Commerce (English Medium) इयत्ता ११ - CBSE Question Bank Solutions

\[\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^3}{x}\] 

\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\] 

\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\] 

`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`

\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\] 

`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`

`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`

\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]

\[\lim_{x \to 0} \frac{a^x - a^{- x}}{x}\]

\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]

\[\lim_{x \to \pi/2} \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)}\]

If the standard deviation of a variable X is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] is

If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be

\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]

\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]

\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]

\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]

\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]

Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is

The standard deviation of first 10 natural numbers is