Free Objective Test 02 Practice Test - 11th and 12th 

Given, x + 2 > $\sqrt{x+4}$  ⇒ $(x+2{)}^2>(x+4)$ but inequality sign will be remained same if both sides are positive or negative.
R.H.S is positive and so L.H.S will also be positive.
Thus x>-2 

⇒$x^2+4x+4>x+4$ ⇒$x^2+3x>0$

⇒ $x(x+3)>0$⇒ $x<-3orx>0$  ⇒ $x>0$

Equation $|3x^2+12x+6|=5x+16$     ......(i)

when $3x^2+12x+6\ge 0$ ⇔ $x^2+4x\ge -2$

⇔ $|x+2{|}^2\ge 4-2$⇔ $|x+2|\ge (\sqrt{2}{)}^2$

⇔ x + 2 ≤ - $\sqrt{2}$ or x + 2 ≥ $\sqrt{2}$.......(ii)

Then (i) becomes $3x^2+12x+6=5x+16$

⇔ $3x^2+7x-10=0$ ⇒ x = 1, -$\frac{10}{3}$

But x = -$\frac{10}{3}$ does not satisfy (ii)

When $3x^2+12x+6<0$ ⇒ $x^2+4x>-2$

$\Rightarrow |x+2|\le \sqrt{2}\Rightarrow -\sqrt{2}-2\le x\le -2+\sqrt{2}$....(iii)

⇒ $3x^2+17x+22=0$ ⇒ $x=-2,-\frac{11}{3}$

But x = - $\frac{11}{3}$ does not satisfy (iii). So, 1 and -2 are the only solutions.

x, y, z, ∈ R and distinct.

Now, u = $x^2+4y^2+9z^2-6yz-3zx-2xy$

$=\frac{1}{2}(2x^2+8y^2+18z^2-12yz-6zx-4xy)$

$=\frac{1}{2}\left\{\right(x^2-4xy+4y^2)+(x^2-6zx+9z^2)+(4y^2-12yz+9z^2\left)\right\}$

$=\frac{1}{2}\left\{\right(x-2y{)}^2+(x-3z{)}^2+(2y-3z{)}^2\}$

Since it is sum of squares. So U is always non-negative.

If the given expression be y, then

$y=2x^{2}y+(3y-1)x+(6y-2)=0$

If y ≠ 0 then  Δ ≥ 0 for real x i.e. $B^2-4AC\ge 0$

or $-39y^2+10y+1\ge 0or(13y+1)(3y-1)\le 0$

⇒  $-\frac{1}{13}\le y\le \frac{1}{3}$

if y = 0 then x = -2 which is real and this value of y is included in the above range.

Let $y=\frac{x^2+14x+9}{x^2+2x+3}$

⇒ $y(x^2+2x+3)-x^2-14x-9=0$

⇒ $(y-1)x^2+(2y-14)x+3y-9=0$

For real x, its discriminant ≥ 0

i.e. $4(y-7{)}^2-4(y-1\left)3\right(y-3)\ge 0$

⇒ $y^2+y-20\le 0or(y-4)(y+5)\le 0$

Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:

Case 1: $y-4\ge 0ory\ge 4y+5\le 0ory\le -5$

This is not possible.

Case 2: $y-4\le 0ory\le 4y+5\ge 0ory\ge -5$ Both of these are satisfied if -5≤y≤4

hence maximum value of y is 4 and minimum value is -5

$x^2-3x+2$ be factor of $x^4-p{x}^2+q=0$

Hence $(x^2-3x+2)=0$  ⇒ (x-2)(x-1) = 0

⇒ x = 2, 1, putting these values in given equation

so, 4p - q -16 = 0     ......(i)

and, p - q - 1 = 0      .......(ii)

Solving (i) and (ii), we get (p, q) = (5, 4)

Exprssions are $x^2-11x+a$ and $x^2-14x+2a$ will have a common factor, then

⇒ $\frac{x^2}{-22a+14a}=\frac{x}{a-2a}=\frac{1}{-14+11}$

⇒ $\frac{x^2}{-8a}=\frac{x}{-a}=\frac{1}{-3}$ ⇒ $x^2=\frac{8a}{3}andx=\frac{a}{3}$

⇒ $(\frac{a}{3}{)}^2=\frac{8a}{3}$ ⇒ $\frac{a^2}{9}=\frac{8a}{3}$ ⇒ a = 0, 24.

Alternate Method : We can check by putting the values of a from the options.

Let α be a common root, then

${\alpha }^2+\alpha \alpha +10=0$    

and ${\alpha }^2+b\alpha -10=0$

from (i) - (ii),

$(a-b)\alpha +20=0$ ⇒ α = $-\frac{20}{a-b}$

Substituting the value of α in (i), we get

$(-\frac{20}{a-b}{)}^2+a(-\frac{20}{a-b})+10=0$

⇒400 - 20 a(a - b) + $10(a-b{)}^2$ = 0

⇒40 -2$a^2$ + 2ab + $a^2$ + $b^2$ -2ab = 0

⇒$a^2$ - $b^2$ = 40

Let α is the common root, so ${\alpha }^2+p\alpha +q=0$

and ${\alpha }^2+q\alpha +p=0$

⇒ (p - q) α + (q - p) = 0 ⇒  α=1

Put the value of α , p + q + 1 = 0

The condition for common roots gives $(bc-a^2{)}^2=(ca-b^2\left)\right(ab-c^2)$ On simplification gives $a^3+b^3+c^3=3abc$

Given equation $e^{sinx}-e^{-sinx}-4=0$

Let  $e^{sinx}=y$, then given equation can be written as

$y^2-4y-1=0$ ⇒ $y=2\pm \sqrt{5}$

But the value of $y=e^{sinx}$ is always positive, so

$y=2+\sqrt{5}$ (∵ 2 < $\sqrt{5}$

⇒$lo{g}_{e}y=lo{g}_e(2+\sqrt{5})$ ⇒ sin x = $lo{g}_e(2+\sqrt{5})>1$

no solution

Given $|x{|}^2-3|x|+2=0$

Here we consider two cases viz.x < 0 and x > 0

Case 1: x < 0 This gives $x^2+3x+2=0$

⇒ $(x+2)(x+1)=0$ ⇒ $x=-2,-1$

Also $x=-1,-2$ satisfy $x<0,$ so $x=-1,-2$ is solution in this case

Case 2: x > 0. This gives $x^2-3x+2=0$

⇒ (x-2)(x-1) = 0 ⇒ x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2

Aliter : $|x{|}^2-3|x|+2=0$

⇒ (|x|-1)(|x|-2)=0

⇒ |x| = 1 and |x| = 2 ⇒ x = ±1, x = ±2.

$x^{lo{g}_x(1-x{)}^2}=9$

⇒ $lo{g}_x\left(9\right)=lo{g}_x(1-x{)}^2($;∵ $a^x=N$ ⇒ $lo{g}_{a}N=x)$

⇒ $9=(1-x^2)$ ⇒ $1+x^2-2x-9=0$

⇒ $x^2-2x-8=0$ ⇒ (x + 2)(x - 4) = 0 ⇒ x = -2, 4
But $x>0;\therefore x=4$

$x^2+y^2=25$ and $xy=12$

⇒ $x^2+(\frac{12}{x}{)}^2=25$ ⇒ $x^4+144-25x^2=0$

⇒ $(x^2-16)(x^2-9)=0$ ⇒ $x^2=16$ and $x^2=9$

⇒ x= ±4 and x = ±3

given equation is $\sqrt{3x+1}+1$ = $\sqrt{x}$

⇒ $\sqrt{3x+1}$  = $\sqrt{x}$ - 1

Squaring both the sides, we get 3$x$ + 1 = $x$ + 1 -2$\sqrt{x}$

⇒ 2$\sqrt{x}$ + 2$x$ = 0            (irrational function)

Thus $x$ ≠ 0 and $x$ ≠ 1, Since equation is non-quadratic equation.