Practice Start Test

Quadratic Equation and Inequalities

JEE Mathematics · 144 questions · Page 15 of 15 · Click an option or "Show Solution" to reveal answer

Q141
If x is a solution of the equation, 2x+1\sqrt {2x + 1} −2x−1=1, - \sqrt {2x - 1} = 1,   (x≥12),\,\,\left( {x \ge {1 \over 2}} \right), then 4x2−1\sqrt {4{x^2} - 1} is equal to :
A 34{3 \over 4}
B 12{1 \over 2}
C 2
D 222\sqrt 2
Correct Answer
Option A
Solution

Given,

2x+1−2x−1=1\sqrt {2x + 1} - \sqrt {2x - 1} = 1

⇒\Rightarrow

2x+1=1+2x−1\sqrt {2x + 1} = 1 + \sqrt {2x - 1}

Squaring both sides, we get 2x + 1 == 1 + 2x −- 1 + 2

2x−1\sqrt {2x - 1}

⇒\Rightarrow 1 == 2

2x−1\sqrt {2x - 1}

⇒\Rightarrow 1 == 4(2x −- 1) ⇒\Rightarrow 8x −- 4 == 1 ⇒\Rightarrow x ==

58{5 \over 8}

So,

4x2−1\sqrt {4{x^2} - 1}
=4(2564)−1= \sqrt {4\left( {{{25} \over {64}}} \right) - 1}
=3664= \sqrt {{{36} \over {64}}}
=68= {6 \over 8}
=34= {3 \over 4}
Q142
Consider the quadratic equation (c – 5)x2 – 2cx + (c – 4) = 0, c ≠ \ne 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is -
A 12
B 18
C 10
D 11
Correct Answer
Option D
Solution

Let f(x) = (c −- 5)x2 −- 2cx + c −- 4 ∴\therefore f(0)f(2) < 0 . . . . .(1) & f(2)f(3) < 0 . . . . .(2) from (1) and (2) (c −- 4)(c −- 24) < 0 & (c −- 24)(4c −- 49) < 0 ⇒\Rightarrow

494{{49} \over 4}

< c < 24 ∴\therefore s = {113, 14, 15, . . . . . 23} Number of elements in set S = 11

Q143
If a∈Ra \in R and the equation −3(x−[x])2+2(x−[x])+a2=0 - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0 (where [xx] denotes the greater integer ≤x \le x) has no integral solution, then all possible values of a lie in the interval :
A (−2,−1)\left( { - 2, - 1} \right)
B (−∞,−2)∪(2,∞)\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)
C (−1,0)∪(0,1)\left( { - 1,0} \right) \cup \left( {0,1} \right)
D (1,2)\left( {1,2} \right)
Correct Answer
Option C
Solution

Given,

−3(x−[x])2+2(x−[x])+a2=0- 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0

As we know,

[x]+{x}=x\left[ x \right] + \left\{ x \right\} = x

where

[x]\left[ x \right]

is integral part and

{x}\left\{ x \right\}

is fractional part. ∴\therefore

{x}=x−[x]\left\{ x \right\} = x - \left[ x \right]

Now put

{x}\left\{ x \right\}

inplace of

x−[x]x - \left[ x \right]

in the equation. The new equation is

−3{x}2+2{x}+a2=0- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0

[Note : Question says this equation has no integral solution, it means

{x}≠\left\{ x \right\} \ne

0. So,

xx

is not a integer.] ∴\therefore

{x}\left\{ x \right\}

=

−2±4−4×(−3)a2−6{{ - 2 \pm \sqrt {4 - 4 \times \left( { - 3} \right){a^2}} } \over { - 6}}

=

−2±4+12a2−6{{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}

As

{x}\left\{ x \right\}

is fractional part so it is lies between 0 to 1(

0≤{x}<10 \le \left\{ x \right\} < 1

). By considering positive sign, we get

0≤−2+4+12a2−6<10 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1

⇒\Rightarrow

0≥−2+4+12a2>−60 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6

⇒\Rightarrow

2≥+4+12a2>−42 \ge + \sqrt {4 + 12{a^2}} > - 4

∵\because

+4+12a2+ \sqrt {4 + 12{a^2}}

is always positive which is greater than any negative no. So can ignore the inequality

+4+12a2>−4+ \sqrt {4 + 12{a^2}} > - 4

Consider this inequality,

2≥+4+12a22 \ge + \sqrt {4 + 12{a^2}}

⇒\Rightarrow

4≥4+12a24 \ge 4 + 12{a^2}

⇒\Rightarrow

12a2≤012{a^2} \le 0

⇒\Rightarrow

a2≤0{a^2} \le 0

⇒\Rightarrow

a2=0{a^2} = 0

⇒\Rightarrow

a=0{a} = 0

If

aa

= 0 then

−3{x}2+2{x}=0- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0

so

{x}\left\{ x \right\}

becomes 0 but question says

{x}\left\{ x \right\}

≠\ne 0. So

aa

can't be 0. Now by considering negative sign, we get

0≤−2−4+12a2−6<10 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1

⇒\Rightarrow

0≥−2−4+12a2>−60 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6

⇒\Rightarrow

2≥−4+12a2>−42 \ge - \sqrt {4 + 12{a^2}} > - 4

As 2 is always greater than

−4+12a2{ - \sqrt {4 + 12{a^2}} }

. Ignore this inequality. Now consider this inequality,

−4+12a2>−4- \sqrt {4 + 12{a^2}} > - 4

⇒\Rightarrow

4+12a2<4\sqrt {4 + 12{a^2}} < 4

⇒\Rightarrow

4+12a2<164 + 12{a^2} < 16

⇒\Rightarrow

12a2<1212{a^2} < 12

⇒\Rightarrow

a2<1{a^2} < 1

⇒\Rightarrow

(a2−1)<0\left( {{a^2} - 1} \right) < 0

⇒\Rightarrow

(a+1)(a−1)<0\left( {a + 1} \right)\left( {a - 1} \right) < 0

⇒\Rightarrow

−1<a<1- 1 < a < 1

But earlier we found that

aa

≠\ne 0. So, the range of

aa

is =

(−1,0)∪(0,1)\left( { - 1,0} \right) \cup \left( {0,1} \right)
Q144
Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x−- 1 and it leaves remainder 6 when divided by x + 1; then :
A p(2) = 11
B p(2) = 19
C p(−- 2) = 19
D p(−- 2) = 11
Correct Answer
Option C
Solution

Let, P(x) = ax2 + bx + c As, P(0) = 1,

∴   \therefore\,\,\,

a(0)2 + b(0) + c = 1 ⇒\Rightarrow

   \,\,\,

c = 1

∴   \therefore\,\,\,

P(x) = ax2 + bx + 1 If P(x) is divided by x −- 1, remainder = 4 ⇒\Rightarrow

   \,\,\,

P

(1)=4\left( 1 \right) = 4
∴   \therefore\,\,\,

a + b + 1 = 4 . . . . . (1) If P(x) is divided by x + 1, remainder = 6 ⇒\Rightarrow

   \,\,\,

P(−- 1) = 6

∴   \therefore\,\,\,

a −- b + 1 = 6 . . . .(2) By solving (1) and (2) we get, a = 4, and b = −-1

∴   \therefore\,\,\,

P(x) = 4x2 −- x + 1 P(2) = 4(2)2 −- 2 + 1 = 15 P(−- 2) = 4 (−-2)2 −- (−- 2) + 1 = 19

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