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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 9 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 9
If $$f(x)$$ is a differentiable function $$\forall \ x\ \epsilon \ R$$ so that, $$f(2)=4,\ f'(x)\ge 5\ \forall \ x\ \epsilon \ [2,6]$$, then, $$f(6)$$ is :
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$$\ge 24$$
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$$\le 24$$
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$$\ge 9$$
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$$\le 9$$
If $$x=\sin^ {-1}\left( \dfrac { 2\theta }{ 1+{ \theta }^{ 2 } } \right),y=\sec^ {-1}\sqrt {1+\theta^ {2}}$$, then $$\dfrac {dy}{dx}=$$
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$$\dfrac {-1}{2}$$
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$$\dfrac {1}{2}$$
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$$-2$$
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$$2$$
If $$f(x)=0$$ for $$x < 0$$ and $$f(x)$$ is differential at $$x=0$$ then for $$x\ge 0, f(x)$$ may be
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$$x^{2}$$
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$$x$$
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$$-x$$
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$$-x^{3/2}$$
If $$f(x)=\begin{cases} mx-1,\quad x\le 5 \\ 3x-5,\quad x>5 \end{cases} $$ is continuous then value of m is:
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$$\dfrac{11}{5}$$
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$$\dfrac{5}{11}$$
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$$\dfrac{5}{3}$$
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$$\dfrac{3}{5}$$
Explanation
$$f{\left( x \right)} \begin{cases} mx-1 & \text{ if } x \le 5 \\ 3x - 5 & \text{ if } x > 5 \end{cases}$$
Since the function is continuous at $$x = 5$$.
Therefore,
$$\lim_{x \rightarrow {5}^{-}}{f{\left( x \right)}} = \lim_{x \rightarrow {5}^{+}}{f{\left( x \right)}} = f{\left( 5 \right)}$$
$$\lim_{x \rightarrow {5}^{-}}{\left( mx - 1 \right)} = \lim_{x \rightarrow {5}^{+}}{\left( 3x - 5 \right)}$$
$$5m - 1 = 15- 5$$
$$5m = 10 + 1$$
$$\Rightarrow m = \cfrac{11}{5}$$
Hence the value of $$m$$ is $$\cfrac{11}{5}$$.
If $$y={ tan }^{ -1 }\left( \frac { ax-b }{ bx+a } \right)$$, the value of $$\frac { dy }{ dx } $$ is
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$$\frac{ a}{1+x^{2}}$$
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$$\frac{1}{1+x^{2}}$$
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$$\frac{-b}{1+x^{2}}$$
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$$\frac{b}{1+x^{2}}$$
Let $$f:[0,2]\rightarrow R$$ be a twice differentiable function such that $$f"(x)>0$$, for all $$x\in (0,2)$$ If $$\phi (x)=f(x)+f(2-x)$$, then $$\phi$$is:
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decreasing on $$(0,2)$$
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decreasing on $$(0,1)$$ and increasing on $$(1,2)$$
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increasing on $$(0,2)$$
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increasing on $$(0,1)$$ and decreasing on $$(1,2)$$
Explanation
$$\phi(x)=f(x)+f(2-x)$$
$$\phi '(x)=f'(x)-f'(2-x)$$...(1)
Since $$f"(x)>0$$
$$\Rightarrow f'(x) is increasing \forall x\in (0,2)$$
case I when $$x>2-x \Rightarrow x>1$$
$$\Rightarrow \phi'(x) >0\forall x \in (1,2)$$
$$\therefore \phi (x) is increasing on (1,2)$$
case II: when $$x<2-x \Rightarrow x<1$$
$$\Rightarrow \phi '(x) <0\forall x\in (0,1)$$
$$\therefore \phi (x) $$ is decreasing on $$(0,1)$$
If $$f(x)=|\cos x-\sin x|$$, then $$f'\left(\dfrac{\pi}{6}\right)$$ equal to?
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$$-\dfrac{1}{2}(1+\sqrt{3})$$
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$$\dfrac{1}{2}(1+\sqrt{3})$$
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$$-\dfrac{1}{2}(1-\sqrt{3})$$
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$$\dfrac{1}{2}(1-\sqrt{3})$$
If $$f(x)=\left\{\begin{matrix} \dfrac{log_ex}{x-1} & x\neq 1\\ k & x=1\end{matrix}\right.$$ continuous at $$x=1$$, then the value of k is?
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$$e$$
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$$1$$
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$$-1$$
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$$0$$
Explanation
Given
$$f(x)=\left\{\begin{matrix} \dfrac{log_ex}{x-1} & x\neq 1\\ k & x=1\end{matrix}\right.$$
we know that at $$x=1$$ is continuous so
$$k=\displaystyle\lim_{x\rightarrow 1}\dfrac{log_ex}{x-1}$$
It is $$\dfrac 00$$ form so by applying L-hospital rule
$$=\displaystyle\lim_{x\rightarrow 1}\dfrac{1/x}{1}=1$$.
Let $$f(x)=15-|x-10|; x\in R$$. Then the set of all values of x, at which the function, $$g(x)=f(f(x))$$ is not differentiable, is?
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$$\{5, 10, 15, 20\}$$
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$$\{10, 15\}$$
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$$\{5, 10, 15\}$$
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$$\{10\}$$
Explanation
$$f(x)=15-|x-10|, x\in R$$
$$f(f(x))=15-|f(x)-10|$$
$$=15-|15-|x-10|-10|$$
$$=15-|5-|x-10||$$
$$x=5, 10, 15$$ are points of non differentiability
Aliter:
At $$x=10$$ $$f(x)$$ is no differentiable
also, when $$15-|x-10|=10$$
$$\Rightarrow x=5, 15$$
$$\therefore$$ non differentiability points are $$\{5, 10, 15\}$$.
If $$f(x)=\begin{vmatrix} \sqrt{1+kx}-\sqrt{1-kx} & if & -1\leq x < 0\\ \begin{matrix} 2x+1\\ x-1\end{matrix} & if & 0\leq x\leq 1\end{vmatrix}$$ is continuous at $$x=0$$, then the value of k is?
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$$k=1$$
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$$k=-1$$
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$$k=0$$
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$$k=2$$
Explanation
$$\displaystyle\lim_{x\rightarrow 0^-}\dfrac{\sqrt{1+Kx}-\sqrt{1-Kx}}{x}=\displaystyle\lim_{x\rightarrow 0^+}\dfrac{mKx}{x(\sqrt{1-Kx}+\sqrt{x})}$$
$$\displaystyle\lim_{x\rightarrow 0^+}\dfrac{2x+1}{x-1}=-1=K$$ (after rationlation)
$$\therefore K=-1$$.
If $$f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}& , &x < 0\\ \quad \quad \quad q&,& x = 0\\ \dfrac{\sqrt{x^2 + x}- \sqrt{x}}{x^{3/2}}&,& x > 0 \end{cases}$$ is continuous at $$x = 0$$ the $$(p, q)$$ is
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$$\left(-\dfrac{1}{2}, - \dfrac{3}{2}\right)$$
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$$\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$$
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$$\left(\dfrac{1}{2}, \dfrac{3}{2}\right)$$
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$$\left(-\dfrac{3}{2}, \dfrac{1}{2}\right)$$
If $$x=3\sin {t} ,\ y=3\cos {t} ,$$ find $$\dfrac {dy}{dx}$$ at $$t=\dfrac { \pi }{ 3 } $$
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$$3$$
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$$0$$
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$$-\sqrt{3}$$
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$$1$$
Explanation
$$x=3\sin t$$
$$\dfrac{dx}{dt}=3\cos t$$
$$y=3\cos t$$
$$\dfrac{dy}{dt}=-3\sin t$$
Thus, $$\dfrac{dy}{dx}=-\tan t$$
At $$t=\dfrac{\pi}{3}$$,
$$\dfrac{dy}{dx}=-\sqrt3$$
Let $$f(x)=(x+|x|)|x|$$. Then, for all x.
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$$f$$ is continuous
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$$f$$ is differentiable for some x
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$$f'$$ is continuous
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$$f''$$ is continuous
Explanation
Consider the given function.
$$f(x)=(x+|x|)|x|$$
$$f(x)=\left\{\begin{matrix} (x+x)x, & x>0\\ (x-x)x, & x<0 \end{matrix}\right.$$
$$f(x)=\left\{\begin{matrix} 2x^2, & x>0\\ 0, & x<0 \end{matrix}\right.$$
$$f'(x)=\left\{\begin{matrix} 4x, & x>0\\ 0, & x<0 \end{matrix}\right.$$
Now,
$$\lim_{x\to\ 0^{+}}\ 2x^2=\lim_{x\to 0^{-}}0=0$$
Since, it is polynomial function.
Hence, it is continuous and differentiable at $$0$$.
The set of points where the function $$f(x)=x|x|$$ is differentiable is?
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$$(-\infty, \infty)$$
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$$(-\infty, 0)\cup (0, \infty)$$
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$$(0, \infty)$$
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$$[0, \infty]$$
Explanation
Consider the given function.
$$f(x)=x|x|$$
$$f(x)=\left\{\begin{matrix} x^2, & x>0\\ -x^2, & x<0 \\ 0 & x =0\end{matrix}\right.$$
Since, it is polynomial function.
Hence, it is differentiable at $$(-\infty, \infty)$$.
Differentiate the following function with respect to x.
If for $$f(x)=\lambda x^2+\mu x+12$$, $$f'(14)=15$$ and $$f'(2)=11$$, then find $$\lambda$$ and $$\mu$$.
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$$\lambda =1, \mu =6$$.
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$$\lambda =1, \mu =7$$.
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$$\lambda =6, \mu =7$$.
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$$\lambda =6, \mu =1$$.
Explanation
Given function
$$f(x)=\lambda x^2 +\mu x +12$$
$$\implies f^{\prime}(x)=\dfrac d{dx}(\lambda x^2+\mu x+12)$$
$$f^{\prime}(x)=2\lambda x+\mu$$
$$f^{\prime}(2)=2\lambda (2)+\mu$$
$$\implies 4\lambda +\mu=11\cdots (1)$$
$$f^{\prime}(4)=2\lambda (4)+\mu $$
$$\implies 8\lambda+\mu=15 \cdots (2)$$
$$(2)-(1)$$
$$8\lambda-4\lambda+\mu-\mu=15-11$$
$$4\lambda =4$$
$$\implies \lambda=1$$
Put it in $$(1)$$
$$\implies 4(4)+\mu =11$$
$$\mu=11-4$$
$$\mu =7$$
If $$f(9) = 9, f'(9) = 0$$, then $$\underset{x\to 9}{\lim} \dfrac{\sqrt{f(x)}-3}{\sqrt{x}-3}$$ is equal to
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$$0$$
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$$f(0)$$
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$$f'(3)$$
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$$f(9)$$
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$$1$$
Explanation
$$\underset{x\to 9}{\lim} \dfrac{\sqrt{f(x)} -3}{\sqrt{x} -3} $$ $$\left[\dfrac{0}{0}from\right]$$
$$= \underset{x\to 9}{\lim} \dfrac{\dfrac{f'(x)}{2\sqrt{f(x)}}}{\dfrac{1}{2\sqrt{x}}}$$
$$= \dfrac{\sqrt{x}f'(x)}{\sqrt{f(x)}}$$
$$= \dfrac{\sqrt{9}f'(9)}{\sqrt{f(9)}}$$
$$= \dfrac{3\times 0}{3} = 0$$
Let $$f : R \rightarrow R$$ be a continuous function such that $$f(x^2) = f(x^3)$$ for all $$ x \in R$$. Consider the following statements.
I. f is an odd function.
II. f is an even function.
III. f is differentiable everywhere
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I is true and III is false
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II is true and III is false
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both I and III are true
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both II and III are true
Explanation
$$f:R\rightarrow\,R$$ be a continuous function such that
$$f\left({x}^{2}\right)=f\left({x}^{3}\right)$$ ..........$$(1)$$ for all $$x\in R$$
Put $$x=-x$$
$$f\left({x}^{2}\right)=f\left(-{x}^{3}\right)$$
From $$(1)$$ we have $$f\left({x}^{3}\right)=f\left(-{x}^{3}\right)$$
Put $${x}^{3}=t$$ we have $$f\left(t\right)=f\left(-t\right)$$
$$\Rightarrow\,f\left(x\right)$$ is an even function.
$$(ii)$$Now take $${x}^{3}=t$$
$$\Rightarrow\,f\left({t}^{\frac{2}{3}}\right)=f\left(t\right)$$
Put $$t={t}^{\frac{2}{3}}\Rightarrow\,f\left({\left({t}^{\frac{2}{3}}\right)}^{2}\right)=f\left(t\right)$$
$$\Rightarrow\,f\left(t\right)=f\left({t}^{\frac{2}{3}}\right)=f\left({\left({t}^{\frac{2}{3}}\right)}^{2}\right)=f\left({\left({t}^{\frac{2}{3}}\right)}^{3}\right)=....=f\left({\left({t}^{\frac{2}{3}}\right)}^{n}\right)$$
This is true for all $$t\in R$$ and any $$n\in I$$
Hence if we take $$n\rightarrow\,\infty,\,{\left(\dfrac{2}{3}\right)}^{n}\rightarrow\,0$$
Then $$f\left(t\right)=f\left({t}^{\circ}\right)=1$$
$$\Rightarrow\,f\left(x\right)$$ is a constant function, hence it is differentiable everywhere.
Let $$f(x + y) = f(x) f(y)$$ for all $$x$$ and $$y$$. If $$f(0) = 1, f(3) = 3$$ and $$f'(0) = 11$$, then $$f'(3)$$ is equal to
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$$11$$
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$$22$$
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$$33$$
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$$44$$
Explanation
We have,
$$f'(x) = \underset{h\to 0}{\lim} \dfrac{f(x+h)-f(x)}{h}$$
$$\Rightarrow f'(3) = \underset{h\to 0}{\lim} \dfrac{f(3+h)-f(3)}{h}$$
$$=\underset{h\to 0}{\lim} \dfrac{f(3)f(h) - f(3+0)}{h}$$
$$=\underset{h\to 0}{\lim} \dfrac{f(3)f(h) - f(3)f(0)}{h}$$
$$=f(3) \underset{h\to 0}{\lim} \dfrac{f(h)-f(0)}{h}$$
$$=f(3) \underset{h\to 0}{\lim} \dfrac{f(0+h)-f(0)}{h}$$
$$= f(3)f'(0)$$
$$=3\times 11$$
$$=33$$
Let $$f(x+y) = f(x) f(y) $$ and $$f(x) = 1+\sin(3x)g(x)$$, where $$g$$ is differentiable. The $$f'(x)$$ is equal to
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$$3f(x)$$
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$$g(0)$$
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$$3f(x) g(0)$$
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$$3g(x)$$
Explanation
$$f'(x) = \underset{h\to 0}{\lim} \dfrac{f(x+h) - f(x)}{h}$$
$$=\underset{h\to 0}{\lim} \dfrac{f(x)f(h) - f(x)}{h}$$
$$= f(x) \underset{h \to 0}{\lim} \left(\dfrac{1+\sin 3h(g(h))-1}{h}\right)$$
$$3f(x) \underset{h\to 0}{\lim} \dfrac{\sin 3h}{3h} \underset{h\to 0}{\lim} g(h)$$
$$=3 f(x) \times 1 \times g(0) = 3f(x) g(0)$$
If $$y=\tan^{-1}\left(\dfrac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{a}{b}$$
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$$\dfrac{-b}{a}$$
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$$1$$
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$$-1$$
Explanation
Let, $$\dfrac{a}{b}=\tan\theta$$
$$y=\tan^{-1}\left(\dfrac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)\\\,\,=\tan^{-1}\left(\dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x}\right)\\\,\,=\tan^{-1}\left(\dfrac{\tan \theta-\tan x}{1+\tan \theta \tan x}\right)$$
$$=\tan^{-1}\tan (\theta-x)\\=\theta-x\\=\left(\tan^{-1}\dfrac{a}{b}-x\right)$$.
$$\therefore \dfrac{dy}{dx}=-1$$.
If $$y=\tan^{-1}\left\{\dfrac{\cos x+\sin x}{\cos x-\sin x}\right\}$$ then $$\dfrac{dy}{dx}=?$$
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$$1$$
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$$-1$$
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$$\dfrac{1}{2}$$
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$$\dfrac{-1}{2}$$
If $$y=\sin^{-1}(3x-4x^3)$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{3}{\sqrt {1-x^2}}$$
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$$\dfrac{-4}{\sqrt {1-x^2}}$$
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$$\dfrac{3}{\sqrt {1+x^2}}$$
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$$none\ of\ these$$
If $$\displaystyle f(x) = sin^{1}\left (\dfrac{2x}{1 + x^{2}} \right ) $$ , then
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$$ f $$ is derivative for all $$ x $$ with ,$$ \left |x \right | < 1 $$
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$$ f $$ is not derivative at $$ x = 1 $$
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$$ f $$ is not derivable at $$ x = -1 $$
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$$ f $$ is derivative for all $$ x $$ with ,$$ \left |x \right | > 1 $$
If $$f(x) = \left\{\begin{matrix} x^2 (sgn [x]) + \{x\}), & 0 \le x < 2 \\ \sin x + | x - 3|, & 2 \le x < 4\end{matrix}\right.$$
were [ ] and { } represent the greatest integer and the fractional part function, respectively.
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f (x) is differentiable at x=1.
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f(x) is continuous but non-differentiable at x=1.
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f(x) is non-differentiable at x=2.
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f(x) is non-differentiable at x=2.
If $$y=\cos^{-1}x^{3}$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{-1}{\sqrt{1-x^{6}}}$$
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$$\dfrac{-3x^{2}}{\sqrt{1-x^{6}}}$$
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$$\dfrac{-3}{x^{2}\sqrt{1-x^{6}}}$$
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$$none\ of\ these$$
If $$y=\cos^{-1}(4x^{3}-3x)$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{3}{\sqrt{1-x^{2}}}$$
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$$\dfrac{-3}{\sqrt{1-x^{2}}}$$
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$$\dfrac{4}{\sqrt{1-x^{2}}}$$
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$$\dfrac{-4}{(3x^{2}-1)}$$
Which of the following statement is always true ([ .] represents the greatest integer function)
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If f(x) is discontinuous, then $$ \left | f(x) \right | $$ is discontinuous
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If f(x) is discontinuous , then f $$ \left ( \left | x \right | \right ) $$ is discontinuous
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f(x) = $$ \left [g \left (x \right ) \right ] $$ is discontinuous when g(x) is an integer
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None of these
The largest value of c such that there exists a differential function $$ h(x) for -c < x < c $$ that is a solution of $$ y_1 = 1 +y^2 $$ with h(0) = 0 is
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$$ 2\pi $$
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$$ \pi $$
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$$ \frac { \pi }{2} $$
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$$ \frac { \pi }{4} $$
Which of the following function is thrice differentiable at x=0?
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$$f\left ( x \right )=\left | x^{3} \right |$$
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$$f\left ( x \right )=x^{3}\left | x \right |$$
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$$f\left ( x \right )=\left | x \right |\sin ^{3}x$$
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$$f\left ( x \right )=x\left | \tan ^{3}x \right |$$
If $$y=\tan^{-1}(\sec x+6\tan x)$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{1}{2}$$
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$$\dfrac{-1}{2}$$
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$$1$$
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$$none\ of\ these$$
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