Sequences & SerieshardComprehensionFree

Sequences & Series: Let Arithmetic Progression Non Zero Common Difference Let

JEE Maths reading comprehension with full step-by-step solutions.

Passage

Let

\{t_n\}_{n \in \mathbb{N}}

be an arithmetic progression with non-zero common difference

d

, and let

S_n = \displaystyle\sum_{i=1}^n t_i

denote the sum of its first

n

terms. The partial sums satisfy the strict inequalities

S_6 > S_7 > S_5

, and the terms satisfy the identity

2(t_6 + t_7) = d

. Let

k

be the uniquely determined positive integer satisfying

S_k S_{k+1} < 0

. Consider a real-valued function

f:\mathbb{R} \to \mathbb{R}

defined as:

f(x) = \begin{cases} \left(\dfrac{\sin\!\left(\dfrac{kx^2}{2}\right)+\cos(ax)}{e^{bx^2}}\right)^{\!\frac{1}{x^2}}, & x \neq 0 \\[8pt] L, & x = 0 \end{cases}

where

a \in (0,\infty)

and

b \in \mathbb{R}

. The function

f(x)

is continuous at

x = 0

. Furthermore, let

g:[1,e] \to \mathbb{R}

be a twice-differentiable function satisfying

g(1) = 0

g(\sqrt{e}) = \dfrac{L}{2}

, and

g(e) = L

Question 1 · Multiple correct

Based on the properties of the arithmetic progression

\{t_n\}

and the integer

k

, which of the following statements is/are CORRECT?

AThe sequence

\{t_n\}

contains exactly

6

strictly positive terms.correct

BThe sum of the first

k

terms is strictly positive.correct

CThe sequence of partial sums

S_n

is monotonically decreasing for all integers

n \geq 6

.correct

D) If the first term

t_1 = 21

, then

\displaystyle\sum_{r=1}^{k} t_r = 11

.correct

Solution

Step 1: Derive the sign of

d

and determine

t_1

From

S_6 > S_7

\quad S_6 - S_7 = -t_7 > 0 \implies t_7 < 0 \implies t_1 + 6d < 0

. From

S_6 > S_5

(which follows since

S_6 > S_7 > S_5

\quad t_6 > 0 \implies t_1 + 5d > 0

. Applying the identity

2(t_6+t_7) = d

2\bigl[(t_1+5d)+(t_1+6d)\bigr] = d \implies 4t_1 + 22d = d \implies t_1 = -\frac{21}{4}d

Since

t_6 = t_1+5d = -\frac{21d}{4}+5d = -\frac{d}{4} > 0

, it follows that

d < 0

. Step 2: General term and general sum

t_n = t_1+(n-1)d = -\frac{21d}{4}+(n-1)d = \frac{d(4n-25)}{4}

Since

d < 0

t_n > 0

when

4n-25 < 0

, i.e.

n \leq 6

, and

t_n < 0

when

n \geq 7

. Therefore the sequence contains exactly

6

strictly positive terms. Statement (1) is correct.

S_n = \frac{n}{2}(2t_1+(n-1)d) = \frac{nd}{2}\!\left(n-\frac{23}{2}\right)

Step 3: Determine

k

S_{11} = \frac{11d}{2}\!\left(11-\frac{23}{2}\right) = \frac{11d}{2}\!\left(-\frac{1}{2}\right) = -\frac{11d}{4}

Since

d < 0

S_{11} > 0

S_{12} = \frac{12d}{2}\!\left(12-\frac{23}{2}\right) = 6d\cdot\frac{1}{2} = 3d

Since

d < 0

S_{12} < 0

. Therefore

S_{11}S_{12} < 0

, giving

k = 11

. Statement (2) holds since

S_{11} = -\frac{11d}{4} > 0

. Statement (2) is correct. Step 4: Monotonicity for

n \geq 6

S_{n+1} - S_n = t_{n+1} = \frac{d(4(n+1)-25)}{4} = \frac{d(4n-21)}{4}

For

n \geq 6

4n-21 \geq 3 > 0

, and

d < 0

, so

t_{n+1} < 0

. Hence

S_{n+1} < S_n

for all

n \geq 6

. Statement (3) is correct. Step 5: Verify statement (4) when

t_1 = 21

t_1 = -\frac{21d}{4} = 21 \implies d = -4

S_{11} = -\frac{11(-4)}{4} = \frac{44}{4} = 11

Statement (4) is correct. Correct answer: (1), (2), (3), (4)

Question 2 · Single correct

If the limit

L

evaluates to unity, then the maximum possible value of the parameter

(a+b)

is:

4

\dfrac{11}{2}

6

correct

\dfrac{13}{2}

Solution

Step 1: Evaluate the limit

L

with

k = 11

L = \lim_{x \to 0}\left(\frac{\sin\!\left(\frac{11x^2}{2}\right)+\cos(ax)}{e^{bx^2}}\right)^{\!\frac{1}{x^2}}

This is a

1^\infty

indeterminate form. Using Maclaurin series to

O(x^2)

\sin\!\left(\frac{11x^2}{2}\right) \approx \frac{11x^2}{2}, \qquad \cos(ax) \approx 1-\frac{a^2x^2}{2}, \qquad e^{bx^2} \approx 1+bx^2

Step 2: Simplify the base

\frac{\sin\!\left(\frac{11x^2}{2}\right)+\cos(ax)}{e^{bx^2}} \approx \frac{1+\dfrac{11-a^2}{2}x^2}{1+bx^2} \approx 1+\left(\frac{11-a^2}{2}-b\right)x^2

Step 3: Evaluate the

1^\infty

limit

L = \exp\!\left(\lim_{x \to 0}\frac{1}{x^2}\cdot\left(\frac{11-a^2}{2}-b\right)x^2\right) = e^{\,\frac{11-a^2}{2}-b}

Step 4: Apply the condition

L = 1

\frac{11-a^2}{2} - b = 0 \implies b = \frac{11-a^2}{2}

Step 5: Maximize

a + b

a+b = a+\frac{11-a^2}{2} = \frac{-(a^2-2a-11)}{2} = \frac{-(a-1)^2+12}{2}

This attains its maximum at

a = 1

. Since

a \in (0,\infty)

, this is admissible.

\max(a+b) = \frac{12}{2} = 6 \qquad \text{(attained at } a=1,\; b=5\text{)}

Correct answer: (3)

Question 3 · Single correct

Based on the given boundary conditions for

g(x)

, there must exist at least one

c \in (1, e)

such that:

c^2 g''(c) - L = 0

c^2 g''(c) + L = 0

correct

cg''(c)+g'(c) = L

c^2 g''(c)+cg'(c)-g(c) = 0

Solution

Step 1: Construct the auxiliary function Define

H(x) = g(x) - L\ln x

for

x \in [1,e]

. Evaluate

H

at the three given points:

H(1) = g(1) - L\ln 1 = 0 - 0 = 0

H(\sqrt{e}) = g(\sqrt{e}) - L\ln e^{1/2} = \frac{L}{2} - \frac{L}{2} = 0

H(e) = g(e) - L\ln e = L - L = 0

Step 2: Apply Rolle's theorem twice to

H

Since

H

is twice-differentiable on

[1,e]

and

H(1) = H(\sqrt{e}) = 0

, Rolle's theorem guarantees an

\alpha \in (1,\sqrt{e})

with

H'(\alpha) = 0

. Since

H(\sqrt{e}) = H(e) = 0

, Rolle's theorem guarantees a

\beta \in (\sqrt{e},e)

with

H'(\beta) = 0

. Step 3: Apply Rolle's theorem to

H'

Since

H'(\alpha) = H'(\beta) = 0

and

[\alpha,\beta] \subset (1,e)

, Rolle's theorem applied to

H'

guarantees a

c \in (\alpha,\beta) \subset (1,e)

such that

H''(c) = 0

. Step 4: Identify the resulting equation

H'(x) = g'(x) - \frac{L}{x}, \qquad H''(x) = g''(x) + \frac{L}{x^2}

Setting

H''(c) = 0

g''(c) + \frac{L}{c^2} = 0 \implies c^2 g''(c) + L = 0

Correct answer: (2)

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