## 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.