$lim_{n \rightarrow \infty} (\frac{1}{n} + \frac{e^{1/n}}{n} + \frac{e^{2/n}}{n} + ….+ \frac{e^{n-1/n}}{n})$ equals

Q: $\large lim_{n \rightarrow \infty} (\frac{1}{n} + \frac{e^{1/n}}{n} + \frac{e^{2/n}}{n} + ….+ \frac{e^{n-1/n}}{n})$ equals

(A) 0

(B) e

(C) e – 1

(D) e + 1

Required limit = $\large lim_{n \rightarrow \infty} \frac{1}{n}. \frac{(e^{1/n})^n – 1}{e^{1/n} – 1}$

$\large (e-1)lim_{n \rightarrow \infty} [ \frac{1}{\frac{e^{1/n} – 1}{1/n}}]$

= (e – 1) × 1 = e – 1.

Hence (C) is the correct answer.

Let f be a function satisfying f(x + y) = f(x) + f(y) and f(x) = x2 g(x) for all x and y, where g(x) is a continuous function …

Q: Let f be a function satisfying f(x + y) = f(x) + f(y) and f(x) = x2 g(x) for all x and y, where g(x) is a continuous function, then f ‘(x) is equal to

(A) g'(x)

(B) g(0)

(C) g(0) + g'(x)

(D) 0

Solution : $\large f'(x) = lim_{h \rightarrow 0 } \frac{f(x+h) – f(x)}{h}$

$\large = lim_{h \rightarrow 0 } \frac{f(x) + f(h) – f(x)}{h}$

$\large = lim_{h \rightarrow 0 } \frac{f(h)}{h}$

$\large = lim_{h \rightarrow 0 } \frac{h^2 g(x)}{h} = 0$

Hence (D) is the correct answer.

lim_ n →∞ (n!/(mn)^n)^(1/n) (m ∈ N) is equal to

Q: $\large lim_{n \rightarrow \infty} (\frac{n!}{(m n)^n})^{1/n}$  (m ∈ N) is equal to

(A) $\large \frac{1}{e m}$

(B) $\large \frac{m}{e }$

(C) e m

(D) $\large \frac{e}{ m}$

Solution :

L.H.S $\large = lim_{n \rightarrow \infty} (\frac{1.2.3…..n}{(m n)^n})^{1/n}$

$\large = lim_{n \rightarrow \infty} (\frac{1.2.3…..n}{(n)^n})^{1/n} \times \frac{1}{m}$

$\large lim_{n \rightarrow \infty} \frac{1}{m}[(\frac{1}{n}) (\frac{2}{n}) (\frac{3}{n})….(\frac{n-1}{n}) (\frac{n}{n})]^{1/n}$ = S (say)

$\large ln S = lim_{n \rightarrow \infty} [ln(\frac{1}{m}) + \frac{1}{n} \Sigma_{r=1}^{n} ln(\frac{r}{n})]$

$\large ln S = – ln m + \int_{0}^{1} ln x dx$

$\large ln S = – ln m + [x ln x – x ]_{0}^{1}$

$\large ln S = – ln m + (-1)$

$\large ln S = – ln m – ln e$

$\large ln S = – ln e m$

$\large ln S = ln\frac{1}{e m}$

$\large S = \frac{1}{e m}$

Hence (A) is the correct answer.

The value of : lim x→ 0 [1^{1/sin^2 x} + 2^{1/sin^2 x} + …. + n^{1/sin^2 x}]^{sin^2 x}

Q: The value of $\large lim_{x \rightarrow 0} [1^{1/sin^2 x} + 2^{1/sin^2 x} + …. + n^{1/sin^2 x}]^{sin^2 x}$

(A) ∞

(B) 0

(C) $\frac{n(n+1)}{2}$

(D) n

Solution : Put $\frac{1}{sin^2 x} = t \ge 1$

$\large lim_{t \rightarrow \infty} [1^t + 2^t + …. + n^t]^{1/t}$

$\large lim_{t \rightarrow \infty} n[(\frac{1}{n})^t + (\frac{2}{n})^t + …. + 1]^{1/t}$

= n[0 + 0 + … + 1]0 = n

Hence (D) is the correct answer.

The function $(x^2 – 1)|x^2 – 3x + 2| + cos|x|$ is not differentiable at

Q: The function $(x^2 – 1)|x^2 – 3x + 2| + cos|x|$ is not differentiable at

(A) –1

(B) 0

(C) 1

(D) 2

Sol: $\large f(x) = \left\{\begin{array}{ll} (x^2 – 1)(x-1)(x-2)+ cosx \; , x < 1 , x > 2\\ -(x^2 – 1)(x-1)(x-2)+ cosx \; , 1 \le x \leq 2 \end{array} \right.$

f(x) is differentiable every where possibly not at x = 1, 2.

After testing the condition of differentiability, we can see that f(x) is not differentiable at x = 2 .

Hence (D) is the correct answer.

$\lim_{x \rightarrow 1} \frac{sec^{-1}(2-x)}{x^2}$ is equal to

Q: $\lim_{x \rightarrow 1} \frac{sec^{-1}(2-x)}{x^2}$ is equal to

(A) 0

(B) 1/2

(C) 4

(D) does not exist

Ans: (D)

The function $f(x) = \frac{log(1 + ax)-log(1 – bx)}{x}$ is not defined at x= 0. The value which should be assigned

Q: The function $f(x) = \frac{log(1 + ax)-log(1 – bx)}{x}$ is not defined at x= 0. The value which should be assigned to f at x = 0, so that it is continuous at x = 0 is

(A) a – b

(B) a + b

(C) log a + log b

(D) none of these

Sol: $f(x) = a[\frac{log(1+ax)}{ax}] + b[\frac{log(1-bx)}{-bx}]$

So , $\large \lim_{x \rightarrow 0}f(x) = a .1 + b.1 = (a+b) = f(0)$

Hence (B) is the correct answer.

$\lim_{x \rightarrow 0} (\frac{a^x + b^x + c^x}{3})^{2/x}$ , a, b, c > 0 is equal to

Q: $\large \lim_{x \rightarrow 0} (\frac{a^x + b^x + c^x}{3})^{2/x}$ , a, b, c > 0 is equal to

(A) 0

(B) (abc)2/3

(C) abc

(D) (abc)1/2

Sol: Let $\large I = \lim_{x \rightarrow 0} (\frac{a^x + b^x + c^x}{3})^{2/x}$ ; (1 form )

$\large logI = \lim_{x \rightarrow 0} \frac{2}{x}log(\frac{a^x + b^x + c^x}{3})$

[Using L’Hospital]

$\large logI = \frac{2}{3} log(abc)$

$\large logI = log(abc)^{2/3}$

I = (abc)2/3

Hence (B) is the correct answer.

$f(x) = [tan^2 x]$ where [.] denotes the greatest integer function. Then

Q: Let $f(x) = [tan^2 x]$ where [.] denotes the greatest integer function. Then

(A) $\lim_{h \rightarrow 0} f(x)$ does not exist

(B) f(x) is continuous at x = 0

(C) f(x) is not differentiable at x = 0

(D) f'(0) = 1

Sol: $\lim_{h \rightarrow 0} [tan^2 (0+h)] = \lim_{h \rightarrow 0} [tan^2 (0-h)]= [tan^2 (0)] = 0$

So, f(x) is continuous at x = 0.

Since f(x) = 0 in the neighbourhood of 0, f'(0) = 0.

Hence (B) is the correct answer.

$f(x) = \left\{\begin{array}{ll} \frac{sin[x]}{[x]} \; , for \; [x] \ne 0 \\ 0 \; , for \; [x] = 0 \end{array} \right.$

Q: $\large If \; f(x) = \left\{\begin{array}{ll} \frac{sin[x]}{[x]} \; , for \; [x] \ne 0 \\ 0 \; , for \; [x] = 0 \end{array} \right.$ where [x] denotes the greatest integer less than or equal to x, then $\lim_{x \rightarrow 0} f(x)$ equals to

(A) 1

(B) 0

(C) –1

(D) None of these

Sol: $\lim_{x \rightarrow 0^-} f(x)= \lim_{x \rightarrow 0^-} \frac{sin[x]}{[x]}$

$\large = \frac{sin(-1)}{(-1)} = sin1$

and , $\lim_{x \rightarrow 0^+} f(x)= 0$ as  it is given that  f(x) = 0 for [x]=0

So , $\lim_{x \rightarrow 0} f(x)$ does not exist

Hence (D) is the correct answer.