Question
The function $\text{f(x)}=|\cos\text{x}|$ is:
  1. Everywhere continuous and differentiable.
  2. Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
  3. Neither continuous nor differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
  4. None of these.

Answer

  1. Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

Solution:

As cos x is even function it is continuous everywhere but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

$\cos\Big[(2\text{n}+1)\frac{\pi}{2}=\cos\Big(\text{n}\pi+\frac{\pi}{2}\Big)=-\sin\text{n}\pi$

For n as an integer $\Rightarrow\sin\text{n}\pi=0$

For n as rational $\Rightarrow\sin\text{n}\pi=-1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is:
  1. $\frac{4}{3}\big(4\pi-\sqrt{3}\big)$
  2. $\frac{4}{3}\big(4\pi+\sqrt{3}\big)$
  3. $\frac{4}{3}\big(8\pi-\sqrt{3}\big)$
  4. $\frac{4}{3}\big(8\pi+\sqrt{3}\big)$
A function $y = f(x)$ is given by $x = \frac{1}{1 + t^2}$ and $y = \frac{1}{t(1 + t^2)}$ for all $t > 0$ then $f$ is :
Let $f : R \rightarrow R$ be defined as $f ( x )= x ^{3}+ x -5$.

If $g ( x )$ is a function such that $f ( g ( x ))= x$, $\forall x \in R$, then $g ^{\prime}(63)$ is equal to

${d \over {dx}}({x^{{{\log }_e}x}}) = $
If $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ and ${A_1},{B_1},{C_1}$denote the co-factors of ${a_1},{b_1},{c_1}$ respectively, then the value of the determinant $\left| {\begin{array}{*{20}{c}}{{A_1}}&{{B_1}}&{{C_1}}\\{{A_2}}&{{B_2}}&{{C_2}}\\{{A_3}}&{{B_3}}&{{C_3}}\end{array}} \right|$ is
If $ \frac{3\text{x}+1}{(\text{x}-1)(\text{x}+3)} = \frac{\text{A}}{\text{x}-1}+\frac{B}{\text{x}+3} $ then $ {\sin}^{-1} \frac{\text{A}}{\text{B}} :$

  1. $ \frac{\pi}{2}$
  2. $ \frac{\pi}{3}$

  3. $ \frac{\pi}{6}$

  4. $ \frac{\pi}{8}$

If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is
The perimeter of a triangle with sides $3i + 4j + 5k,\,$ $4i - 3j - 5k$ and $7i +j$ is
Solve: $\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
  1. $\frac{1}{2}$
  2. $1$
  3. $2$
  4. $\frac{3}{2}$
Let f(x) = |x| + |x - 1|, then:
  1. f(x) is continuous at x = 0, as well as at x = 1
  2. f(x) is continuous at x = 0, but not at x = 1
  3. f(x) is continuous at x = 0, but not at x = 0
  4. none of these