गणित (Mathematics) – Mathematics MCQ
गणित (Mathematics) – इन्जिनियरिङ प्रवेश
Mathematics · Engineering Entrance – Mathematics, Physics, Chemistry & English · १,३०८ प्रश्न
१,३०८ प्रश्न उपलब्ध। मोड छान्नुहोस्:
गणित (Mathematics): प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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Simpson's one-third rule with h = 1 applied to ∫₀² x³ dx (points x = 0, 1, 2) gives:
सही उत्तर: ख) 4
(h/3)[y₀ + 4y₁ + y₂] = (1/3)[0 + 4(1) + 8] = 4, which equals the exact value because Simpson's rule is exact for cubics.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Using the trapezoidal rule with step h = 1 (points x = 0, 1, 2), the approximate value of ∫₀² x² dx is:
सही उत्तर: ग) 3
Trapezoidal rule: (h/2)[y₀ + 2y₁ + y₂] = ½[0 + 2(1) + 4] = 3 (the exact value is 8/3).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The equation x³ − x − 1 = 0 has a root in [1, 2]. After one bisection step, the root is known to lie in:
सही उत्तर: ख) [1, 1.5]
f(1) = −1 < 0 and f(1.5) = 3.375 − 1.5 − 1 = 0.875 > 0, so the sign change (and the root) lies in [1, 1.5].
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The Newton–Raphson iteration formula for finding a root of f(x) = 0 is:
सही उत्तर: ख) xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ)
The method follows the tangent at xₙ to where it meets the x-axis, giving xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Applying one step of the Newton–Raphson method to f(x) = x² − 2 with starting value x₀ = 1 gives x₁ =
सही उत्तर: घ) 1.5
x₁ = x₀ − f(x₀)/f′(x₀) = 1 − (1 − 2)/(2 × 1) = 1 + 0.5 = 1.5.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In the simplex method, the non-negative variable added to the left side of a '≤' constraint to turn it into an equation is called a:
सही उत्तर: ख) slack variable
A slack variable takes up the unused amount in a ≤ constraint; a surplus variable is subtracted from a ≥ constraint.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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When the objective function line of an LPP is parallel to the edge of the feasible region on which the optimum occurs, the problem has:
सही उत्तर: घ) infinitely many optimal solutions
Every point of that edge gives the same optimal value of Z, so there are infinitely many optimal solutions (alternative optima).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Maximise Z = 5x + 4y subject to 6x + 4y ≤ 24, x + 2y ≤ 6, x ≥ 0, y ≥ 0. The maximum value of Z is:
सही उत्तर: ग) 21
6x + 4y = 24 and x + 2y = 6 intersect at (3, 1.5). Corner values: (4,0) → 20, (3,1.5) → 21, (0,3) → 12. Maximum = 21.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Maximise Z = x + y subject to 2x + y ≤ 10, x + 2y ≤ 8, x ≥ 0, y ≥ 0. The maximum value of Z is:
सही उत्तर: ग) 6
The lines 2x + y = 10 and x + 2y = 8 meet at (4, 2). Corner values: (0,0) → 0, (5,0) → 5, (4,2) → 6, (0,4) → 4. Maximum = 6.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Minimise Z = 2x + 5y subject to x + y ≥ 4, x ≥ 1, y ≥ 0. The minimum value of Z is:
सही उत्तर: घ) 8
The (unbounded) region has corner points (4, 0) and (1, 3), where Z = 8 and 17. Since both coefficients are positive, Z only grows further out, so the minimum is 8 at (4, 0).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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