Differentiability MCQ Quiz in తెలుగు - Objective Question with Answer for Differentiability - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Mar 14, 2025

పొందండి Differentiability సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Differentiability MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Differentiability MCQ Objective Questions

Differentiability Question 1:

यदि u = exyz, तब \(\rm \frac{\partial^3 u}{\partial x \partial y \partial z}\) पर (1, 1, 1) _____ है।

  1. 5e
  2. 3e
  3. 2e
  4. 4e

Answer (Detailed Solution Below)

Option 1 : 5e

Differentiability Question 1 Detailed Solution

\(\rm \frac{\partial u}{\partial z} = xye^{xyz}\)

\(\rm \frac{\partial }{\partial y} \left( \rm \frac{\partial u}{\partial z} \right)= \rm \frac{\partial }{\partial y}(xye^{xyz})\)

\(\rm \frac{\partial^2 u}{\partial y \partial z} = xy \rm \frac{\partial }{\partial y}(e^{xyz}) + e^{xyz} \rm \frac{\partial }{\partial y}(xy)\)

= xy(xz)exyz + xexyz

\(\rm \frac{\partial }{\partial x} \left(\frac{\partial^2 u}{\partial y \partial z} \right) = \rm \frac{\partial }{\partial x}(x^2 yz + x) e^{xyz}\)

\(\rm \frac{\partial^3 u}{\partial x \partial y \partial z} = (x^2 yz + x) yze^{xyz} + e^{xyz} (2xyz + 1)\)

\(\rm = e^{xyz} (x^2 y^2 z^2 + xyz + 2xyz + 1)\)

= (1 + 3xyz + x2y2z2) exyz

x, y, z = 1, 1, 1 रखने पर हमें प्राप्त होता है

\(\rm \frac{\partial^3 u}{\partial x \partial y \partial z} = (1 + 3 + 1) e\)

= 5e

Top Differentiability MCQ Objective Questions

Differentiability Question 2:

यदि u = exyz, तब \(\rm \frac{\partial^3 u}{\partial x \partial y \partial z}\) पर (1, 1, 1) _____ है।

  1. 5e
  2. 3e
  3. 2e
  4. 4e

Answer (Detailed Solution Below)

Option 1 : 5e

Differentiability Question 2 Detailed Solution

\(\rm \frac{\partial u}{\partial z} = xye^{xyz}\)

\(\rm \frac{\partial }{\partial y} \left( \rm \frac{\partial u}{\partial z} \right)= \rm \frac{\partial }{\partial y}(xye^{xyz})\)

\(\rm \frac{\partial^2 u}{\partial y \partial z} = xy \rm \frac{\partial }{\partial y}(e^{xyz}) + e^{xyz} \rm \frac{\partial }{\partial y}(xy)\)

= xy(xz)exyz + xexyz

\(\rm \frac{\partial }{\partial x} \left(\frac{\partial^2 u}{\partial y \partial z} \right) = \rm \frac{\partial }{\partial x}(x^2 yz + x) e^{xyz}\)

\(\rm \frac{\partial^3 u}{\partial x \partial y \partial z} = (x^2 yz + x) yze^{xyz} + e^{xyz} (2xyz + 1)\)

\(\rm = e^{xyz} (x^2 y^2 z^2 + xyz + 2xyz + 1)\)

= (1 + 3xyz + x2y2z2) exyz

x, y, z = 1, 1, 1 रखने पर हमें प्राप्त होता है

\(\rm \frac{\partial^3 u}{\partial x \partial y \partial z} = (1 + 3 + 1) e\)

= 5e

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