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

Answer: (d)

$(\text"log"_{1/2}2)(\text"log"_{1/3}3)(\text"log"_{1/4}4).......(\text"log"_{1/{1000}}1000)$

=$({\text"log" 2}/{\text"log"{1/2}})({\text"log" 3}/{\text"log"{1/3}})({\text"log" 4}/{\text"log"{1/4}})......({\text"log" 1000}/{\text"log"{1/{1000}}})$ $(∵ \text"log"_b a = {\text"log" a}/{\text"log" b})$

= $({\text"log" 2}/{- \text"log" 2})({\text"log" 3}/{-\text"log" 3})({\text"log" 4}/{-\text"log" 4}) ...... ({\text"log" 1000}/{\text"log" 1000})$

= (-1) × (-1) × (-1) × .....× (-1)

(∵ number of terms is odd) = -1

Answer: (a)

We have, y = $x^{logx}$

Taking log on both sides log

y = log $(x^{logx})$

= logx. logx = $(1.25)^2$ = 1.5625

Answer: (d)

abc = ${\text"log" 12}/{\text"log" 24} . {\text"log" 24}/{\text"log" 36} . {\text"log" 36}/{\text"log" 48} = {\text"log" 12}/{\text"log" 48}$

∴ 1 + abc = ${\text"log" 48 + \text"log"12}/{\text"log" 48} = {\text"log" (48.12)}/{\text"log" 48}$

= ${\text"log"24}^2/{\text"log" 48} = 2.{\text"log" 24}/{\text"log" 48}$ = 2bc

Answer: (d)

Consider $ax^2 - (a^2 + 1)$ x + a .......(1)

⇒ $ax^2 - a^2 x$ - x + a

⇒ $a^x$ (x - a) - 1 (x - a)

= (x - a) (ax - 1)

Given p and q are two linear factors of (1)

∴ p = x + a and q = ax - 1

⇒ p + q = x - a + ax - 1

= x(a + 1) - 1 (a + 1)

= (x - 1) (a + 1)

∴ Option (d) is correct.

Answer: (d)

Let &alpha, β be the roots of the equation $lx^2$ + mx + m = 0

Given ${α}/{β} = p/q$

Now α + β (sum of roots) = ${- m}/l$

and α β (product of roots) = $m/l$

Consider $√{p/q} + √{q/p} + √{m/l}$

Using (1)

= $√{{α}/{β}} = √{{β}/{α}} + √{m/l}$

= ${α + β}/{√{α β}} + √{m/l}$

= ${- m/l}/{√{m/l}} + √{m/l} = - √{m/l} + √{m/l}$ = 0

∴ option (d) is correct.

Answer: (d)

Given equation, $2x^2$ - 7x + 3 = 0

∴ $2x^2$ - 6x - x + 3 = 0

⇒ 2x(x - 3) - 1(x - 3) = 0

⇒ (2x - 1) (x - 3) = 0

Both equation have a common root.

So, we put x = $1/2 ⇒ 4(1/2)^2 + a(1/2)$ - 3 = 0

⇒ 1 + $a/2 - 3 = 0 ⇒ a/2$ = 2 ⇒ a = 4

Again, we put x = 3

$4(3)^2 + a(3) - 3 = 0$

⇒ 36 + 3a - 3 = 0 ⇒ a = - 11

a = - 11 or 4.