5}{\log_2 10} =\frac{3pq}{1+3pq}.

Therefore, the correct answer is D.

19.

In the adjoining figure

ABCD

is a square and

CMN

is an equilateral triangle. If the area of

ABCD

is one square inch, then the area of

CMN

in square inches is

2\sqrt3-3

1-\frac{\sqrt3}{3}

\frac{\sqrt3}{4}

\frac{\sqrt2}{3}

4-2\sqrt3

Concepts:equilateral triangle Pythagorean Theorem triangle area

Difficulty rating: 2070

Small hint:

The square has side

1;

set

DM=NB=x

by comparing the equal sides

CM

and

CN

Big hint:

Equate

CM^2=1+x^2

with

MN^2=2(1-x)^2

Solution:

The square has side

1.

Since

CM=CN,

symmetry of their squared lengths gives

DM=NB=x.

Also,

\begin{aligned} CM^2&=1+x^2,\\ MN^2&=2(1-x)^2. \end{aligned}

Equating these because

CMN

is equilateral gives

x^2-4x+1=0.

The root in

(0,1)

is

x=2-\sqrt3.

The triangle’s side has square

1+x^2,

so its area is

\begin{aligned} A_\triangle&=\frac{\sqrt3}{4}(1+x^2)\\ &=\frac{\sqrt3}{4} \bigl(1+(2-\sqrt3)^2\bigr)\\ &=2\sqrt3-3. \end{aligned}

Therefore, the correct answer is A.

20.

Let

\begin{aligned} T={}&\frac1{3-\sqrt8} -\frac1{\sqrt8-\sqrt7}\\ &+\frac1{\sqrt7-\sqrt6} -\frac1{\sqrt6-\sqrt5}\\ &+\frac1{\sqrt5-2}; \end{aligned}

then

T\lt1

T=1

1\lt T\lt2

T\gt2

\displaystyle T=\frac1{\substack{(3-\sqrt8)(\sqrt8-\sqrt7)(\sqrt7-\sqrt6)\\ {}\cdot(\sqrt6-\sqrt5)(\sqrt5-2)}}

Concepts:rationalizing denominator telescoping

Difficulty rating: 1740

Small hint:

Rationalize each denominator separately

Big hint:

Every denominator is a difference of square roots whose squares differ by

1

Solution:

Rationalizing the five terms gives

\begin{aligned} T={}&(3+\sqrt8)-(\sqrt8+\sqrt7)\\ &+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)\\ &+(\sqrt5+2)=5. \end{aligned}

Thus

T\gt2.

Therefore, the correct answer is D.

21.

In a geometric series of positive terms the difference between the fifth and fourth terms is

576,

and the difference between the second and first terms is

9.

What is the sum of the first five terms of this series?

1061

1023

1024

768

none of these

Concepts:geometric sequence summation

Difficulty rating: 1900

Small hint:

Write the first term as

a

and the common ratio as

r

Big hint:

Divide

ar^3(r-1)=576

by

a(r-1)=9

Solution:

Let the first term be

a

and the ratio be

r.

The conditions give

\begin{aligned} ar^3(r-1)&=576,\\ a(r-1)&=9. \end{aligned}

Their quotient gives

r^3=64,

so

r=4.

Then

3a=9,

hence

a=3.

The sum is

\begin{aligned} S&=3(1+4+16+64+256)\\ &=3\cdot341\\ &=1023. \end{aligned}

Therefore, the correct answer is B.

22.

The minimum value of

\sin\frac A2-\sqrt3\cos\frac A2

is attained when

A

is

-180^\circ

60^\circ

120^\circ

0^\circ

none of these

Concepts:trigonometric identity trigonometry

Difficulty rating: 1780

Small hint:

Combine the sine and cosine into one shifted sine with amplitude

2

Big hint:

Use

\sin u-\sqrt3\cos u=2\sin(u-60^\circ)

Solution:

Let

u=\frac{A}{2}.

Then

\sin u-\sqrt3\cos u

equals

2\sin(u-60^\circ).

Its minimum occurs when

u-60^\circ=270^\circ+360^\circ n,

so

A=660^\circ+720^\circ n

for an integer

n.

None of the first four choices has this form.

Therefore, the correct answer is E.

23.

In the adjoining figure

TP

and

T'Q

are parallel tangents to a circle of radius

r,

with

T

and

T'

the points of tangency.

PT''Q

is a third tangent with

T''

as point of tangency. If

TP=4

and

T'Q=9

then

r

is

\frac{25}{6}

6

\frac{25}{4}

a number other than

\frac{25}{6},

6,

or

\frac{25}{4}

not determinable from the given information

Concepts:tangent line right triangle similarity

Difficulty rating: 2070

Small hint:

The center lies on each angle bisector formed by two tangents from the same external point

Big hint:

Show that

\angle POQ=90^\circ

and use the altitude-to-hypotenuse theorem at

T''

Solution:

The radii to the tangency points make congruent right triangles from each external point, so

OP

and

OQ

bisect the two tangent angles. Because the other two tangents are parallel, these half-angles add to

90^\circ;

hence

\angle POQ=90^\circ.

Thus

OT''=r

is the altitude to the hypotenuse

PQ.

Equal tangent segments give

PT''=PT=4

and

QT''=QT'=9.

The altitude theorem yields

r^2=(PT'')(QT'')=4\cdot9=36,

so

r=6.

Therefore, the correct answer is B.

24.

A fair die is rolled six times. The probability of rolling at least a five at least five times is

\frac{13}{729}

\frac{12}{729}

\frac{2}{729}

\frac{3}{729}

none of these

Concepts:binomial probability dice (probability)

Difficulty rating: 1900

Small hint:

A roll is a success with probability

\frac{2}{6}=\frac{1}{3}

Big hint:

Add the disjoint cases of exactly five successes and exactly six successes

Solution:

Let a success be a

5

or

6,

so its probability is

\frac{1}{3}.

The desired probability is

\begin{aligned} &\binom65\left(\frac13\right)^5 \left(\frac23\right) +\left(\frac13\right)^6\\ &\qquad=\frac{12}{729}+\frac1{729}\\ &\qquad=\frac{13}{729}. \end{aligned}

Therefore, the correct answer is A.

25.

In parallelogram

ABCD

of the accompanying diagram, line

DP

is drawn bisecting

BC

at

N

and meeting

AB

(extended) at

P.

From vertex

C,

line

CQ

is drawn bisecting side

AD

at

M

and meeting

AB

(extended) at

Q.

Lines

DP

and

CQ

meet at

O.

If the area of parallelogram

ABCD

is

k,

then the area of triangle

QPO

is equal to

k

\frac{6k}{5}

\frac{9k}{8}

\frac{5k}{4}

2k

Concepts:parallelogram coordinate geometry triangle area

Difficulty rating: 2100

Small hint:

Use

A=(0,0),

B=(1,0),

D=(u,v),

and

C=(u+1,v)

Big hint:

The midpoint lines meet the baseline at

Q=(-1,0)

and

P=(2,0)

; then find

O

’s height

Solution:

Choose coordinates

A=(0,0),

B=(1,0),

D=(u,v),

and

C=(u+1,v),

so

k=v.

The midpoints are

\begin{aligned} M&=(\frac{u}{2},\frac{v}{2}),\\ N&=(1+\frac{u}{2},\frac{v}{2}). \end{aligned}

Extending

CM

and

DN

to

y=0

gives

Q=(-1,0)

and

P=(2,0).

Parameterizing

DP

and

CQ

shows their intersection is

O=(\frac{3u}{4}+\frac{1}{2},\frac{3v}{4}).

Thus

QP=3

and the height from

O

is

\frac{3v}{4},

so

\begin{aligned} [QPO]&=\frac12(3)\left(\frac{3v}{4}\right)\\ &=\frac{9v}{8}=\frac{9k}{8}. \end{aligned}

Therefore, the correct answer is C.

26.

The number of distinct positive integral divisors of

(30)^4

excluding

1

and

(30)^4

is

100

125

123

30

none of these

Concepts:prime factorization factor counting

Difficulty rating: 1790

Small hint:

Factor

30^4

into powers of

2,

3,

and

5

Big hint:

Choose each prime exponent independently from

0

through

4,

then remove two divisors

Solution:

Because

30^4=2^4\cdot3^4\cdot5^4,

a divisor independently chooses each of three exponents from

0,

1,

2,

3,

4.

There are

5^3=125

divisors. Excluding

1

and

30^4

leaves

125-2=123.

Therefore, the correct answer is C.

27.

If

f(x)=3x+2

for all real

x,

then the statement:

“

|f(x)+4|\lt a

whenever

|x+2|\lt b

and

a\gt0

and

b\gt0

”

is true when

b\le \frac{a}{3}

b\gt \frac{a}{3}

a\le \frac{b}{3}

a\gt \frac{b}{3}

The statement is never true.

Concepts:absolute value function inequality

Difficulty rating: 1670

Small hint:

Factor

f(x)+4

in terms of

x+2

Big hint:

The condition becomes

3|x+2|\lt a

whenever

|x+2|\lt b

Solution:

We have

|f(x)+4|=|3x+6|=3|x+2|.

If

|x+2|\lt b

and

b\le \frac{a}{3},

then

3|x+2|\lt3b\le a,

as required. If

b\gt \frac{a}{3},

one can choose

|x+2|

strictly between

\frac{a}{3}

and

b,

so the statement fails. Thus the exact condition is

b\le \frac{a}{3}.

Therefore, the correct answer is A.

28.

Which of the following is satisfied by all numbers

x

of the form

x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}},

where

a_1

is

0

or

2,

a_2

is

0

or

2,

\ldots,

a_{25}

is

0

or

2?

0\le x\lt\frac{1}{3}

\frac{1}{3}\le x\lt\frac{2}{3}

\frac{2}{3}\le x\lt1

0\le x\lt\frac{1}{3}

or

\frac{2}{3}\le x\lt1

\frac{1}{2}\le x\le\frac{3}{4}

Concepts:number base geometric sequence bounding to limit cases

Difficulty rating: 2090

Small hint:

Separate the cases

a_1=0

and

a_1=2

Big hint:

Bound the remaining terms by the infinite geometric tail

\sum_{n=2}^{\infty}\frac{2}{3^n}

Solution:

If

a_1=0,

then

0\le x\le\sum_{n=2}^{25}\frac2{3^n} \lt\sum_{n=2}^{\infty}\frac2{3^n}=\frac13.

If

a_1=2,

then

x\ge\frac{2}{3},

while

x\le\sum_{n=1}^{25}\frac2{3^n} \lt\sum_{n=1}^{\infty}\frac2{3^n}=1.

Hence every such

x

lies in one of the two stated outer thirds.

Therefore, the correct answer is D.

29.

For

p=1,

2,

\ldots,

10

let

S_p

be the sum of the first

40

terms of the arithmetic progression whose first term is

p

and whose common difference is

2p-1;

then

S_1+S_2+\cdots+S_{10}

is

80{,}000

80{,}200

80{,}400

80{,}600

80{,}800

Concepts:arithmetic sequence summation

Difficulty rating: 1740

Small hint:

Use the

40

-term arithmetic-series formula for a fixed

p

Big hint:

Simplify

S_p=20\bigl(2p+39(2p-1)\bigr)

before summing over

p

Solution:

For each

p,

\begin{aligned} S_p&=\frac{40}{2} \bigl(2p+39(2p-1)\bigr)\\ &=1600p-780. \end{aligned}

Since

1+2+\cdots+10=55,

therefore

\begin{aligned} \sum_{p=1}^{10}S_p &=1600(55)-7800\\ &=80{,}200. \end{aligned}

Therefore, the correct answer is B.

30.

A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If

R

is the ratio of the lesser part to the greater part, then the value of

R^{\left[R^{\left(R^2+R^{-1}\right)}+R^{-1}\right]}+R^{-1}

is

2

2R

R^{-1}

2+R^{-1}

2+R

Concepts:quadratic exponent substitution

Difficulty rating: 2100

Small hint:

Translate the division condition into

R=\frac{1}{1+R}

Big hint:

From

R^2+R=1,

derive

R^{-1}=R+1

and simplify from the innermost exponent outward

Solution:

Scale the greater part to

1,

making the lesser part

R.

The condition gives

R=\frac{1}{1+R},

so

R^2+R=1,\qquad R^{-1}=R+1.

Consequently

R^2+R^{-1}=R^2+R+1=2.

Working outward, the exponent of the outer

R

becomes

\begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2. \end{aligned}

The whole expression is therefore

R^2+R^{-1}=2.

Therefore, the correct answer is A.