1

### JEE Main 2018 (Online) 15th April Evening Slot

If f(x) is a quadratic expression such that f (1) + f (2) = 0, and $-$ 1 is a root of f (x) = 0, then the other root of f(x) = 0 is :
A
$-$ ${5 \over 8}$
B
$-$ ${8 \over 5}$
C
${5 \over 8}$
D
${8 \over 5}$

## Explanation

Let $\alpha$ and $\beta$ = - 1 are the roots of the polynomial, then we get

f(x) = x2 + (1 - $\alpha$)x - $\alpha$

$\therefore$ f(1) = 2 - 2$\alpha$

and f(2) = 6 - 3$\alpha$

Also given,

f (1) + f (2) = 0

$\therefore$ 2 - 2$\alpha$ + 6 - 3$\alpha$ = 0

$\Rightarrow$ $\alpha$ = ${8 \over 5}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let p, q and r be real numbers (p $\ne$ q, r $\ne$ 0), such that the roots of the equation ${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :
A
${{{p^2} + {q^2}} \over 2}$
B
p2 + q2
C
2(p2 + q2)
D
p2 + q2 + r2

## Explanation

Given,

${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$

$\Rightarrow $$\,\,\, {{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r} \Rightarrow$$\,\,\,$ (2x + p + q) r = x2 + px + qx + pq

$\Rightarrow $$\,\,\, x2 + (p + q - 2r) x + pq - pr - qr = 0 Let \alpha and \beta are the roots, \therefore\,\,\, \alpha + \beta = - (p + q - 2r) and \alpha \beta = pq - pr - qr Given that, \alpha = - \beta \Rightarrow \alpha + \beta = 0 \therefore\,\,\, - (p + q - 2r) = 0 Now, \alpha 2 + \beta 2 = (\alpha + \beta )2 - 2\alpha \beta = (- (p + q - 2r))2 - 2 (pq - pr - qr) = p2 +q2 + 4r2 + 2pq - 4pr - 4qr - 2pq + 2pr + 2qr = p2 + q2 + 4r2 - 2pr - 2qr = p2 + q2 - 2r (p + q - 2r) = p2 + q2 - 2r (0) = p2 + q2 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 16th April Morning Slot If an angle A of a \Delta ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are : A secA, cotA B sinA, secA C secA, tanA D tanA, cosA ## Explanation Here, 9x2 + 27x + 20 = 0 \therefore\,\,\, x = {{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}} \Rightarrow$$\,\,\,$ x = ${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$

$\Rightarrow$$\,\,\,$ x = $-$ ${4 \over 3}$, $-$ ${5 \over 3}$

Given, cosA = $-$ ${3 \over 5}$

$\therefore\,\,\,$ sec A = ${1 \over {\cos A}}$ = $-$ ${5 \over 3}$

Here, A is an obtuse angle.

$\therefore\,\,\,$ tan A = $-$ $\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.$

Hence, roots of the equation are sec A and tan A.
4

### JEE Main 2019 (Online) 9th January Evening Slot

If both the roots of the quadratic equation x2 $-$ mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval :
A
($-$5, $-$4)
B
(4, 5)
C
(5, 6)
D
(3, 4)

## Explanation

x2 $-$mx + 4 = 0 Case-I :

D > 0

m2 $-$ 16 > 0

$\Rightarrow$  m $\in$ ($-$ $\infty$, $-$ 4) $\cup$ (4, $\infty$)

Case-II :

$\Rightarrow \,\,1 < {{ - b} \over {2a}} < 5$

$\Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)$

Case-III :

f(1) > 0   and f(5) > 0

1 $-$ m + 4 > 0   and 25 $-$ 5m + 4 > 0

m < 5   and m < ${{29} \over 5}$

Case-IV :

Let one root is x = 1

1 $-$ m + 4 = 0

m = 5

Now equation x2 $-$ 5x + 4 = 0

(x $-$ 1) (x $-$ 4) = 0

x = 1 i.e. m = 5 is also included

hence m $\in$ (4, 5]

So given option is (4, 5)