First Principles of Derivatives MCQ Quiz in தமிழ் - Objective Question with Answer for First Principles of Derivatives - இலவச PDF ஐப் பதிவிறக்கவும்

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

Latest First Principles of Derivatives MCQ Objective Questions

Top First Principles of Derivatives MCQ Objective Questions

First Principles of Derivatives Question 1:

If f (x) = 1 – x + x2 – x3 ... – x99 + x100, then f′(1) is euqal to

  1. 150
  2. –50
  3. –150
  4. 50

Answer (Detailed Solution Below)

Option 4 : 50

First Principles of Derivatives Question 1 Detailed Solution

Concept:

1.Derivative of a function with respect to a variable is the rate of change

of a function with respect to the variable.

2. Sum of n terms of an A.P is n2[2a+(n1)d] where n is the number of terms,

a is the first term, and d is the common difference.

Calculation:

f(x) = 1 – x + x2 – x3 ... – x99 + x100

Differentiating the function with respect to x.

⇒ f'(x) = -1 + 2x - 3x2 + ..+.. - 99x98 + 100 x99.

Put x = 1 in the above equation.

⇒ f'(1) = -1 + 2 - 3 + 4 - 5 + ... - 99 + 100.

⇒ f'(1) = (- 1 - 3 - 5 .. - 99) + (2 + 4 + 6 +...+ 100)

Here we have two A.P.

We know sum of n terms of an A.P is n2[2a+(n1)d] where n is the number of terms,

a is the first term, and d is the common difference.

∴ f'(1) = 502[2(1)+(501)×2] + 502[2×2+(501)×2]

⇒ f'(1) = 25 × (- 100) + 25 × 102

⇒ f'(1) = 25 [ - 100 + 102]

⇒ f'(1) = 25 × 2 = 50

Therefore, the required value of f'(1) is 50.

First Principles of Derivatives Question 2:

If f(x) = x3sinx, Find f'(x)

  1. (x cos x + 3sin x)
  2. x2sin x)
  3. x2(x cos x + 3sin x)
  4. x2 cos x

Answer (Detailed Solution Below)

Option 3 : x2(x cos x + 3sin x)

First Principles of Derivatives Question 2 Detailed Solution

Concept:

d(xn)dx=nxn1

d(uv)dx=udvdx+vdudx

d(sinx)dx=cosx

Calculation:

Given: f(x) = x3sinx

f'(x) = d(x3sinx)dx

⇒ d(x3sinx)dx=x3d(sinx)dx+sinxdx3dx

= x3 cos x + sin x 3x2

= x2(x cos x + 3sin x)

∴ The required value is x2(x cos x + 3sin x).

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