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1

### JEE Main 2021 (Online) 27th August Morning Shift

If $$\alpha$$, $$\beta$$ are the distinct roots of x2 + bx + c = 0, then

$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$ is equal to :
A
b2 + 4c
B
2(b2 + 4c)
C
2(b2 $$-$$ 4c)
D
b2 $$-$$ 4c

## Explanation

$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$

$$= \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$

$$= \mathop {\lim }\limits_{x \to \beta } {{2{{({x^2} + bx + 1)}^2}} \over {{{(x - \beta )}^2}}}$$

$$= \mathop {\lim }\limits_{x \to \beta } {{2{{(x - \alpha )}^2}{{(x - \beta )}^2}} \over {{{(x - \beta )}^2}}}$$

$$= 2{(\beta - \alpha )^2} = 2({b^2} - 4c)$$
2

### JEE Main 2021 (Online) 26th August Evening Shift

$$\mathop {\lim }\limits_{x \to 2} \left( {\sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} } \right)$$ is equal to :
A
$${9 \over {44}}$$
B
$${5 \over {24}}$$
C
$${1 \over 5}$$
D
$${7 \over {36}}$$

## Explanation

$$S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}}$$

$$S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)}$$

$$S = {1 \over 2}\left( {{1 \over 2} - {1 \over {11}}} \right) = {9 \over {44}}$$
3

### JEE Main 2021 (Online) 26th August Morning Shift

The value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}}$$ is :
A
$${1 \over 2}{\tan ^{ - 1}}(2)$$
B
$${1 \over 2}{\tan ^{ - 1}}(4)$$
C
$${\tan ^{ - 1}}(4)$$
D
$${1 \over 4}{\tan ^{ - 1}}(4)$$

## Explanation

$$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}}$$

$$\Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx}$$

$$\Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4$$
4

### JEE Main 2021 (Online) 27th July Morning Shift

Let f : R $$\to$$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is equal to :
A
4
B
8
C
16
D
12

## Explanation

Apply L' Hospital Rule

$$\mathop {\lim }\limits_{x \to 2} \left( {{{2xf(2) - 4f'(x)} \over 1}} \right)$$

$$= {{4(4) - 4} \over 1} = 12$$

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