JEE Mathematics Sequence and Series MCQs Set 14

Answer: (c) \( 5x^2-16x+7=0 \)

Explanation:
1. Sum of roots \( =\frac{16}5 \). The A.M. of the reciprocals gives \( \frac{\text{sum}}{2\cdot\text{product}}=\frac87 \), so product \( =\frac75 \).
2. The equation is \( x^2-\frac{16}5x+\frac75=0 \), that is \( 5x^2-16x+7=0 \).

Teacher's Note:
1. Use sum and product of roots.

Answer: (b) \( \{1,-1\} \)

Explanation:
1. \( 5^{1+x}+5^{1-x}=26 \). Put \( t=5^x \): \( 5t+\frac5t=26 \), so \( 5t^2-26t+5=0 \).
2. \( t=5 \) or \( \frac15 \), so \( x=1 \) or \( -1 \).

Teacher's Note:
1. Substitute t = 5 to the x.

Answer: (c) \( 5 \)

Explanation:
1. \( 2(x+8)=2x+3x+1 \) gives \( 2x+16=5x+1 \), so \( x=5 \).

Teacher's Note:
1. The middle term is the mean of the other two.

Answer: (a) \( \frac{a_{m+k}+a_{m-k}}2 \)

Explanation:
1. \( a_{m+k}+a_{m-k}=2a+(2m-2)d=2a_m \).

Teacher's Note:
1. Terms equidistant from a_m average to a_m.

Answer: (d) all of \( z^3=x,\ x=y^{-1} \) and \( z^{-3}=y \)

Explanation:
1. Let \( p=\log_yx,q=\log_zy,r=\log_xz \); then \( pqr=1 \). With common difference d, \( p=1+d,q=1+2d,-15r=1+3d \).
2. \( -15=(1+d)(1+2d)(1+3d) \) gives \( (d+2)(6d^2-d+8)=0 \), so \( d=-2 \) is the only real value.
3. Then \( p=-1,q=-3,r=\frac13 \): \( x=y^{-1} \), \( y=z^{-3} \) and \( z=x^{1/3} \). All three hold.

Teacher's Note:
1. Use the product of the three logarithms equals 1.

Answer: (b) \( p^2,q^2,r^2 \) are in A.P.

Explanation:
1. \( \frac2{r+p}=\frac1{p+q}+\frac1{q+r} \) gives \( 2(p+q)(q+r)=(r+p)(p+r+2q) \).
2. Expanding and simplifying: \( 2q^2=p^2+r^2 \).

Teacher's Note:
1. Cross multiply and expand.

Answer: (a) A.P.

Explanation:
1. \( a_p=a+(p-1)d \) is linear in p, so \( a_p,a_q,a_r \) are in A.P. exactly when p, q, r are.

Teacher's Note:
1. A linear function maps an A.P. to an A.P.

Answer: (a) A.P.

Explanation:
1. \( -\frac ba=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{b^2-2ac}{c^2} \).
2. So \( -bc^2=ab^2-2a^2c \), that is \( 2a^2c=ab^2+bc^2 \).
3. This says \( 2(ca^2)=ab^2+bc^2 \), so \( bc^2,ca^2,ab^2 \) are in A.P.

Teacher's Note:
1. Use the sum and product of roots.

Answer: (c) A.P.

Explanation:
1. \( b^2-a^2=c^2-b^2 \) gives \( (b-a)(b+a)=(c-b)(c+b) \).
2. Dividing by \( (a+b)(b+c) \): \( \frac{b-a}{b+c}=\frac{c-b}{a+b} \).
3. Add and rearrange: \( \frac1{c+a}-\frac1{b+c}=\frac{b-a}{(c+a)(b+c)} \) and \( \frac1{a+b}-\frac1{c+a}=\frac{c-b}{(a+b)(c+a)} \) are equal, so the reciprocals are in A.P.

Teacher's Note:
1. Subtract the reciprocals in pairs.

Answer: (c) 3, 4, 5

Explanation:
1. Let the sides be \( a-d,a,a+d \). Then \( (a-d)^2+a^2=(a+d)^2 \) gives \( a=4d \).
2. The sides are \( 3d,4d,5d \).

Teacher's Note:
1. Use the Pythagoras theorem with the largest side as hypotenuse.

Answer: (b) \( (1,-2) \)

Explanation:
1. \( c=2b-a \), so the line is \( a(x-1)+b(y+2)=0 \).
2. It passes through the intersection of \( x-1=0 \) and \( y+2=0 \), that is \( (1,-2) \).

Teacher's Note:
1. Substitute c = 2b - a and group by a and b.

Answer: (c) \( 2(d-c) \)

Explanation:
1. Let the common difference be \( \delta \). Then \( e-c=2\delta \).
2. \( 2(d-c)=2\delta \), while \( 2(c-a)=4\delta \) and \( 2(f-d)=4\delta \).

Teacher's Note:
1. Express everything in the common difference.

Answer: (a) \( \left|\frac rp-7\right|\ge4\sqrt3 \)

Explanation:
1. The discriminant condition \( q^2\ge4pr \) with \( q=\frac{p+r}2 \) gives \( (p+r)^2\ge16pr \).
2. Dividing by \( p^2 \) with \( t=\frac rp \): \( t^2-14t+1\ge0 \), that is \( |t-7|\ge4\sqrt3 \).

Teacher's Note:
1. Use q = (p + r) / 2 in the discriminant.

Answer: (d) \( 4abc \)

Explanation:
1. With \( a+c=2b \): \( a+2b-c=2a \), \( 2b+c-a=2c \) and \( c+a-b=b \).
2. The product is \( 2a\cdot2c\cdot b=4abc \).

Teacher's Note:
1. Replace 2b by a + c.

Answer: (c) \( \pm3 \)

Explanation:
1. The sum of roots is 12, so the middle root is 4. Check: \( 64-192+156-28=0 \).
2. Dividing, \( (x-4)(x^2-8x+7)=0 \) gives roots 1, 4, 7.
3. The common difference is \( \pm3 \).

Teacher's Note:
1. The middle root is one third of the sum of the roots.

Answer: (b) \( 1-\log_34 \)

Explanation:
1. \( 2\log_9(3^{1-x}+2)=1+\log_3(4\cdot3^x-1) \) gives \( 3^{1-x}+2=3(4\cdot3^x-1) \).
2. With \( t=3^x \): \( 3+2t=12t^2-3t \), so \( 12t^2-5t-3=0 \) and \( t=\frac34 \) (the other root is negative).
3. \( x=\log_3\frac34=1-\log_34 \).

Teacher's Note:
1. Convert log base 9 to base 3.

Answer: (a) \( \frac{n(n+1)}2\cdot\frac{a_2-a_1}{a_{n+1}} \)

Explanation:
1. Each denominator equals \( 2a_{n+1} \) (terms equidistant from the middle).
2. The numerators are \( 2nd,(2n-2)d,\dots,2d \), whose sum is \( n(n+1)d \).
3. The total is \( \frac{n(n+1)d}{2a_{n+1}} \) with \( d=a_2-a_1 \).

Teacher's Note:
1. Use the symmetry about the middle term.

Answer: (a) \( x=y=z \)

Explanation:
1. \( 2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z \) gives \( \frac{2y}{1-y^2}=\frac{x+z}{1-xz}=\frac{2y}{1-xz} \).
2. So \( y^2=xz \) (as \( y\ne0 \)). With \( 2y=x+z \): \( (x-z)^2=0 \).
3. Hence \( x=y=z \).

Teacher's Note:
1. Take tan of both sides.

Answer: (d) None of these

Explanation:
1. Since 1 is the middle term, \( \sin^2\alpha+\sin^4\alpha=2 \), so \( \sin^2\alpha=1 \).
2. Also \( \sin\alpha+\sin^5\alpha=2 \) forces \( \sin\alpha=1 \), so \( \alpha=\frac\pi2 \).
3. This is not inside any of the three open intervals.

Teacher's Note:
1. The middle term is the mean of the terms on both sides.

Answer: (b) 4 : 5 : 6

Explanation:
1. For the sides 4, 5, 6: \( \cos C=\frac{16+25-36}{40}=\frac18 \) and \( \cos A=\frac{25+36-16}{60}=\frac34 \).
2. \( \cos2A=2\cdot\frac9{16}-1=\frac18=\cos C \), so \( C=2A \). The other ratios do not satisfy this.

Teacher's Note:
1. Test the angle relation with the cosine rule.

Answer: (a) \( 3\tan\frac A2 \)

Explanation:
1. For sides in A.P., \( \cot\frac A2\cot\frac C2=3 \) (from \( 2b=a+c \) and the half angle formulas).
2. So \( \cot\frac C2=\frac3{\cot\frac A2}=3\tan\frac A2 \).

Teacher's Note:
1. Use the identity for sides in A.P.

Answer: (c) \( 12 \)

Explanation:
1. Let \( u=5^x+5^{-x}\ge2 \). Then \( 25^x+25^{-x}=u^2-2 \).
2. \( a=5u+u^2-2 \), which is increasing for \( u\ge2 \), so the least value is \( 10+4-2=12 \).

Teacher's Note:
1. Use A.M. is at least G.M. for 5 to the x and 5 to the minus x.

Answer: (c) \( 1 \) or \( -1 \)

Explanation:
1. Expanding the determinant (using \( \beta=\alpha+d \), \( \gamma=\alpha+2d \), \( \delta=\alpha+3d \)) gives the constant \( f(x)=-2d^2 \).
2. \( \int_0^2f\,dx=-4d^2=-4 \), so \( d^2=1 \).
3. Hence \( d=1 \) or \( -1 \).

Teacher's Note:
1. Simplify by column and row operations.

Answer: (a) A and D only

Explanation:
1. By A.M. and G.M., \( y=\frac{x+z}2\ge\sqrt{xz} \), so (A) is true. (B) fails for \( x=y=z=1 \).
2. Since \( 2y-x=z \) and \( 2y-z=x \), the expression is \( \frac{x+y}z+\frac{y+z}x \), which equals \( \frac{3x+z}{2z}+\frac{x+3z}{2x}=1+\frac32\left(\frac xz+\frac zx\right)\ge1+3=4 \). So (D) is true and (C) is false.
3. Hence (A) and (D) only.

Teacher's Note:
1. Replace 2y - x by z and 2y - z by x.