As inverse of Differentiation MCQ Quiz in मराठी - Objective Question with Answer for As inverse of Differentiation - मोफत PDF डाउनलोड करा

Last updated on Apr 10, 2025

पाईये As inverse of Differentiation उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा As inverse of Differentiation एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest As inverse of Differentiation MCQ Objective Questions

Top As inverse of Differentiation MCQ Objective Questions

As inverse of Differentiation Question 1:

The anti - derivative of the function u(x)v”(x) - u”(x)v(x) is:

  1. f(x)g’(x) + g(x)f’(x) + C
  2. f”(x)g’(x) + g”(x)f’(x) + C
  3. f(x)g’(x) – g(x)f’(x) + C
  4. f”(x)g’(x) – g”(x)f’(x) + C

Answer (Detailed Solution Below)

Option 3 : f(x)g’(x) – g(x)f’(x) + C

As inverse of Differentiation Question 1 Detailed Solution

\(\smallint \left\{ {{\rm{u}}\left( {\rm{x}} \right){\rm{v''}}\left( {\rm{x}} \right) - {\rm{u''}}\left( {\rm{x}} \right){\rm{v}}\left( {\rm{x}} \right)} \right\}dx = \smallint {\rm{u}}\left( {\rm{x}} \right){\rm{v''}}\left( {\rm{x}} \right){\rm{dx}} - \smallint {\rm{u''}}\left( {\rm{x}} \right){\rm{v}}\left( {\rm{x}} \right){\rm{dx}}\)

\(= \left\{ {u\left( x \right)v'\left( x \right) - \smallint u'\left( x \right)v'\left( x \right)dx} \right\} - \left\{ {u\left( x \right)v'\left( x \right) - \smallint u'\left( x \right)v'\left( x \right)dx} \right\}\; + \;C\)

= f(x)g’(x) – g(x)f’(x) + C

As inverse of Differentiation Question 2:

If \(f\left( x \right) = \frac{1}{{\sqrt {ax + b} }}\), then its antiderivative is 

  1. \(\frac{1}{a}{\left( {ax + b} \right)^{\frac{{ - 1}}{2}}} + C\)
  2. \(\frac{2}{a}{\left( {ax + b} \right)^{\frac{1}{2}}} + C\)
  3. \(\frac{2}{a}{\left( {ax + b} \right)^{\frac{3}{2}}} + C\)
  4. \(\frac{3}{{2a}}{\left( {ax + b} \right)^{\frac{3}{2}}} + C\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{2}{a}{\left( {ax + b} \right)^{\frac{1}{2}}} + C\)

As inverse of Differentiation Question 2 Detailed Solution

Given function is \(f\left( x \right) = \frac{1}{{\sqrt {ax + b} }}\)

Its antiderivative is \(F\left( x \right) = \smallint f\left( x \right)dx + C\)

\(F\left( x \right) = \smallint \frac{1}{{\sqrt {ax + b} }}dx = \smallint \frac{1}{{{{\left( {ax + b} \right)}^{1/2}}}}dx = \smallint {\left( {ax + b} \right)^{ - 1/2}}dx = \frac{{{{\left( {ax + b} \right)}^{\frac{1}{2}}}}}{{\frac{1}{2}a}} + C = \frac{2}{a}{\left( {ax + b} \right)^{\frac{1}{2}}} + C\)

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