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

Last updated on Apr 8, 2025

பெறு Nonlinear Equations in One Variable பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Nonlinear Equations in One Variable MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Nonlinear Equations in One Variable MCQ Objective Questions

Top Nonlinear Equations in One Variable MCQ Objective Questions

Nonlinear Equations in One Variable Question 1:

Solve for the negative value of \(y\) in the equation \(4y^2 = 36\).

  1. 3
  2. -3
  3. 6
  4. -6

Answer (Detailed Solution Below)

Option 2 : -3

Nonlinear Equations in One Variable Question 1 Detailed Solution

To solve the equation \(4y^2 = 36\), we first divide both sides by \(4\) to isolate the \(y^2\) term: \(y^2 = 9\). Taking the square root of both sides, we get \(y = \pm 3\). The negative value of \(y\) is \(-3\). Therefore, the correct answer is \(-3\), which corresponds to option 2. Options 1 and 3 provide positive values, and option 4 is incorrect as it does not satisfy the original equation.

Nonlinear Equations in One Variable Question 2:

What is the negative solution for \(z\) in the equation \(7z^2 = 63\)?

  1. 7
  2. -7
  3. -3
  4. 3

Answer (Detailed Solution Below)

Option 3 : -3

Nonlinear Equations in One Variable Question 2 Detailed Solution

Start by dividing both sides of the equation \(7z^2 = 63\) by \(7\), resulting in \(z^2 = 9\). Solving for \(z\), we take the square root of both sides: \(z = \pm 3\). The negative solution for \(z\) is \(-3\), thus the correct answer is option 3. The other options do not provide the correct negative solution as option 1 and 4 provide positive values, and option 2 provides an incorrect negative value.

Nonlinear Equations in One Variable Question 3:

Given \(m = 5n^2 - 4p\), express \(n\) in terms of \(m\) and \(p\).

  1. \(n = \sqrt{\frac{m + 4p}{5}}\)
  2. \(n = \sqrt{\frac{m + 4p}{5}}\)
  3. \(n = \frac{m - 4p}{5}\)
  4. \(n = 5m - 4p\)

Answer (Detailed Solution Below)

Option 2 : \(n = \sqrt{\frac{m + 4p}{5}}\)

Nonlinear Equations in One Variable Question 3 Detailed Solution

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.

Nonlinear Equations in One Variable Question 4:

If \(x^2 + 3y = 27\), express \(y\) in terms of \(x\).

  1. \(y = 27 - x^2\)
  2. \(y = \frac{27 + x^2}{3}\)
  3. \(y = \frac{x^2 - 27}{3}\)
  4. \(y = \frac{27 - x^2}{3}\)

Answer (Detailed Solution Below)

Option 4 : \(y = \frac{27 - x^2}{3}\)

Nonlinear Equations in One Variable Question 4 Detailed Solution

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.

Nonlinear Equations in One Variable Question 5:

The function \(g(x) = -x^2 + 4x + 7\) represents a parabola. What is the maximum value of \(g(x)\)?

  1. 11
  2. 7
  3. 9
  4. 10

Answer (Detailed Solution Below)

Option 1 : 11

Nonlinear Equations in One Variable Question 5 Detailed Solution

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.

Nonlinear Equations in One Variable Question 6:

A projectile is launched with its height in meters given by \(h(t) = -5t^2 + 20t + 15\), where \(t\) is time in seconds. What is the maximum height reached by the projectile?

  1. 15 meters
  2. 20 meters
  3. 25 meters
  4. 35 meters

Answer (Detailed Solution Below)

Option 4 : 35 meters

Nonlinear Equations in One Variable Question 6 Detailed Solution

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.

Nonlinear Equations in One Variable Question 7:

Determine the axis of symmetry for the quadratic function \(y = 3x^2 - 12x + 5\).

  1. \(x = 2\)
  2. \(x = -2\)
  3. \(x = 0\)
  4. \(x = 3\)

Answer (Detailed Solution Below)

Option 1 : \(x = 2\)

Nonlinear Equations in One Variable Question 7 Detailed Solution

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.

Nonlinear Equations in One Variable Question 8:

The graph of \(y = x^2 - 4x + 4\) is a parabola. At what point does it touch the x-axis?

  1. \((2, -4)\)
  2. \((2, 0)\)
  3. \((0, 2)\)
  4. \((4, 0)\)

Answer (Detailed Solution Below)

Option 2 : \((2, 0)\)

Nonlinear Equations in One Variable Question 8 Detailed Solution

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.

Nonlinear Equations in One Variable Question 9:

Find the vertex of the parabola represented by the equation \(y = -2x^2 + 12x - 18\).

  1. \((2, -18)\)
  2. \((3, 0)\)
  3. \((3, 0)\)
  4. \((3, 18)\)

Answer (Detailed Solution Below)

Option 3 : \((3, 0)\)

Nonlinear Equations in One Variable Question 9 Detailed Solution

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 \((3, 0)\).

Nonlinear Equations in One Variable Question 10:

If \(\left|3x + 5\right| = 47\), what is the positive value of \(x - 2\)?

  1. 12
  2. 15
  3. 10
  4. 17

Answer (Detailed Solution Below)

Option 3 : 10

Nonlinear Equations in One Variable Question 10 Detailed Solution

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.
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