Limits and Derivatives Test

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Limits and Derivatives Test - 36

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Question 1
1 / -0

Directions For Questions

Consider the function $$f(x)=x+\cos x-a$$

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Which of the following is not true about $$y=f(x)$$?

A
It is an increasing function

B
It is a monotonic function

C
It has infinite points of inflections

D
None of these

Solution

$$f(x)=x+\cos x -a\Rightarrow f'(x)=1-\sin x \ge 0 \forall x\in R$$.
Thus $$f(x)$$ is increasing in $$(-\infty , \infty)$$, as for $$f'(x)=0, x$$ is not forming an interval.
Also $$f"(x)=-\cos x=0$$
$$\Rightarrow x=(2n+1)\dfrac {\pi}{2}, n\in Z$$
Hence infinite points of inflection
Now $$f(0)=1-a$$.
For positive root $$1-a < 0 \Rightarrow a > 1$$. For negative root
$$1-a > 0\Rightarrow a < 1$$.
Question 2
1 / -0

Differential coefficient of $$\sec (\tan^{-1} x)$$ w.r.t. x is

A
$$\dfrac{x}{\sqrt{1+x^2}}$$

B
$$\dfrac{x}{1+x^2}$$

C
$$x\sqrt{1+x^2}$$

D
$$\dfrac{1}{\sqrt{1+x^2}}$$

Solution

a
Question 3
1 / -0

Given that $$f (x)$$
is a differentiable function of $$ x$$ and that $$f(x)$$ . $$f (y)$$ = $$f (x) $$+ $$f (y)$$ + $$f (xy) -2$$ and that
$$f (2) =5$$.
Then $$f (3)$$ is equal to?

A
$$6$$

B
$$24$$

C
$$15$$

D
$$19$$

Solution

Given, $$f\left( x \right) .f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2\quad -(1)$$
$$f\left( 2 \right) =5$$
$$f\left( x \right) ={ x }^{ 2 }+1$$
Equation $$1 \Rightarrow $$ $$\left( { x }^{ 2 }+1 \right) \left( { y }^{ 2 }+1 \right) ={ x }^{ 2 }+1+{ y }^{ 2 }+1+{ x }^{ 2 }{ y }^{ 2 }+1-2$$
$${ x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }+1={ x }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }+1$$
$$\therefore f\left( 3 \right) ={ x }^{ 2 }+1={ 3 }^{ 2 }+1=10$$
Question 4
1 / -0

lf $${f}'({x})={g}({x})$$ and $${g}'({x})=-{f}({x})$$ for all $$x$$ and $${f}(2)=4= {g}(2)$$, then $${f}^{2}(24)+{g}^{2}(24)$$ is

A
$$32$$

B
$$24$$

C
$$64$$

D
$$48$$

Solution

$$f(x) = -g'(x)$$ (given)
By multiply both sides with $$f'(x)$$, we get
$$f(x)f'(x) = -g'(x)f'(x)$$ ...(1)
$$\because f'(x) = g(x)$$
$$\therefore$$ Eqn (1) becomes
$$f(x)f'(x) = -g(x)g'(x)$$
Now integrating both sides, we get
$$f^2(x) + g^2(x) = c$$
Where, $$c$$ is the constant of integration.
Now, it is given that $$f(2) = g(2) = 4$$
$$\therefore c = 32$$
$$\implies f^2(24) + g^2(24) = 32$$

Hence, option A is correct.
Question 5
1 / -0

Assertion(A): Let $${ f }({ x })$$ be twice differentiable function such that $$f^{ '' }(x)=-{ f }({ x })$$ and $$f^{ ' }(x)={ g }({ x })$$. lf $${ h }({ x })=[{ f }({ x })]^{ 2 }+[{ g }({ x })]^{ 2 }$$ and $${ h }(1)=8$$, then $${ h }(2)=8$$

Reason (R): Derivative of a constant function is zero.

A
Both A and R are true R is correct reason of A

B
Both A and R are true R is not correct reason of A

C
A is true but R is false

D
A is false but R is true

Solution

Let $$f(x)=c$$
$${f}'(x)=0$$ So Reason is True

Now, $$h(x) = [f(x)]^2 + [g(x)]^2$$
$${h}'(x)=2f(x){f}'(x)+2g(x){g}'(x)$$
$$=2f(x)g(x)+2g(x){f}''(x)$$ ...since $$f'(x) = g(x) \Rightarrow f''(x) = g'(x)$$
$$=2g(x)[f(x) + f''(x)]$$
$$=2g(x)(f(x)-f(x))$$
$$\Rightarrow {h}'(x)=0$$
means $$h(x)=C$$ (a constant function)
Given $$h(1)=8$$
$$\Rightarrow h(x)=8$$
Hence, $$f(2)= 8$$
So, both A and R are correct and R is the correct explanation of A.
Question 6
1 / -0

Assertion (A): lf $$f(x)=\cos^{2}x+\cos^{2}\left(x+\dfrac{\pi}3\right)- \cos x \cos \left(x+\dfrac{\pi}3\right)$$ then $$f'(x)=0$$

Reason(R): Derivative of constant function is zero

A
Both A & R are true, R is correct explanation for A

B
Both A & R are true,R is not correct explanation for A

C
A is true but R is false

D
A is false but R is true

Solution

$$f(x)=\cos ^{2}x+\cos ^{2}(x+\cfrac{\pi }{3})-2\cos x \cos (x+\cfrac{\pi }{3})+\cos x \cos (x+\cfrac{\pi }{3})$$
$$=(\cos x - \cos (x+\cfrac{\pi }{3}))^{2} + \cos x \cos (x+\cfrac{\pi }{3}) $$
$$=(\cos x - (\cfrac{1}{2}\cos x - \cfrac{\sqrt{3}}{2} \sin x ))^{2} + \cos x \cos (x+\cfrac{\pi }{3})$$
$$=(\cfrac{\cos x }{2} + \cfrac{\sqrt{3}}{2} \sin x )^{2} + \cos x \cos (x+\cfrac{\pi }{3})$$
$$=\cfrac{\cos^2x}4 + \cfrac{3\sin^2x}4 +\cfrac{\sqrt3}2 \sin x\cos x + \cos x\left(\cfrac{\cos x}2 - \cfrac{\sqrt3}2\sin x\right)$$
$$f(x)=\cfrac{3}{4}$$
$$f{}'(x)=0$$
Question 7
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lf $$ \displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}.\sin \frac{1}{x}$$ for $$x \neq 0,\ \mathrm{f}(\mathrm{0})=0$$ then?

A
$$f$$ is continuous at $$ x=0$$

B
$$f$$ is differentiable at $$ x=0$$

C
$${f}'(0^{-})$$ exists but $${f}'(0^{+})$$ does not exist

D
$$f$$ is discontinuous at $$x=0$$

Solution

If $$f(x)$$ is continuous at $$x=0$$, then $$ \displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =f(0)$$
$$ \displaystyle LHL = \lim _{ h\rightarrow 0 }{ \left[ (0-h)\sin { \left( \frac { 1 }{0- h } \right) } \right] } =\lim _{ h\rightarrow 0 }{ h\sin { \frac { 1 }{ h } } } =0\times \; number\; bet.\; 0 \; and \; 1=0$$
$$ \displaystyle RHL= \lim _{ h\rightarrow 0 }{ h\sin { \frac { 1 }{ h } } } =0\times \; number=0$$
Hence $$LHL=RHL=f(0)$$. The given function is continuous at $$x=0$$
.
$$ \displaystyle f'\left( x \right) =\lim _{ h\rightarrow 0 }{ \frac { h\sin { \frac { 1 }{ h } } -f\left( 0 \right) }{ h } } =\lim _{ h\rightarrow 0 }{ \sin { \frac { 1 }{ h } } } -0=\lim _{ h\rightarrow 0 }{ \sin { \frac { 1 }{ h } } } $$
Since this limit does not exist, the given function is not differentiable at $$x=0$$
Question 8
1 / -0

The graph of the function $$y = f (x)$$ has a unique tangent at the point $$(e^{a} ,0)$$ through which the graph passes then $$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-sinf(x)}{3f(x)}$$ is

A
$$1$$

B
$$2$$

C
$$0$$

D
$$-1$$

Solution

Given $$y=f(x)$$ has a unique tangent at the point $$(e^a,0)$$.
So, as $$x\rightarrow e^{a}$$, $$f(x)\rightarrow 0$$

Now, $$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}$$

$$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) } $$

$$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) } $$

$$=\dfrac{7-1}{3}=2$$
Question 9
1 / -0

$$f(x)=|x-1|+|x-3|$$ then $$f^{'}(2)=$$

A
$$-2$$

B
$$2$$

C
$$0$$

D
$$1$$

Solution

$$f(x) = |x-1|+|x-3| =\begin{cases} -(x-1)-(x-3), x\le 1\\(x-1)-(x-3), 1< x\le 3\\x-1+(x-3), x>3\end{cases}$$
$$\Rightarrow f(x) =\begin{cases} 4-2x, x\le 1\\2, 1< x\le 3\\2x-4, x>3\end{cases}$$
$$\therefore f'(x) =\begin{cases} -2, x\le 1\\0, 1< x\le 3\\2, x>3\end{cases}$$
Hence $$f'(2) =0$$
Question 10
1 / -0

Let $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{x}^{\mathrm{n}}\sin\frac{1}{\mathrm{x}},\quad \mathrm{x}\neq 0\\0, \quad \mathrm{x}=0\end{array}\right.$$ , then f(x) is continuous but not differentiable at x=0 if

A
$$\mathrm{n}\in(0, 1 ] $$

B
$$ \mathrm{n}\in[1, \infty)$$

C
$$\mathrm{n}\in(-\infty,0)$$

D
$$\mathrm{n}=0$$

Solution

Since, $$f\left( x \right)$$ is continuous at $$x=0$$, therefore, $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =0$$

$$\displaystyle\Rightarrow \lim _{ x\rightarrow 0 }{ { x }^{ n } } \sin { \left( \frac { 1 }{ x } \right) } =0\Rightarrow n>0$$

$$f(x)$$ is differential at $$x=0$$ if $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { f\left( x \right)-f\left( 0 \right) }{ x-0 } } $$ exists finitely

$$\displaystyle \Rightarrow \lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }\sin { \left( \frac { 1 }{ x } \right) } -0 }{ x } } $$ exists finitely

$$\displaystyle \Rightarrow { x }^{ n-1 }\sin { \left( \frac { 1 }{ x } \right) } $$ exists finitely.

$$\Rightarrow n-1>0 \Rightarrow n>1$$

If $$n\le1$$, then $$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ n-1 }\sin { \left( \frac { 1 }{ x } \right) } } $$ does not exists and hence, $$f(x)$$ is not differentiable at $$x=0$$

Hence, $$f(x)$$ is continuous but not differentiable at $$x=0$$ for $$0<n\le1$$

i.e., $$n\in (0,1]$$

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Limits and Derivatives Test - 36

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Limits and Derivatives Test - 36

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SHARING IS CARING

Question 1

It is an increasing function

It is a monotonic function

It has infinite points of inflections

None of these

Solution

Question 2

$$\dfrac{x}{\sqrt{1+x^2}}$$

$$\dfrac{x}{1+x^2}$$

$$x\sqrt{1+x^2}$$

$$\dfrac{1}{\sqrt{1+x^2}}$$

Solution

Question 3

$$6$$

$$24$$

$$15$$

$$19$$

Solution

Question 4

$$32$$

$$24$$

$$64$$

$$48$$

Solution

Question 5

Both A and R are true R is correct reason of A

Both A and R are true R is not correct reason of A

A is true but R is false

A is false but R is true

Solution

Question 6

Both A & R are true, R is correct explanation for A

Both A & R are true,R is not correct explanation for A

A is true but R is false

A is false but R is true

Solution

Question 7

$$f$$ is continuous at $$ x=0$$

$$f$$ is differentiable at $$ x=0$$

$${f}'(0^{-})$$ exists but $${f}'(0^{+})$$ does not exist

$$f$$ is discontinuous at $$x=0$$

Solution

Question 8

$$1$$

$$2$$

$$0$$

$$-1$$

Solution

Question 9

$$-2$$

$$2$$

$$0$$

$$1$$

Solution

Question 10

$$\mathrm{n}\in(0, 1 ] $$

$$ \mathrm{n}\in[1, \infty)$$

$$\mathrm{n}\in(-\infty,0)$$

$$\mathrm{n}=0$$

Solution

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