If $a =\displaystyle\lim _{ n \rightarrow \infty} \sum_{ k =1}^{ n } \frac{2 n }{ n ^2+ k ^2}$ and $f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}}, x \in(0,1)$, then :
- $2 \sqrt{2} f \left(\frac{ a }{2}\right)= f ^{\prime}\left(\frac{ a }{2}\right)$
- $f\left(\frac{a}{2}\right) f^{\prime}\left(\frac{a}{2}\right)=\sqrt{2}$
- $\sqrt{2} f \left(\frac{ a }{2}\right)= f ^{\prime}\left(\frac{ a }{2}\right)$
- $f \left(\frac{ a }{2}\right)=\sqrt{2} f ^{\prime}\left(\frac{ a }{2}\right)$
The Correct Option is C
Solution and Explanation
The correct option is (C): $\sqrt{2} f \left(\frac{ a }{2}\right)= f ^{\prime}\left(\frac{ a }{2}\right)$
Top Questions on limits and derivatives
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.
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Questions Asked in JEE Main exam
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Concepts Used:
Limits And Derivatives
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.
Limits Formula:
Derivatives of a Function:
A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.
Properties of Derivatives:
Read More: Limits and Derivatives