Advanced Math MCQ Quiz in हिन्दी - Objective Question with Answer for Advanced Math - मुफ्त [PDF] डाउनलोड करें

Last updated on Mar 12, 2025

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Latest Advanced Math MCQ Objective Questions

Top Advanced Math MCQ Objective Questions

Advanced Math Question 1:

Consider a function \(g\) that halves in value for every unit increase in \(x\). If \(g(1) = 10\), what is the equation for \(g(x)\)?

  1. \(g(x) = 10(0.5)^{x-1}\)
  2. \(g(x) = 10(2)^{x}\)
  3. \(g(x) = 10(0.5)^{x}\)
  4. \(g(x) = 5(0.5)^{x}\)

Answer (Detailed Solution Below)

Option 1 : \(g(x) = 10(0.5)^{x-1}\)

Advanced Math Question 1 Detailed Solution

The function \(g(x)\) halves in value for each increase in \(x\) by \(1\). Therefore, the function can be expressed as \(g(x) = a(0.5)^x\). Given \(g(1) = 10\), we can substitute \(10\) for \(g(1)\) to find \(a\). Hence, \(g(x) = 10(0.5)^{x-1}\) represents the function correctly. Option 1 correctly uses the initial condition at \(x = 1\). Option 2 and Option 3 misrepresent the decay process, while Option 4 uses incorrect scaling.

Advanced Math Question 2:

A quantity \(h\) decreases by \(30\%\) each time \(x\) increases by \(1\), and \(h(0) = 20\). Which equation represents \(h\)?

  1. \(h(x) = 20(0.7)^{x}\)
  2. \(h(x) = 20(1.3)^{x}\)
  3. \(h(x) = 20(0.3)^{x}\)
  4. \(h(x) = 20(1.7)^{x}\)

Answer (Detailed Solution Below)

Option 1 : \(h(x) = 20(0.7)^{x}\)

Advanced Math Question 2 Detailed Solution

The function \(h(x)\) decreases by \(30\%\) with every unit increase in \(x\). This indicates an exponential decay model \(h(x) = a(1 - 0.3)^x\). Given \(h(0) = 20\), \(a = 20\). Therefore, \(h(x) = 20(0.7)^x\). Option 1 is correct. Option 2 suggests growth rather than decay. Option 3 incorrectly models the decay rate, and Option 4 misrepresents the exponential factor.

Advanced Math Question 3:

Sarah invests \( x \) dollars in a savings account with a function \( V(x) = 1.05x^2 + 20 \). If she invests \( 100 \) dollars, what is \( V(100) \)?

  1. 1000
  2. 2020
  3. 10520
  4. 5020

Answer (Detailed Solution Below)

Option 3 : 10520

Advanced Math Question 3 Detailed Solution

To find \( V(100) \), substitute \( x = 100 \) into the function: \( V(100) = 1.05(100)^2 + 20 \). First, calculate \( (100)^2 = 10000 \). Then \( 1.05(10000) = 10500 \). Add \( 20 \): \( 10500 + 20 = 10520 \). Thus, \( V(100) = 10520 \). Option 3 is correct, as it reflects the correct computation of the investment value.

Advanced Math Question 4:

If \( (9x^3 - 7) \) represents the sales in dollars and \( (5x^3 + 3) \) represents the expenses, what is the profit in terms of \( x^3 \)?

  1. 4x^3 - 10
  2. 5x^3 - 4
  3. 6x^3 - 8
  4. 7x^3 - 6

Answer (Detailed Solution Below)

Option 1 : 4x^3 - 10

Advanced Math Question 4 Detailed Solution

Profit is calculated as sales minus expenses. Therefore, the expression for profit is \( (9x^3 - 7) - (5x^3 + 3) \). Simplifying this, we have: \( 9x^3 - 5x^3 = 4x^3 \) for the \( x^3 \) terms, and \( -7 - 3 = -10 \) for the constants. Thus, the profit expression is \( 4x^3 - 10 \), making option 1 the correct answer.

Advanced Math Question 5:

For which value of \(b\) does the equation \(25x^2 + bx + 36 = 0\) have two distinct real solutions?

  1. 100
  2. 60
  3. -30
  4. 10

Answer (Detailed Solution Below)

Option 1 : 100

Advanced Math Question 5 Detailed Solution

The quadratic \(25x^2 + bx + 36 = 0\) requires the discriminant \(b^2 - 4ac\) to be positive for two distinct real solutions. Here, \(a = 25\) and \(c = 36\). Calculating, \(b^2 - 4(25)(36) > 0\) simplifies to \(b^2 - 3600 > 0\), or \(b^2 > 3600\). Solving gives \(b > 60\) or \(b < -60\). Therefore, the value \(100\) satisfies \(b > 60\), ensuring two distinct real solutions. Thus, \(100\) is the correct option.

Advanced Math Question 6:

Find the value of \(k\) such that the system of equations \(y = 2x + k\) and \(y = x^2 - 3x + 4\) has exactly one solution.

  1. 5
  2. 8
  3. 3
  4. 6.25

Answer (Detailed Solution Below)

Option 4 : 6.25

Advanced Math Question 6 Detailed Solution

To find the value of \(k\) where the line \(y = 2x + k\) intersects the parabola \(y = x^2 - 3x + 4\) at one point, equate the equations: \(2x + k = x^2 - 3x + 4\). Rearrange to form a quadratic: \(x^2 - 5x + 4 - k = 0\). For the quadratic to have exactly one solution, the discriminant must be zero. The discriminant is \((-5)^2 - 4(1)(4 - k) = 25 - 16 + 4k = 0\). Simplifying gives \(9 + 4k = 0\), solving for \(k\) gives \(k = 6.25\). Therefore, the value of \(k\) is \(6.25\).

Advanced Math Question 7:

A function \(j(x) = m^x + n\) passes through the points \((0, 5)\) and \((2, 13)\). Determine \(mn\) if \(m\) and \(n\) are positive constants.

  1. 8
  2. 10
  3. 12
  4. 14

Answer (Detailed Solution Below)

Option 1 : 8

Advanced Math Question 7 Detailed Solution

Start with the point \((0, 5)\): \(j(0) = m^0 + n = 5\), so \(1 + n = 5\), which simplifies to \(n = 4\). Substitute into the second point \((2, 13)\): \(m^2 + 4 = 13\), which simplifies to \(m^2 = 9\). Therefore, \(m = 3\) (since \(m\) is positive). Now, calculate \(mn = 3 \times 4 = 12\). The correct option is 3. Each step involves solving the exponential and linear equations that describe the behavior of nonlinear functions. The problem tests understanding of substitution and solving quadratics in the context of exponential growth. The final solution confirms that multiplication of these constants yields the correct value of \(mn\).

Advanced Math Question 8:

The profit \( P \) in dollars for a company is modeled by the function \( P(x) = -2x^2 + 24x - 36 \), where \( x \) is the number of units sold. What is the maximum profit?

  1. \$100
  2. \$48
  3. \$36
  4. \$64

Answer (Detailed Solution Below)

Option 2 : \$48

Advanced Math Question 8 Detailed Solution

The profit function \( P(x) = -2x^2 + 24x - 36 \) is a downward-opening parabola, indicating a maximum profit at the vertex. Using \( x = \frac{-b}{2a} \), where \( a = -2 \) and \( b = 24 \), we find \( x = \frac{-24}{2(-2)} = 6 \). Substituting \( x = 6 \) into the function gives \( P(6) = -2(6)^2 + 24(6) - 36 = -72 + 144 - 36 = 36 \). Therefore, the maximum profit is \$48, making option 2 correct.

Advanced Math Question 9:

If the height \(h(t)\) of a projectile is given by \(h(t) = -4(t - 5)^2 + 50\), at what time \(t\) does it reach its maximum height?

  1. 0
  2. 4
  3. 5
  4. 10

Answer (Detailed Solution Below)

Option 3 : 5

Advanced Math Question 9 Detailed Solution

The function \(h(t) = -4(t - 5)^2 + 50\) is in the form \(a(t - h)^2 + k\) with \(a < 0\), indicating the parabola opens downwards, and the vertex is the maximum point. The time \(t = 5\) is where the projectile reaches its maximum height, as it is the vertex. The other times do not correspond to the vertex, hence they are incorrect.

Advanced Math Question 10:

A computer's value decreases exponentially by \(15\%\) annually. If the initial value is \(\$1,200\), what is the value after \(3\) years?

  1. \(1,200(0.85)^3\)
  2. \(1,200(0.15)^3\)
  3. \(1,200(1.15)^3\)
  4. \(1,200(0.9)^3\)

Answer (Detailed Solution Below)

Option 1 : \(1,200(0.85)^3\)

Advanced Math Question 10 Detailed Solution

The exponential decay function is \(V(t) = 1,200(0.85)^t\), representing a \(15\%\) annual decrease, where \(0.85\) signifies the remaining value after each year. Calculating \(V(3)\), we have \(1,200(0.85)^3 = 1,200 \times 0.614125 = 736.95\). Option 1 correctly models this scenario. Option 2, using \(0.15\), suggests a different base that reflects \(85\%\) depreciation, not \(15\%\). Option 3 uses \(1.15\), indicating appreciation, not depreciation. Option 4 uses \(0.9\), which suggests a \(10\%\) decrease, not \(15\%\). Thus, option 1 is correct.
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