Q76.Let us consider a curve, y = f(x) passing through the point (−2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf ′(x) = x2. Then (1) x3 −3xf(x) −4 = 0 (2) x2 + 2xf(x) −12 = 0 (3) x3 + xf(x) + 12 = 0 (4) x2 + 2xf(x) + 4 = 0
What This Question Tests
This question requires recognizing the given equation as an exact differential or simplifying it into a standard linear first-order form and then solving it using integration.
Concepts Tested
Formulas Used
d/dx (y*x) = x dy/dx + y
∫ x^n dx = x^(n+1)/(n+1)
📚 NCERT Sections This Tests
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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📋 Question Details
- Chapter
- Differential Equations
- Topic
- First Order Linear Differential Equation
- Year
- 2021
- Shift
- 27 Aug Shift 1
- Q Number
- Q76
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 9: Differential Equations
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