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To master calculus problems in competitive exams, understanding how to compute the value of −(ur+u)′ at x=0 is essential. This article walks through the full differentiation process, shows how to plug in the given function values, and then checks each multiple-choice option against the correct result −10.

Problem Statement

If \( u(x) \) and \( r(x) \) are differentiable at \( x = 0 \), and if

\( u(0) = 5, \quad u'(0) = -3, \quad r(0) = -1, \quad r'(0) = 2 \)

then the value of \( −(ur + u)’ \) at \( x = 0 \) is

Note: The symbol that looks like “g” in scans is the derivative sign; the expression is \( −(ur+u)’ \) evaluated at \( x=0 \).

Step-by-Step Solution

1. Simplify the expression:
   \( ur + u = u(x)r(x) + u(x) = u(x)(r(x) + 1) \)
2. Differentiate using product rule:
   Let \( f(x) = u(x) \), \( g(x) = r(x) + 1 \).
   \( (ur + u)'(x) = [u(x)(r(x) + 1)]’ = u'(x)(r(x) + 1) + u(x)r'(x) \)
3. Evaluate at \( x = 0 \):
   \( (ur + u)'(0) = u'(0)(r(0) + 1) + u(0)r'(0) \)
4. Substitute values:
   \( u'(0) = -3 \), \( r(0) = -1 \Rightarrow r(0) + 1 = 0 \),
   \( u(0) = 5 \), \( r'(0) = 2 \).
   \( (ur + u)'(0) = (-3) \cdot 0 + 5 \cdot 2 = 10 \)
5. Final expression:
   \( −(ur + u)'(0) = -10 \)

The mathematically correct value is −10, suggesting a mismatch between printed options and the actual answer.

Explanation of Each Option

Key Takeaways for CSIR NET

• Always factor expressions before differentiating: \( ur + u = u(r + 1) \).
• Product rule: \( (fg)’ = f’g + fg’ \).
• Check for printing errors when calculated answer doesn’t match options.
• Practice similar problems with given values at specific points.