JEE Mathematics Progressions Related Inequalities and Series MCQs Set 11

Answer: (b) \( -(p+q) \)

Explanation:
1. Let the AP have first term \( A \) and difference \( D \). Then \( S_p=\tfrac p2[2A+(p-1)D]=q \) and \( S_q=\tfrac q2[2A+(q-1)D]=p \).
2. Dividing and subtracting gives \( 2A+(p+q-1)D=-2 \).
3. Hence \( S_{p+q}=\tfrac{p+q}2\cdot(-2)=-(p+q) \).

Teacher's Note:
1. Subtract the two relations to get the common value of 2A + (p + q - 1)D.

Answer: (a) \( 2^{105}-2^{121} \)

Explanation:
1. There are 16 factors \( (1-2^kx) \), \( k=0,\dots,15 \). The coefficient of \( x^{15} \) is \( (-1)^{15}\sum_{j}\prod_{k\ne j}2^k \).
2. \( \prod_{k}2^k=2^{120} \), so the sum is \( 2^{120}\sum_{j=0}^{15}2^{-j}=2^{120}\left(2-2^{-15}\right)=2^{121}-2^{105} \).
3. With the negative sign the coefficient is \( 2^{105}-2^{121} \).

Teacher's Note:
1. Choose 15 of the 16 factors to supply x; each term omits exactly one factor.

Answer: (a) AP

Explanation:
1. Let the common difference be \( d \), so \( b-a=c-b=d \).
2. \( \left(b+\tfrac1{ca}\right)-\left(a+\tfrac1{bc}\right)=d+\dfrac{b-a}{abc}=d\left(1+\tfrac1{abc}\right) \).
3. Similarly \( \left(c+\tfrac1{ab}\right)-\left(b+\tfrac1{ca}\right)=d\left(1+\tfrac1{abc}\right) \). The differences are equal, so the terms are in AP.

Teacher's Note:
1. Bring each difference to a common denominator abc.

Answer: (a) \( \tfrac32 \)

Explanation:
1. Let the numbers be \( a\gt b \) with \( a+b=4 \). After increasing \( a \) by 1, \( \sqrt{(a+1)b}=2 \), so \( ab+b=4 \).
2. Substituting \( a=4-b \): \( b^2-5b+4=0 \), so \( b=1 \) (then \( a=3 \)) or \( b=4 \) (rejected as \( b\lt a \)).
3. HM \( =\dfrac{2ab}{a+b}=\dfrac64=\tfrac32 \).

Teacher's Note:
1. Use a + b = 4 to reduce the GM condition to a quadratic in b.

Answer: (a) \( \dfrac{2n}{h} \)

Explanation:
1. In the AP, terms equidistant from the ends have the same sum: \( a_k+a_{2n+1-k}=a+b \).
2. In the GP, \( g_kg_{2n+1-k}=ab \).
3. Each of the \( n \) fractions equals \( \dfrac{a+b}{ab}=\dfrac2h \), so the total is \( \dfrac{2n}h \).

Teacher's Note:
1. Pair terms equidistant from the two ends of each progression.

Answer: (c) \( A\ge-\tfrac12 \)

Explanation:
1. Squaring, \( a_i^2=a_{i-1}^2+2a_{i-1}+1 \). Adding for \( i=2,\dots,n \) gives \( a_n^2=2\sum_{i=1}^{n-1}a_i+(n-1) \).
2. So \( \sum_{i=1}^na_i=\dfrac{a_n^2-n+1}2+a_n \) and \( A=\dfrac{(a_n+1)^2-n}{2n} \).
3. Since \( (a_n+1)^2\ge0 \), \( A\ge-\tfrac12 \).

Teacher's Note:
1. Square the relation and add it over all indices so the squares telescope.

Answer: (a) \( a^2r^{n-1} \)

Explanation:
1. \( A=\dfrac{a(r^n-1)}{n(r-1)} \).
2. \( H=\dfrac{n}{\sum\frac1{ar^k}}=\dfrac{n\,a\,r^{n-1}(r-1)}{r^n-1} \).
3. Multiplying, \( AH=a^2r^{n-1} \).

Teacher's Note:
1. Write A and H separately using the geometric series formula.

Answer: (a) 0

Explanation:
1. We need \( (a^n+b^n)(a+b)=2ab(a^{n-1}+b^{n-1}) \).
2. Expanding gives \( a^{n+1}+b^{n+1}=a^nb+ab^n \), i.e. \( (a-b)(a^n-b^n)=0 \).
3. For all \( a,b \) this holds only when \( a^n=b^n \) identically, i.e. \( n=0 \) (check: \( \tfrac{2}{1/a+1/b}=\tfrac{2ab}{a+b} \)).

Teacher's Note:
1. Cross-multiply and factorise the difference.

Answer: (b) 2

Explanation:
1. For numbers \( a,b \): \( y=a^{2/3}b^{1/3} \), \( z=a^{1/3}b^{2/3} \), \( yz=ab \), \( x=\tfrac{a+b}2 \).
2. \( y^3+z^3=a^2b+ab^2=ab(a+b) \).
3. The ratio is \( \dfrac{ab(a+b)}{\frac{a+b}2\cdot ab}=2 \).

Teacher's Note:
1. Insert two geometric means as a to the 2/3 times b to the 1/3 and so on.

Answer: (a) 4 : 1

Explanation:
1. Let \( \text{GM}=\sqrt{ab}=5k \) and \( \text{HM}=\dfrac{2ab}{a+b}=4k \). Then \( ab=25k^2 \) and \( a+b=\tfrac{25k}2 \).
2. \( \dfrac{a+b}{\sqrt{ab}}=\tfrac52 \), i.e. \( t+\tfrac1t=\tfrac52 \) where \( t=\sqrt{a/b} \).
3. So \( t=2 \) (or \( \tfrac12 \)) and \( a:b=t^2:1=4:1 \).

Teacher's Note:
1. Use GM squared equals AM times HM to find the AM.

Answer: (c) \( -2 \)

Explanation:
1. Let the terms be \( a,ar,ar^2 \). The condition is \( 2a=ar+ar^2 \), so \( r^2+r-2=0 \).
2. \( r=1 \) or \( r=-2 \).
3. Terms alternate in sign, so \( r\lt0 \) and \( r=-2 \).

Teacher's Note:
1. Write the AM condition for three consecutive terms.

Answer: (c) \( p^2-q^2=p'^2-q'^2 \)

Explanation:
1. \( p^2-p'^2=\left(\tfrac{a+b}2\right)^2-ab=\tfrac{(a-b)^2}4 \).
2. Similarly \( q^2-q'^2=\tfrac{(b-c)^2}4 \).
3. In an AP \( a-b=b-c \), so \( p^2-p'^2=q^2-q'^2 \), i.e. \( p^2-q^2=p'^2-q'^2 \).

Teacher's Note:
1. AM squared minus GM squared equals a quarter of the squared difference.

Answer: (b) 2

Explanation:
1. In this range \( \cos\theta\gt0 \), so AM-GM applies.
2. \( \cos^3\theta+\sec^3\theta\ge2\sqrt{\cos^3\theta\sec^3\theta}=2 \).
3. Equality at \( \cos\theta=1 \), i.e. \( \theta=0 \).

Teacher's Note:
1. Both terms are positive reciprocals, so AM-GM gives the bound.

Answer: (c) 2

Explanation:
1. Both logarithms are positive and are reciprocals of each other.
2. Hence \( \log_ba+\log_ab\ge2 \) by AM-GM.
3. Equality holds at \( a=b \), so the minimum is 2.

Teacher's Note:
1. The two terms are reciprocals of each other.

Answer: (b) 4

Explanation:
1. By AM-GM, \( 4^x+4^{1-x}\ge2\sqrt{4^x\cdot4^{1-x}} \).
2. \( 4^x\cdot4^{1-x}=4 \), so the bound is \( 2\cdot2=4 \).
3. Equality at \( 4^x=4^{1-x} \), i.e. \( x=\tfrac12 \).

Teacher's Note:
1. The product of the two terms does not depend on x.

Answer: (d) none of these

Explanation:
1. Each term is positive and the product of the \( n \) terms is \( \dfrac{\ln a_1}{\ln a_2}\cdot\dfrac{\ln a_2}{\ln a_3}\cdots\dfrac{\ln a_n}{\ln a_1}=1 \).
2. By AM-GM the sum is at least \( n \), with equality when all \( a_i \) are equal.
3. The minimum value is \( n \), which is not any of 1, 2 or 0 for general \( n \).

Teacher's Note:
1. The product of all the terms telescopes to 1.

Answer: (d) greater than or equal to \( n \)

Explanation:
1. By AM-GM, \( \dfrac{\text{sum}}n\ge(\text{product})^{1/n}=1 \).
2. So the sum is at least \( n \).
3. The sum need not be an integer or a multiple of \( n \) (for example \( 2,\tfrac12 \) has sum \( \tfrac52 \)).

Teacher's Note:
1. Apply AM-GM with a product equal to 1.

Answer: (c) 4

Explanation:
1. Let \( L=\log_{10}a\gt0 \). \( 2\log_{100}a=L \) and \( \log_a10^{-4}=-\dfrac4L \).
2. The expression is \( L+\dfrac4L \).
3. By AM-GM \( L+\dfrac4L\ge4 \), attained at \( L=2 \) (that is \( a=100 \)).

Teacher's Note:
1. Change all logarithms to base 10.

Answer: (d) none of these

Explanation:
1. Put \( t=\sin x+\cos x\in(1,\sqrt2] \), so \( \sin2x=t^2-1 \) and the expression is \( \left(t+\dfrac1{t^2-1}\right)^3 \).
2. \( g(t)=t+\dfrac1{t^2-1} \) has \( g'(t)=1-\dfrac{2t}{(t^2-1)^2}\lt0 \) on \( (1,\sqrt2] \), so \( g \) decreases and is least at \( t=\sqrt2 \) (\( x=\tfrac\pi4 \)).
3. The minimum is \( (\sqrt2+1)^3=7+5\sqrt2\approx14.07 \), which is not listed.

Teacher's Note:
1. Reduce to a single variable t and check the sign of the derivative.

Answer: (a) 3

Explanation:
1. Let \( X=\log x,\ Y=\log y,\ Z=\log z \). The terms are \( e^{X(Y-Z)},e^{Y(Z-X)},e^{Z(X-Y)} \).
2. The sum of the exponents is \( XY-XZ+YZ-XY+ZX-YZ=0 \), so the product of the terms is 1.
3. By AM-GM the sum is at least 3, attained at \( x=y=z \).

Teacher's Note:
1. Take logarithms to see that the product of the three terms is 1.

Answer: (b) \( a=b=\tfrac14,\ c=\tfrac12 \)

Explanation:
1. Write \( a+b+\tfrac c2+\tfrac c2=1 \). By AM-GM, \( \dfrac{a\cdot b\cdot\frac c2\cdot\frac c2}{1}\le\left(\tfrac14\right)^4 \), so \( abc^2\le\tfrac4{256}=\tfrac1{64} \).
2. Equality needs \( a=b=\tfrac c2 \).
3. With \( a+b+c=1 \): \( a=b=\tfrac14,\ c=\tfrac12 \).

Teacher's Note:
1. Split c into two equal halves so that four equal parts can be used in AM-GM.