As inverse of Differentiation MCQ Quiz in தமிழ் - Objective Question with Answer for As inverse of Differentiation - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Mar 21, 2025

பெறு As inverse of Differentiation பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் As inverse of Differentiation MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

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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