Linear Equations Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Here m and n are the roots of the equation

$x^2$ + ax + b = 0.

m + n = - a .......(i)

mn = b ......(ii)

Also m $m^2 and n^2$ are the roots of the equation of

$x^2$ - cx + d = 0.

$m^2 + n^2$ = c...(iii)

$m^2 n^2$ = d .....(iv)

by squaring Eq. (i) both sides, we get

$m^2 + n^2 + 2mn = a^2$ [from Eqs. (i) and (ii)]

⇒ c + 2b = $a^2 ⇒ c = a^2 - 2b$

⇒ 2b - $a^2$ = -c

Therefore, Statement 1 is incorrect.

From Eq. (ii)

$m^2 n^2 = b^2 ⇒ b^2$ = d

Therefore, Statement 2 is correct.

Answer: (b)

We know that two equation

$a_1 x^2 + b_1 x + c_1$ = 0

$a_2 x^2 + b_2 + c_2$ = 0

have common root when

$(c_1 a_2 - a_1 c_2)^2 = (b_1 c_2 - c_1 b_2) (a_1 b_2 - b_1 a_2)$

So, for $x^2 + 5x + 6 = 0 and x^2 + kx = 1 = 0$

we have $(5)^2$ = (5 - 6x) (x - 5)

⇒ 25 = $-6x^2$ + 35x - 25

⇒ $6x^2$ - 35x + 50 = 0

⇒ x = $5/2 or {10}/3$

Answer: (b)

Given equation,

$2x^2$ - 3x - 4 = 0

For a reciprocal roots, we replace x by $1/x$, we get

$2(1/x)^2 - 3(1/x) - 4 = 0$

⇒ $-4x^2$ - 3x + 2 = 0

⇒ $4x^2$ + 3x - 2 = 0

Answer: (b)

${170 - 70}/{250/50} = {100}/5$ = 20

Answer: (b)

Given,

${5x - 7y + 10}/1 = {3x + 2y + 1}/8 = {11x + 4y - 10}/9$

= ${(5x + 10 - 7y) + (3x + 2y + 1) - (11x + 4y - 10)}/{(1 + 8) - 9}$

= ${- 3x - 9y + 21}/0$

⇒x + 3y = 7 ... (i)

On taking first two terms,

8(5x – 7y + 10) = 3x + 2y + 1

37x – 58y + 79 = 0 ... (ii)

From equation (i), on putting the value of x in equation (ii), we get

37(7 – 3y) – 58y + 79 = 0

⇒259 – 111y – 58y + 79 = 0

⇒169y = 338

⇒y = 2

From equation (i),

x = 7 – 3(2) = 1

∴ x + y = 1 + 2 = 3

Answer: (c)

Given that x + $1/x$ = 2 ... (i)

Squaring both sides, we get

$(x + 1/x)^2$ = 4

⇒ $x^2 + 1/{x^2}$ + 2 = 4

⇒ $x^2 + 1/{x^2}$ = 2 ... (ii)

Now, $(x - 1/x)^2 = (x^2 + 1/{x^2})$ – 2

= 2 – 2 = 0 [from equation (ii)]

∴ x – $1/x$ = 0