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1

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

If 0 < x < 1 and $$y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$$, then the value of e1 + y at $$x = {1 \over 2}$$ is :
A
$${1 \over 2}{e^2}$$
B
2e
C
$${1 \over 2}\sqrt e$$
D
2e2

## Explanation

$$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$

$$= ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$

$$= {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ....} \right)$$

$$= {x \over {1 - x}} + \ln (1 - x)$$

$$x = {1 \over 2} \Rightarrow y = 1 - \ln 2$$

$${e^{1 + y}} = {e^{1 + 1 - \ln 2}}$$

$$= {e^{2 - \ln 2}} = {{{e^2}} \over 2}$$
2

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

If for x, y $$\in$$ R, x > 0, y = log10x + log10x1/3 + log10x1/9 + ...... upto $$\infty$$ terms

and $${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$$, then the ordered pair (x, y) is equal to :
A
(106, 6)
B
(104, 6)
C
(102, 3)
D
(106, 9)

## Explanation

$${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$$

$$\Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}$$

Now,

$$y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty$$

$$= \left( {1 + {1 \over 3} + {1 \over 9} + ....\infty } \right){\log _{10}}x$$

$$= \left( {{1 \over {1 - {1 \over 3}}}} \right){\log _{10}}x = 9$$

So, (x, y) = (106, 9)
3

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

If 0 < x < 1, then $${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$$, is equal to :
A
$$x\left( {{{1 + x} \over {1 - x}}} \right) + {\log _e}(1 - x)$$
B
$$x\left( {{{1 - x} \over {1 + x}}} \right) + {\log _e}(1 - x)$$
C
$${{1 - x} \over {1 + x}} + {\log _e}(1 - x)$$
D
$${{1 + x} \over {1 - x}} + {\log _e}(1 - x)$$

## Explanation

Let $$t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty$$

$$= \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty$$

$$= 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + .....\infty } \right)$$

$$= {{2{x^2}} \over {1 - x}} - (\ln (1 - x) - x)$$

$$\Rightarrow t = {{2{x^2}} \over {1 - x}} + x - \ln (1 - x)$$

$$\Rightarrow t = {{x(1 + x)} \over {1 - x}} - \ln (1 - x)$$
4

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

If the sum of an infinite GP a, ar, ar2, ar3, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, ....... is :
A
$${5 \over 2}$$
B
$${1 \over 2}$$
C
$${25 \over 2}$$
D
$${9 \over 2}$$

## Explanation

Sum of infinite terms :

$${a \over {1 - r}} = 15$$ ..... (i)

Series formed by square of terms :

a2, a2r2, a2r4, a2r6 .......

Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$

$$\Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150$$

$$\Rightarrow {a \over {1 + r}} = 10$$ ...... (ii)

by (i) and (ii), a = 12; r = $${1 \over 5}$$

Now, series : ar2, ar4, ar6

Sum = $${{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}$$

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