Nonlinear Equations in One Variable MCQ Quiz in हिन्दी - Objective Question with Answer for Nonlinear Equations in One Variable - मुफ्त [PDF] डाउनलोड करें

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 \(y = ax^2 + bx + c\) 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 \(\left|3x + 5\right| = 47\) 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 \(18 + 3 = 21\).