1

### JEE Main 2016 (Online) 10th April Morning Slot

If   A > 0, B > 0   and    A + B = ${\pi \over 6}$,

then the minimum value of tanA + tanB is :
A
$\sqrt 3 - \sqrt 2$
B
$2 - \sqrt 3$
C
$4 - 2\sqrt 3$
D
${2 \over {\sqrt 3 }}$

## Explanation

Given,

A + B = ${\pi \over 6}$

$\therefore$   tan(A + B) = tan$\left( {{\pi \over 6}} \right)$ = ${1 \over {\sqrt 3 }}$

We know,

tan(A + B) = ${{\tan A + \tan B} \over {1 - \tan A\tan B}}$

$\Rightarrow$  ${1 \over {\sqrt 3 }}$ = ${y \over {1 - \tan A\tan B}}$

where y = tan A + tan B

$\Rightarrow$    tanA tanB = 1 $-$ $\sqrt 3$ y

Also AM   $\ge$  GM

$\Rightarrow$    ${{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B}$

$\Rightarrow$   y $\ge$ 2$\sqrt {1 - \sqrt 3 y}$

$\Rightarrow$   y2 $\ge$ 4 $-$ 4${\sqrt 3 y}$

$\Rightarrow$    y2 + 4${\sqrt 3 y}$ $-$ 4 $\ge$ 0

$\Rightarrow$   y  $\le$ $-$ 2$\sqrt 3$ $-$ 4

or   y $\ge$ $-$ 2$\sqrt 3$ + 4

(y $\le$ $-$ 2$\sqrt 3$ $-$ 4 is not possible as tan B > 0)
2

### JEE Main 2017 (Offline)

Let $a$, b, c $\in R$. If $f$(x) = ax2 + bx + c is such that
$a$ + b + c = 3 and $f$(x + y) = $f$(x) + $f$(y) + xy, $\forall x,y \in R,$

then $\sum\limits_{n = 1}^{10} {f(n)}$ is equal to
A
165
B
190
C
255
D
330

## Explanation

f(x) = ax2 + bx + c

f(1) = a + b + c = 3 $\Rightarrow$ f (1) = 3

Now f(x + y) = f(x) + f(y) + xy ...(1)

Put x = y = 1 in eqn (1)

f(2) = f(1) + f(1) + 1

= 2f(1) + 1

$\Rightarrow$ f(2) = 7

Similarly f(3) = 12

f(4) = 18

$\sum\limits_{n = 1}^{10} {f(n)}$ = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 + 63 + 75 = 330
3

### JEE Main 2017 (Offline)

For any three positive real numbers a, b and c,

9(25${a^2}$ + b2) + 25(c2 - 3$a$c) = 15b(3$a$ + c).
Then
A
b, c and $a$ are in G.P.
B
b, c and $a$ are in A.P.
C
$a$, b and c are in A.P.
D
$a$, b and c are in G.P.

## Explanation

9(25${a^2}$ + b2) + 25(c2 - 3$a$c) = 15b(3$a$ + c)

$\Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc$

$\Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc$

$\Rightarrow$ ${1 \over 2}\left[ {{{\left( {15a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0$

it is possible when 15a – 3b = 0, 3b – 5 c = 0 and 5c – 15a = 0

$\Rightarrow$ 15a = 3b = 5c

$\Rightarrow$ b = ${{5c} \over 3}$, a = ${c \over 3}$

$\Rightarrow$ a + b = ${c \over 3} + {{5c} \over 3}$ = ${{6c} \over 3}$ = 2c

$\therefore$ b, c, a are in A.P.
4

### JEE Main 2017 (Online) 8th April Morning Slot

If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then ${{a + b} \over {a - b}}$ is equal to :
A
${{\sqrt 6 } \over 2}$
B
${{3\sqrt 2 } \over 4}$
C
${{7\sqrt 3 } \over {12}}$
D
${{5\sqrt 6 } \over {12}}$

## Explanation

A.T.Q.,

A.M. = 5G.M.

${{a + b} \over 2} = 5\sqrt {ab}$

${{a + b} \over {\sqrt {ab} }}$ $= 10$

$\therefore$   ${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$

Use componendo and Dividendo

${{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}$