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

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.

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.

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.

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.

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.

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 \(6.25\).

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 \(mn\).

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.

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.

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.