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Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $f\left( x \right) =\sqrt { { x }^{ 2 }-2x+1 } $, then

  1. $f^{ ' }\left( x \right) =1,\forall x$
  2. $f^{ ' }\left( x \right) =1, \forall x\ge 1$
  3. $f^{ ' }\left( x \right) =1, \forall x\le 1$
  4. $f^{ ' }\left( x \right) =1,if\quad x>1\quad and\quad f^{ ' }\left( x \right) =-1\quad if\quad x<1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

f(x) = sqrt((x-1)^2) = |x-1|. The derivative of |x-1| is 1 for x > 1 and -1 for x < 1. The derivative is undefined at x = 1.

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

The value of sin $ 2^o $ is approximately

  1. $ 2^o $
  2. $0.035$
  3. $ \frac {\pi}{180} $
  4. $0.017$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For small angles in radians, sin(x) is approximately x. 2 degrees = 2 * (pi/180) radians = pi/90 radians. pi/90 is approximately 3.14159 / 90 = 0.0349, which rounds to 0.035.

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

Derivative of $(\sin x)^x + \sin^{-1} \sqrt{x}$ with respect to $x$ is

  1. $(x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$
  2. $(x \cot x + \log \sin x) + \dfrac{1}{\sqrt{x - x^2}}$
  3. $(\sin x)^x (x \cot x + \log \,x) + \dfrac{1}{\sqrt{x - x^2}}$
  4. $(\sin x)^x (x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $y=(\sin x)^x$

$\Rightarrow \log y=x \log (\sin x)$
Now differentiating both sides with respect to $x$ 
$\dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \sin x$
or, $\dfrac{dy}{dx}=(\sin x )^x{x\cot x+\log \sin x}$........(1).
Again let $z=(\sin x)^x+\sin^{-1}\sqrt{x}$
Now differentiating both sides with respect to $x$.
$\dfrac{dz}{dx}=\dfrac{dy}{dx}+\dfrac{1}{\sqrt{1-x}}.\dfrac{1}{2\sqrt{x}}$
$\dfrac{dz}{dx}=(\sin x)^x{x\cot x+\log \sin x}+\dfrac{1}{2\sqrt{x-x^2}}$ [Using (1)]

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

Let f(x) be a differentiable function satisfying $f(x+y)=f(x)+f(y)\forall x, y \in R$ and $f(0)=1$ then $\displaystyle\lim _{x\rightarrow 0}\dfrac{2^{f(\tan^2x)}-2^{f(\sin^2x)}}{x^3f(\sin x)}$ equals to?

  1. $\dfrac{1}{2} ln2$
  2. $ln 2$
  3. $\dfrac{1}{4}ln 2$
  4. $\dfrac{1}{8} ln2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $t={ \sin {  }  }^{ -1 }{ 2 }^{ s }$ Then $\dfrac { ds }{ dt }$ is equal to

  1. $\dfrac { \log { 2 } }{ \sqrt { 1-t^{ 2 } } }$
  2. $\dfrac { \sin { t } }{ \log { 2 } }$
  3. $\dfrac { \cot { t } }{ \log { 2 } }$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given t = sin^(-1)(2^s), we can write sin(t) = 2^s, which means s = log2(sin(t)). Differentiating s with respect to t gives ds/dt = (1 / (sin(t) * ln(2))) * cos(t) = cot(t) / ln(2).

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $u=e^{x}(xcosy-ysiny)$ then $\frac{d^{2}y}{dx^{2}}+\frac{d^{2}u}{dy^{2}}=0$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function u = e^x(x cos y - y sin y) is a standard form where the second partial derivatives satisfy the Laplace-like condition d^2u/dx^2 + d^2u/dy^2 = 0. The statement is true.

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $y=\sqrt{x}-\dfrac{1}{\sqrt{x}}$, then $2x\dfrac{dy}{dx}+y$=

  1. $\sqrt{x}$
  2. $2\sqrt{x}$
  3. $3\sqrt{x}$
  4. $\dfrac{\sqrt{x}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$y=\sqrt x-\dfrac{1}{\sqrt x}\ \dfrac{dy}{dx}=\dfrac{d}{dx}(x)^{\dfrac{1}{2}}-\dfrac{d}{dx}(\dfrac{1}{\sqrt x})=\dfrac{1}{2\sqrt x}-(-\dfrac{1}{2})x^{-\dfrac{3}{2}}=\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}\ \dfrac{2x dy}{dx}+y=2x[\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}]+\sqrt x-\dfrac{1}{\sqrt x}\ \quad =\sqrt x+\dfrac{1}{\sqrt x}+\sqrt x-\dfrac{1}{\sqrt x}\ \quad=2\sqrt x$


Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $x\sqrt {1+y}+y\sqrt {1+x}=0$ then  $\dfrac {dy}{dx}=\dfrac {1}{(1+x)^{2}}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Implicit differentiation of x*sqrt(1+y) + y*sqrt(1+x) = 0 leads to a derivative dy/dx that is generally not equal to 1/(1+x)^2. Testing with algebra or symmetric properties shows the statement is false.

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

Let $f(x)$ be a function continuous on $[1, 2]$ and differentiable on $(1, 2)$ satisfying $f(1)=2, f(2)=3$ and $f'(x)\ge 1\forall x\in (1, 2)$. Define $g(x)=\displaystyle \int _{1}^{x}{f(t)dt}\forall x\in [1, 2]$ then the greatest value of  $g(x)$ on $[1, 2]$ is-

  1. $3$
  2. $5$
  3. $\dfrac{5}{2}$
  4. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since f'(x) >= 1, f(x) is strictly increasing. Given f(1)=2 and f(2)=3, the function g(x) = integral from 1 to x of f(t) dt is also increasing, so its maximum occurs at the upper bound x=2. The integral of f(t) from 1 to 2 is bounded by the trapezoid area (f(1)+f(2))/2 * (2-1) = 2.5.