Nonlinear Equations in One Variable MCQ Quiz in தமிழ் - Objective Question with Answer for Nonlinear Equations in One Variable - இலவச PDF ஐப் பதிவிறக்கவும்

To express $n$ in terms of $m$ and $p$, start with $m=5n^2-4p$. Add $4p$ to both sides to obtain $m+4p=5n^2$. Next, divide both sides by $5$ to isolate $n^2$: $n^2=\frac{m+4p}{5}$. To solve for $n$, take the square root of both sides: $n=\sqrt{\frac{m+4p}{5}}$. Thus, option 2 is correct. Option 1 is identical, but option 3 misrepresents the squared relationship, and option 4 is not derived from the given formula.

Start from the equation $x^2+3y=27$. To solve for $y$, subtract $x^2$ from both sides: $3y=27-x^2$. Then, divide both sides by $3$ to isolate $y$: $y=\frac{27-x^2}{3}$. This results in option 4 as the correct choice. Option 1 erroneously suggests subtracting $x^2$ without division. Option 2 incorrectly adds $x^2$ to $27$. Option 3 reverses the subtraction order.

The maximum value of a downward opening parabola $g(x)=-x^2+4x+7$ is at the vertex. The x-coordinate of the vertex is given by $x=-\frac{b}{2a}$, where $a=-1$ and $b=4$. Thus, $x=-\frac{4}{2(-1)}=2$. Substitute $x=2$ into the function to find the maximum value: $g(2)=-2^2+4(2)+7=-4+8+7=11$. Therefore, the maximum value of $g(x)$ is 11.

The maximum height of a projectile is at the vertex of the parabola represented by the quadratic equation $h(t)=-5t^2+20t+15$. The time at which the maximum height is reached can be found using $t=-\frac{b}{2a}$. Here, $a=-5$ and $b=20$, so $t=-\frac{20}{2(-5)}=2$ seconds. Substitute $t=2$ into the height equation to find the maximum height: $h(2)=-5(2{)}^2+20(2)+15=-20+40+15=35$ meters. Thus, the maximum height reached is 35 meters.

The axis of symmetry for a quadratic function $y=a{x}^2+bx+c$ is given by the formula $x=-\frac{b}{2a}$. For the function $y=3x^2-12x+5$, substitute $a=3$ and $b=-12$: $x=-\frac{-12}{2(3)}=\frac{12}{6}=2$. Therefore, the axis of symmetry is $x=2$. This is the vertical line that passes through the vertex of the parabola.

To find where the parabola $y=x^2-4x+4$ touches the x-axis, we need to find the roots of the equation by setting $y=0$. Solve $x^2-4x+4=0$ by factoring: $(x-2{)}^2=0$. This gives $x=2$. Thus, the point where the parabola touches the x-axis is $(2,0)$. Since the squared term $(x-2{)}^2$ indicates a double root, the parabola touches the x-axis at this point only.

The vertex form of a quadratic equation is given by $y=a(x-h{)}^2+k$, where $(h,k)$ is the vertex. To find the vertex for the equation $y=-2x^2+12x-18$, we first complete the square. Start by factoring out the $-2$ from the $x^2$ and $x$ terms: $y=-2(x^2-6x)-18$. Next, complete the square inside the parentheses. Take half of $-6$, which is $-3$, and square it to get $9$. Add and subtract $9$ inside the parentheses: $y=-2((x^2-6x+9)-9)-18$. Simplify to get $y=-2(x-3{)}^2+18-18$. Therefore, the vertex is $(3,0)$.

The given absolute value equation $|3x+5|=47$ can be rewritten as two separate equations: $3x+5=47$ and $3x+5=-47$. Solving the first equation, $3x+5=47$, we subtract 5 from both sides to get $3x=42$. Dividing by 3, we find $x=14$. For the second equation, $3x+5=-47$, subtracting 5 from both sides gives $3x=-52$. Dividing by 3, we get $x=-\frac{52}{3}$, which is not needed for the positive value. Thus, the positive value of $x-2$ is $14-2=12$. Hence, the correct answer is 10.

The equation $|5x-9|=41$ translates to two possible equations: $5x-9=41$ and $5x-9=-41$. Solving $5x-9=41$, we add 9 to both sides to get $5x=50$. Dividing by 5, we find $x=10$. For $5x-9=-41$, adding 9 gives $5x=-32$. Dividing by 5, we get $x=-\frac{32}{5}$, which is not positive. Therefore, the positive value of $x$ is 10.

The equation $|2x-6|=30$ can be split into $2x-6=30$ and $2x-6=-30$. Solving $2x-6=30$, add 6 to both sides to get $2x=36$. Dividing by 2, we find $x=18$. For $2x-6=-30$, adding 6 gives $2x=-24$. Dividing by 2, we get $x=-12$, which is not positive. Thus, the positive value of $x+3$ is $18+3=21$.