Question

Assertion: For the function
$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\dots +\frac{x^2}{2}+x+1,f^{\prime }(1)=100f^{\prime }(0)$
Reason: $\frac{d}{dx}\left(x^n\right)=n\cdot x^{n-1}$.

• A. Assertion is correct, reason is correct; reason is a correct explanation for assertion
• B. Assertion is correct, reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, reason is incorrect
• D. Assertion is incorrect, reason is correct

Answer 1

Answer: (A)

We know that $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$
$\therefore$ For $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\dots ..+\frac{x^2}{2}+x+1$
$f^{\prime }(x)=\frac{100x^{99}}{100}+99\frac{x^{98}}{99}+\dots ..+\frac{2x}{2}+1$
$=x^{99}+x^{98}+\dots \dots \dots +x+1$
Now, $f^{\prime }(1)=1+1+\dots \dots \dots .$ to 100 term $=100$
$f^{\prime }(0)=1\therefore f^{\prime }(1)=100\times 1=100f^{\prime }(0)$
Hence, $f^{\prime }(1)=100f^{\prime }(0)$