LimitshardFree

Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

a=\min[x^{2}+2x+3,\;x\in\mathbb{R}]

and

b=\displaystyle\lim_{x\to 0}\dfrac{\sin x\cos x}{e^{x}-e^{-x}}

. Then the value of

\displaystyle\sum_{r=0}^{n}a^{r}b^{n-r}

\dfrac{2^{n+1}+1}{3\cdot 2^{n}}

\dfrac{2^{n+1}-1}{3\cdot 2^{n}}

\dfrac{2^{n}-1}{3\cdot 2^{n}}

\dfrac{4^{n+1}-1}{3\cdot 2^{n}}

correct

Solution

Step 1: Find

a

x^{2}+2x+3=(x+1)^{2}+2

, minimum value is

2

x=-1

. So

a=2

. Step 2: Find

b

b=\lim_{x\to 0}\dfrac{\sin x\cos x}{e^{x}-e^{-x}}

Use

\sin x\cos x=\dfrac{1}{2}\sin 2x

and

e^{x}-e^{-x}=2\sinh x

b=\lim_{x\to 0}\dfrac{\sin 2x/2}{2\sinh x}=\lim_{x\to 0}\dfrac{\sin 2x}{4\sinh x}

Use

\dfrac{\sin 2x}{2x}\to 1

and

\dfrac{\sinh x}{x}\to 1

b=\lim_{x\to 0}\dfrac{2x}{4x}=\dfrac{1}{2}

Step 3: The sum

\displaystyle\sum_{r=0}^{n}a^{r}b^{n-r}

is a geometric-like sum. With

a=2,b=\dfrac{1}{2}

\sum_{r=0}^{n}2^{r}\left(\dfrac{1}{2}\right)^{n-r}=\sum_{r=0}^{n}\dfrac{2^{r}}{2^{n-r}}=\sum_{r=0}^{n}\dfrac{2^{2r}}{2^{n}}=\dfrac{1}{2^{n}}\sum_{r=0}^{n}4^{r}

Step 4: Sum the geometric series

\displaystyle\sum_{r=0}^{n}4^{r}=\dfrac{4^{n+1}-1}{4-1}=\dfrac{4^{n+1}-1}{3}

. Step 5:

\dfrac{1}{2^{n}}\cdot\dfrac{4^{n+1}-1}{3}=\dfrac{4^{n+1}-1}{3\cdot 2^{n}}

Correct answer: (4)

\dfrac{4^{n+1}-1}{3\cdot 2^{n}}

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