First Principles of Derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for First Principles of Derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 11, 2025
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
Answer (Detailed Solution Below)
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 \(\rm\frac{n}{2}\left[2a+(n-1)d\right]\) 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 \(\rm\frac{n}{2}\left[2a+(n-1)d\right]\) where n is the number of terms,
a is the first term, and d is the common difference.
∴ f'(1) = \(\frac{50}{2}\left[2(-1)+(50-1)× -2\right]\) + \(\frac{50}{2}\left[2×2+(50-1)× 2\right]\)
⇒ 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)
Answer (Detailed Solution Below)
First Principles of Derivatives Question 2 Detailed Solution
Concept:
\(\rm \frac{\mathrm{d} (x^{n})}{\mathrm{d} x} = nx^{n-1} \)
\(\rm \frac{\mathrm{d} (uv)}{\mathrm{d} x} = u\frac{\mathrm{d} v}{\mathrm{d} x} + v \frac{\mathrm{d} u}{\mathrm{d} x}\)
\(\rm \frac{\mathrm{d} (\sin x)}{\mathrm{d} x} = \cos x\)
Calculation:
Given: f(x) = x3sinx
f'(x) = \(\rm \frac{\mathrm{d} (x^{3}sinx)}{\mathrm{d} x}\)
⇒ \(\rm \frac{\mathrm{d} (x^{3}sinx)}{\mathrm{d} x} = x^{3}\frac{\mathrm{d} (sinx)}{\mathrm{d} x} + sinx \frac{\mathrm{d} x^{3}}{\mathrm{d} x}\)
= x3 cos x + sin x 3x2
= x2(x cos x + 3sin x)
∴ The required value is x2(x cos x + 3sin x).