Science and Education
Calculus
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.
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.
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.
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.
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.
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)
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.
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.
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.
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.