Arts (English Medium) कक्षा ११ - CBSE Question Bank Solutions

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

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

`\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

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is