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1

### JEE Main 2019 (Online) 12th January Morning Slot

Let  Sk = $${{1 + 2 + 3 + .... + k} \over k}.$$ If   $$S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}$$A,  then A is equal to :
A
283
B
156
C
301
D
303

## Explanation

Sk = $${{K + 1} \over 2}$$

$$\sum {S_k^2} = {5 \over {12}}$$ A

$$\sum\limits_{K = 1}^{10} {{{\left( {{{K + 1} \over 2}} \right)}^2}} = {{{2^2} + {3^2} + - - + {{11}^2}} \over 4} = {5 \over {12}}$$ A

$${{11 \times 12 \times 23} \over 6} - 1 = {5 \over 3}$$ A

505 $$= {5 \over 3}$$ A,   A = 303
2

### JEE Main 2019 (Online) 11th January Evening Slot

Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression $${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}$$ is :
A
$${1 \over 2}$$
B
$${1 \over 4}$$
C
$${{m + n} \over {6mn}}$$
D
1

## Explanation

$${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}} = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}} \le {1 \over 4}$$

using AM $$\ge$$ GM
3

### JEE Main 2019 (Online) 11th January Evening Slot

If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is :
A
2 : 1
B
4 : 1
C
1 : 3
D
3 : 1

## Explanation

a + 18d = 0        . . . . .(1)

$${{a + 48d} \over {a + 28d}} = {{ - 18d + 48d} \over { - 18d + 28d}} = {3 \over 1}$$
4

### JEE Main 2019 (Online) 11th January Morning Slot

Let a1, a2, . . . . . ., a10 be a G.P.    If $${{{a_3}} \over {{a_1}}} = 25,$$ then $${{{a_9}} \over {{a_5}}}$$ equals
A
53
B
2(52)
C
4(52)
D
54

## Explanation

a1, a2, . . . . ., a10 are in G.P.,

Let the common ratio be r

$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$

$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$

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