Limits, Continuity and Differentiability

$\alpha $ and $\beta $ are the two root of 375x² - 25x - 2 = 0

Both of the roots are lie in (-1, 1) hence sum of given series is finite

$\mathop {\lim }\limits_{n \to \infty } \left( {{\alpha \over {1 - \alpha }} + {\beta \over {1 - \beta }}} \right) = {{\alpha (1 - \beta ) + \beta (1 - \alpha )} \over {(1 - \alpha )(1 - \beta )}}$

$ \Rightarrow {{\left( {\alpha + \beta } \right) - 2\alpha \beta } \over {1 - (\alpha + \beta ) + \alpha \beta }} = {{25 - 2( - 2)} \over {375 - 25 - 2}} = {{29} \over {348}}$

$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$

$ \Rightarrow $ ${(1)^2} - a(1) + b = 0$

$ \Rightarrow $$1 - a + b = 0$

$ \Rightarrow $$a - b = 1\,\,......(1)$

Now 'L' hospital rule

2x - a = 5

$ \Rightarrow $2 - a = 5 ($ \because $ x = 1)

$ \Rightarrow $ a = - 3

By putting a = -3 in (1)

$ \Rightarrow $ b = -4

$ \therefore $ a + b = -7

$f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \cr q & {x = 0} \cr {{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \cr } } \right.$

is continuous at x = 0

So f(0^–) = f(0) = f (0⁺) ... (1)

$f({0^ - }) = \mathop {lt}\limits_{h \to 0} f(0 - h)$

$ \Rightarrow \mathop {lt}\limits_{h \to 0} {{\sin (p + 1)( - h) + \sin ( - h)} \over { - h}}$

$ \Rightarrow \mathop {lt}\limits_{h \to 0} \left[ {{{ - \sin (p + 1)h} \over { - h}} + {{\sinh } \over h}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sin (p + 1)h} \over {h(p + 1)}} \times (p + 1) + \mathop {\lim }\limits_{h \to 0} {{\sinh } \over h}$

= (p + 1) + 1 = p + 2 ...... (2)

Now $f({0^ + }) = \mathop {\lim }\limits_{h \to 0} (0 + h) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {{h^2} + h} - \sqrt h } \over {{h^{3/2}}}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(h)}^{{1 \over 2}}}\left[ {\sqrt {h + 1} - 1} \right]} \over {h\left( {{h^{{1 \over 2}}}} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sqrt {h + 1} - 1} \over h} \times {{\sqrt {h + 1} + 1} \over {\sqrt {h + 1} + 1}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{h + 1 - 1} \over {h\left( {\sqrt {h + 1} + 1} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {1 \over {\sqrt {h + 1} + 1}} = {1 \over {1 + 1}} = {1 \over 2}$ ..... (3)

Now, from equation (1)
f(0^–) = f(0) = f(0⁺)

p + 2 = q = 1/2

So, $q = {1 \over 2}$ and $p = {1 \over 2} - 2 = {{ - 3} \over 2}$

$(p,q) \equiv \left( { - {3 \over 2},{1 \over 2}} \right)$

$g'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}$

$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}$

$ \therefore $ f(c) = 0

$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}$

$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}$ if f(x) > 0

and $g'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}$ if f(x) < 0

$ \Rightarrow g'(c) = f'(c) = - f'(c)$

$ \Rightarrow $ 2f'(c) = 0

$ \Rightarrow $ f'(c) = 0

If $\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$

L·H·S·

$\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)$

$ \Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4$

Now, $\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$ = 4

$ \Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4$

$ \Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}$

$f(x) = [x] - \left[ {{x \over 4}} \right]$

Here check continuty at x = 4

LHL = $\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)$

= $\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]$

= $\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]$

= 3 - 0 = 3

h $ \to $ 0 means h > 0 (very small positive value)

So 4 - h = 3.something (means less than 4)

$ \therefore $ [4 - h] = [3.something] = 3

and $\left[ {{{4 - h} \over 4}} \right]$ = [0.something] = 0

RHL = $\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)$

= $\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]$

= $\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]$

= 4 - 1 = 3

Here [4 + h] = [4.something] = 4

and $\left[ {{{4 + h} \over 4}} \right]$ = [1.something] = 1

And f(4) = $\left[ 4 \right] - \left[ {{4 \over 4}} \right]$ = 4 - 1 = 3

As LHL = RHL = f(4)

$ \therefore $ f is continuous at x = 4.

As f(x) is continuous at x = 5 then

$\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right)$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1$

= $\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)$

$ \Rightarrow $ ${a\left| {\pi - 5} \right| + 1}$ = $a\left| {\pi - 5} \right| + 1$

= ${b\left| {5 - \pi } \right| + 3}$

Note : As $\pi $ = 3.14, then $\pi $ - 5 = 3.14 - 5 = -1.84 < 0

$ \therefore $ ${\left| {\pi - 5} \right|}$ = - ($\pi $ - 5) and ${\left| {5 - \pi } \right|}$ = (5 - $\pi $)

$ \Rightarrow $ $ - a\left( {\pi - 5} \right) + 1$ = $ - a\left( {\pi - 5} \right) + 1$ = ${b\left( {5 - \pi } \right) + 3}$

$ \Rightarrow $ $ - a\left( {\pi - 5} \right) + 1$ = ${b\left( {5 - \pi } \right) + 3}$

$ \Rightarrow $ $\left( {a - b} \right)\left( {5 - \pi } \right) = 2$

$ \Rightarrow $ $\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}$

ƒ(x) = 15 – |x – 10|

g(x) = ƒ(ƒ(x)) = 15 – |ƒ(x) – 10|

= 15 – |15 – |x – 10| – 10|

= 15 – |5 – |x – 10||

As this is a linear expression so it is non differentiable when value inside the modulus is zero.

So non differentiable when

x – 10 = 0 $ \Rightarrow $ x = 10

and 5 – |x – 10| = 0

$ \Rightarrow $ |x – 10| = 5

$ \Rightarrow $ x - 10 = $ \pm $ 5

$ \Rightarrow $ x = 5, 15

$ \therefore $ g(x) is not differentiable at x = 5, 10, 15.

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$ = f(${{\pi \over 4}}$) = k

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$ (${0 \over 0}$ form) = k

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - \sqrt 2 \sin x} \over {-\cos e{c^2}x}}$ (Using L Hospital Rule)

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {\pi \over 4}} \sqrt 2 {\sin ^3}x$ = k

$ \Rightarrow $ k = $\sqrt 2 {\left( {{1 \over {\sqrt 2 }}} \right)^3}$ = ${1 \over 2}$

The general formula for indeterminate form 1^{$\infty $} is

$\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right) - 1} \right)}}$

I = $\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$

Here I is in 1^{$\infty $} form.

$ \therefore $ I = ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}} - 1} \right){1 \over x}}}$

= ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right) - 1 - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}$

= ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}$

Here ${{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over x}}$ is in ${0 \over 0}$ form.

So using L'Hopital rule we get

= ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)} \over 1}} \right).\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)}}$

= ${e^{\left( {f'\left( 3 \right) + f'\left( 2 \right)} \right).1}}$

= e⁰ [ as given ƒ'(3) + ƒ'(2) = 0 ]

= 1

$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.$

= $ = \left\{ {\matrix{ { - x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {2x,} & {1 \le x < 2} \cr {x + 2,} & {2 \le x < 3} \cr {6,} & {x = 3} \cr } } \right.$

f(-1) = $\mathop {\lim }\limits_{x \to - 1} \left( { - x - 1} \right)$ = -( -1) - 1 = 0

f(-1⁺) = $\mathop {\lim }\limits_{x \to - {1^ + }} \left( { - x - 1} \right)$ = 0

$ \therefore $ f(x) is continuous at x = - 1

f(0^-) =$\mathop {\lim }\limits_{x \to {0^ - }} \left( { - x - 1} \right)$

= -(0) - 1 = - 1

f(0) = $\mathop {\lim }\limits_{x \to 0} \left( x \right)$ = 0

f(0⁺) = $\mathop {\lim }\limits_{x \to {0^ + }} \left( x \right)$ = 0

$ \therefore $ f(x) is discontinuous at x = 0

f(1^-) = $\mathop {\lim }\limits_{x \to {1^ - }} \left( x \right)$ = 1

f(1) = $\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)$ = 2

f(1⁺) = $\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)$ = 2

$ \therefore $ f(x) is discontinuous at x = 1

f(3^-) = $\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)$ = 3 + 2 = 5

f(3) = 6

$ \therefore $ f(x) is discontinuous at x = 3

So, f(x) is discontinuous at x = {0, 1, 3}.

$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$

= $\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right)\left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)$

= $\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {1 - \cos x}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$

= $\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$

= $\mathop {\lim }\limits_{x \to 0} {1 \over 2}{\left( {{{\sin x} \over x}} \right)^2}{x^2}{\left( {{{{x \over 2}} \over {\sin {x \over 2}}}} \right)^2}{1 \over {{{{x^2}} \over 4}}}\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$

= ${1 \over 2} \times 4 \times 2\sqrt 2 $

= $4\sqrt 2 $

${{f'(x)} \over {f(x)}} = 1\forall x \in R$

Intergrate & use f(1) = 2

f(x) = 2e^x-1 $ \Rightarrow $ f '(x) = 2e^x$-$1

h(x) = f(f(x)) $ \Rightarrow $ h'(x) = f '(f(x)) f'(x)

h'(1) = f '(f(1)) f'(1)

= f '(2) f '(1)

= 2e . 2 = 4e

$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}$

$\mathop {\lim }\limits_{x \to {1^ - }} {{2\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

$\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}.{1 \over {2\sqrt \pi }}$

Put  $x = \cos \theta $

$\mathop {\lim }\limits_{x \to {0^ + }} {{2\theta } \over {\sqrt 2 \sin \left( {{\theta \over 2}} \right)}}.{1 \over {2\sqrt \pi }}$

$ = \sqrt {{2 \over \pi }} $

$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{1 - {{\tan }^4}x} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}$

=$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}{\pi \over 4}} \right)} \over {\left( {{{\tan }^3}{\pi \over 4}} \right)\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + 1} \right)} \over {1.\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\left( {1 - {{\tan }^2}x} \right)} \over {\left( {{{\cos x - \sin x} \over {\sqrt 2 }}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {1 - {{{{\sin }^2}x} \over {{{\cos }^2}x}}} \right)} \over {\left( {\cos x - \sin x} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)} \over {{{\cos }^2}x}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \over {{{\cos }^2}{\pi \over 4}}}$

= ${{2\sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$

= ${{2\sqrt 2 \left( {{2 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$

= 8

Other Method :

$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

This is in ${0 \over 0}$ form so L' Hospital rule is applicable.

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\cot }^3}x\left( { - \cos e{c^2}x} \right) - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$

= ${{3 \times 1\left( { - {{\left( {\sqrt 2 } \right)}^2}} \right) - {{\left( {\sqrt 2 } \right)}^2}} \over { - 1}}$

= ${{ - 6 - 2} \over { - 1}}$ = 8

$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$

f(x) = sin$\left| x \right| - \left| x \right|$ + 2(x $-$ $\pi $) cosx

$ \because $  sin$\left| x \right|$ $-$ $\left| x \right|$ is differentiable function at c = 0

$ \therefore $  k = $\phi $

R.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$

(as x $ \to $ 0⁺ $ \Rightarrow $  [x] $=$ 0)

$=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$

$=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$

L.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$

(as x $ \to $ 0^$-$ $ \Rightarrow $  [x] $=$ $-$1)

$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $

R.H.L.  $ \ne $  L.H.L.

$\left| {f\left( x \right)} \right| = \left\{ {\matrix{ 1 & , & { - 2 \le x < 0} \cr {1 - {x^2}} & , & {0 \le x < 1} \cr {{x^2} - 1} & , & {1 \le x \le 2} \cr } } \right.$

and  $f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$

Hence  $g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$

It is not differentiable at x = 1

f : ($-$ 1, 1) $ \to $ R

f(x) = max {$-$ $\left| x \right|, - \sqrt {1 - {x^2}} $}


Non-derivable at 3 points in ($-$1, 1)

$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}$

$=$ $\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - x} \right) + \sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( { - 1} \right)}}$ $\sin \left( {{\pi \over 2}\left( { - 1} \right)} \right)$

$=$ $\mathop {\lim }\limits_{x \to {1^ + }} \left( {1 - {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right)\left( { - 1} \right) = \left( {1 - 1} \right)\left( { - 1} \right) = 0$

$f\left( x \right) = \left\{ {\matrix{ {8 + 2x,} & { - 4 \le x \le - 2} \cr {{x^2},} & { - 2 \le x \le - 1} \cr {\left| x \right|,} & { - 1 < x < 1} \cr {{x^2},} & {1 \le x \le 2} \cr {8 - 2x,} & {2 < x \le 4} \cr } } \right.$


f(x) is not differentiable at

x = $\left\{ { - 2, - 1,0,1,2} \right\}$

$ \Rightarrow $  S = {$-$2, $-$ 1, 0, 1, 2}

$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$

$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$

$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$

$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$

If you put y = 0 at ${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$ it is in ${0 \over 0}$ form. So we can use L' Hospital's Rule.

= $\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$

= $\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$

= ${1 \over {4\sqrt 2 }}$

Checking

if f(x) is continuous at x = 1 :

f(1^$-$) = 5

f(1) = 5

f(1⁺) = a + b

if f(x) is continuous at x = 1,

then

f(1^$-$) = f(1) = f(1⁺)

$ \Rightarrow $  5 = 5 = a + b

$ \therefore $  a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3^$-$) = a + 3b

f(3) = b + 15

f(3⁺) = b + 15

if   f(x) = is continuous at x = 3

then,

f(3^$-$) = f(3) = f(3⁺)

$ \Rightarrow $  a + 3b = b + 15 = b + 15

$ \Rightarrow $  a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5^$-$) = b + 25

f(5) = 30

f(5⁺) = 30

if f(x) is continuous at x = 5 then,

f(5^$-$) = f(5) = f(5⁺)

$ \Rightarrow $   b + 25 = 30 = 30

$ \Rightarrow $   b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$ \Rightarrow $  a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$ \therefore $  f is not continuous for any value of a and b.

$\mathop {\lim }\limits_{n \to \infty } \left[ {{{1 + \root 3 \of 2 + \root 3 \of 3 + ...\root 3 \of n } \over {{n^{7/3}}\left( {{1 \over {{{(an + 1)}^2}}} + {1 \over {{{(an + 2)}^2}}} + ... + {1 \over {{{(an + n)}^2}}}} \right)}}} \right]$, $a \in R,\,|a|\, > 1$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {({r^{1/3}})} } \over {{n^{7/3}}\sum\limits_{r = 1}^n {{1 \over {{{(an + r)}^2}}}} }}$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^{1/3}}{1 \over n}} } \over {\sum\limits_{r = 1}^n {{1 \over {{{(a + {r \over n})}^2}}}{1 \over n}} }}$

$ = {{\int\limits_0^1 {{x^{1/3}}dx} } \over {\int\limits_0^1 {{{dx} \over {{{(a + x)}^2}}}} }} = 54$, (given)

$ \Rightarrow {{{3 \over 4}[{x^{4/3}}]_0^1} \over {\left[ { - {1 \over {x + a}}} \right]_0^1}} = 54$

$ \Rightarrow {{{3 \over 4}} \over { - {1 \over {a + 1}} + {1 \over a}}} = 54$

$ \Rightarrow {3 \over {4 \times 54}} = {1 \over {a(a + 1)}} $

$\Rightarrow {a^2} + a = 72$

$ \Rightarrow {a^2} + 9a - 8a - 72 = 0$

$ \Rightarrow a(a + 9) - 8(a + 9) = 0$

$ \Rightarrow (a - 8)(a + 9) = 0$

$ \Rightarrow a = 8$ or $ - 9$

Hence, options (c) and (d) are correct.

It is given, that f : R $ \to $ R and

Property 1 : $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {\sqrt {|h|} }}$ exists and finite, and


Property 2 : $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$ exists and finite.

Option A,

P2 : $\mathop {\lim }\limits_{h \to 0} {{\sin h - \sin 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {1 \over h}\left( {{{\sin \,h} \over h}} \right)$ = doesn't exist.

Option B,

P1 : $\mathop {\lim }\limits_{h \to 0} {{{h^{2/3}} - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} {h^{2/3 - 1/2}} = \mathop {\lim }\limits_{h \to 0} {h^{1/6}} = 0$ exists and finite.

Option C,

P1 : $\mathop {\lim }\limits_{h \to 0} {{|h| - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} \sqrt {|h|} = 0$, exists and finite.

Option D,

P2 : $\mathop {\lim }\limits_{h \to 0} {{h|h| - 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {{|h|} \over h} = \left\{ {\matrix{ 1 & {if\,h \to {0^ + }} \cr { - 1} & {if\,h \to {0^ - }} \cr } } \right.$

So, $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$ does not exist.

Hence, options (b) and (c) are correct.

Given function f : R $ \to $ R is

$f(x) = \left\{ {\matrix{ {{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 3x + 1,} & {x < 0;} \cr {{x^2} - x + 1,} & {0 \le x < 1;} \cr {{2 \over 3}{x^3} - 4{x^2} + 7x - {8 \over 3},} & {1 \le x < 3;} \cr {(x - 2){{\log }_e}(x - 2) - x + {{10} \over 3},} & {x \ge 3;} \cr } } \right\}$

So,

$f'(x) = \left\{ {\matrix{ {{5x^4} + 20{x^3} + 30{x^2} + 20x + 3,} & {x < 0;} \cr {2x - 1,} & {0 \le x < 1;} \cr {2{x^2} + 8x + 7,} & {1 \le x < 3;} \cr {{{\log }_e}(x - 2),} & {x \ge 3;} \cr } } \right\}$

At x = 1, f"(1^-) = 2 > 0 and f"(1⁺) = 4$ - $8 = $ - $4 < 0

$ \therefore $ f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.

For x $ \in $ ($ - $$\infty $, 0)

f'(x) = 5x⁴ + 20x³ + 30x² + 20x + 3

and since

f'($ - $1) = 5$ - $20 + 20 + 30 $ - $ 20 + 3 = $ - $2 < 0

So, f(x) is not increasing on x $ \in $($ - $$\infty $, 0).

Now, as the range of function f(x) is R, so f is onto function.

Hence, options (b), (c) and (d) are correct.

Given,

$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$

=   $\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$

If the function is continuous at x = 0, then
$\mathop {\lim }\limits_{x \to 0} $ f(x) will exist and f(0) = $\mathop {\lim }\limits_{x \to 0} $ f(x)

Now, $\mathop {\lim }\limits_{x \to 0} $ f(x) = $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$

= $\mathop {\lim }\limits_{x \to 0} \left( {{{{e^{2x}} - 1 - kx + x} \over {\left( x \right)\left( {{e^{2x}} - 1} \right)}}} \right)$

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ....} \right) - 1 - kx + x} \over {\left( x \right)\left( {\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ...} \right) - 1} \right)}}} \right]$

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

$ \Rightarrow $  3 $-$ k = 0

$ \Rightarrow $  k = 3

So the limit reduces to

$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$

= $\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$ = 1

Hence, f(0) = 1

Given,

$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. + $ $\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$

as we know that

${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$

$ \Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$

$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$

$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$

We know $\left\{ {{1 \over x}} \right\}$ is fractional part of ${1 \over x}.$

So, the range of $\,\left\{ {{1 \over x}} \right\}$ is $0 \le \left\{ {{1 \over x}} \right\} < 1$

So, $\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$ (finite no) $=0$

Similarly $\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$

$ = \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$

$ = \,\,\,\,{{15 \times 16} \over 2}$

$ = \,\,\,\,120$

Check differtiability at x = $\pi $ and x = 0

at x = 0 :

We have L. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0$

R. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {0 + h} \right) - f\left( 0 \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}$

= 0

$\therefore,\,\,$ LHD = RHD

Therefore, function is differentiable at x = $\pi $.

at x = $\pi $ :

L. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}$

= 0

RH. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}$

= 0

$\therefore\,\,\,$ L. H. D = R H D

Therefore, function is differentiable at x = $\pi $,

Since, the function f(x) is differentiable at all the points including 0 and $\pi $.

So, f(x) is differentiable everywhere.

Therefore, set S is empty set.

S = $\phi $

$ \because $    f(x) has extremum values at x = 1, and x = 2

$ \because $  f'(1) = 0 and f'(2) = 0

As, f(x) is a polynomial of degree 4.

Suppose f(x) = Ax⁴ + Bx³ + cx² + Dx + E

$ \because $   $\mathop {\lim }\limits_{x \to 0} $ $\left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right)$ = 3

$ \Rightarrow $$\,\,\,$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)$ = 3

$ \Rightarrow $   $\mathop {\lim }\limits_{x \to 0} $ $\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)$ = 3

As limit has finite value, so D = 0 and E = 0

Now A(0)² + B(0) + C + 0 + 0 + 1 = 3

$ \Rightarrow $   c + 1 = 3 $ \Rightarrow $ c = 2

f'(x) = 4Ax³ + 3Bx² + 2Cx + D

f'(1) = 0 $ \Rightarrow $ 4A(1) + 3B(1) + 2C(1) + D = 0

$ \Rightarrow $$\,\,\,$ 4A + 3B = $-$ 4     . . . .(1)

f'(2) = 0 $ \Rightarrow $ 4A(8) + 3B(4) + 2C(2) + D = 0

$ \Rightarrow $ 8A + 3B = $-$ 2      . . . . .(2)

From equations (1) and (2), we get

A = ${1 \over 2}$ and B = $-$ 2

So, f(x) = ${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$

Therefore, f($-$ 1) = ${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$

= ${1 \over 2} + 2 + 2 = {9 \over 2}$

Hence f($-$1) = ${9 \over 2}$

Since f(x) is continuous at x = 2.

$\therefore\,\,\,$$\mathop {\lim }\limits_{x \to 2} $ f(x) = f(2)

$ \Rightarrow $  $\mathop {\lim }\limits_{x \to 2} $ ${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$ = k    (${{1^\infty }}$  form)

$ \therefore $  ${{e^l}}$ = k

Where $l$ = $\mathop {\lim }\limits_{x \to 2} $(x $-$ $l$ $-$ 1) $ \times $ ${1 \over {2 - x}}$ = $\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$

= $\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$

$ \Rightarrow $    k = e^$-$1

Let, L = $\mathop {\lim }\limits_{x \to 0} {{\left( {x\tan 2x - 2x\tan x} \right)} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$ = $\mathop {\lim }\limits_{x \to 0} K$ (say)

$ \Rightarrow $ K = ${{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}$

= ${{2x\tan x - \left[ {2x\tan x - 2x{{\tan }^3}x} \right]} \over {4{{\sin }^4}x \times (1 - {{\tan }^2}x)}}$

= ${{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {1 - {{\tan }^2}x} \right)}}$

= ${{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}$

= ${{2x{{{{\sin }^3}x} \over {{{\cos }^3}x}}} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}$

$ \Rightarrow $K = ${x \over {2\sin x \times ({{\cos }^2}x - {{\sin }^2}x)\cos x}}$

$\therefore\,\,\,$ L = $\mathop {\lim }\limits_{x \to 0} $ ${x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos x({{\cos }^2}x - {{\sin }^2}x)}}$

= $\mathop {\lim }\limits_{x \to 0} $ ${x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}$

= ${1 \over 2}$

S = {($\lambda $, $\mu $) $ \in $ R $ \times $ R : f(t) = $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$ sin $\left( {2\left| t \right|} \right),$ t $ \in $ R

f(t) = $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$

= $\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr } } \right.$

f'(t) = $\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr } } \right.$

As, f(t) is differentiable

$ \therefore $  LHD = RHD at t = 0

$ \Rightarrow $   $\left| \lambda \right|$ . sin2(0) + $\left( {\left| \lambda \right|{e^0} - \mu } \right)$2cos(0)

      = $\left| \lambda \right|{e^{ - 0}}\,$ . sin 2(0) $-$ 2 cos (0) $\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$

$ \Rightarrow $$\,\,\,$ 0 + $\left( {\left| \lambda \right| - \mu } \right)$2 = 0 $-$ 2$\left( {\left| \lambda \right| - \mu } \right)$

$ \Rightarrow $  4$\left( {\left| \lambda \right| - \mu } \right)$ = 0

$ \Rightarrow $  $\left| \mu \right|$ = $\mu $

So, S $ \equiv $ ($\lambda $, $\mu $) = {$\lambda $ $ \in $ R & $\mu $ $ \in $ [0, $\infty $)}

Therefore set S is subset of R $ \times $ [0, $\infty $)

(i) Given,

f₁ : R $ \to $ R and f₁(x) = sin$(\sqrt {1 - {e^{ - {x^2}}}} )$

$ \therefore $ f₁(x) is continuous at x = 0

Now,

${f_1}'(x) = \cos \sqrt {1 - {e^{ - {x^2}}}} .{1 \over {2\sqrt {1 - {e^{ - {x^2}}}} }}(2x{e^{ - {x^2}}})$

At x = 0

f₁'(x) does not exists.

$ \therefore $ f₁(x) is not differential at x = 0

Hence, option (2) for P.

(ii) Given, ${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {{{\tan }^{ - 1}}x'}}\,if\,x \ne 0 \hfill \cr 1,\,if\,x = 0 \hfill \cr} \right\}$

$ \Rightarrow {f_2}(x) = \left\{ \matrix{ {{ - \sin x} \over {{{\tan }^{ - 1}}x}}x < 0 \hfill \cr {{\sin x} \over {{{\tan }^{ - 1}}x}}x > 0 \hfill \cr 1\,x = 0 \hfill \cr} \right.$

Clearly, f₂(x) is not continuous at x = 0.

$ \therefore $ Option (1) for Q.

(iii) Given, f₃(x) = [sin(log_e(x + 2))], where [ ] is G.I.F.

and f₃ : ($-$1, e^{$\pi $/2} $-$ 2) $ \to $ R

It is given,

$ - 1 < x < {e^{\pi /2}} - 2$

$ \Rightarrow - 1 + 2 < x + 2 < {e^{\pi /2}} - 2 + 2$

$ \Rightarrow 1 < x + 2 < {e^{\pi /2}}$

$ \Rightarrow {\log _e}1 < {\log _e}(x + 2) < {\log _e}{e^{\pi /2}}$

$ \Rightarrow 0 < {\log _e}(x + 2) < {\pi \over 2}$

$ \Rightarrow \sin 0 < \sin {\log _e}(x + 2) < \sin {\pi \over 2}$

$ \Rightarrow 0 < \sin {\log _e}(x + 2) < 1$

$ \therefore $ $[\sin {\log _e}(x + 2)] = 0$

$ \therefore $ ${f_3}(x) = 0,\,f{'_3}(x) = {f_3}''(x) = 0$

It is differentiable and continuous at x = 0.

$ \therefore $ Option (4) for R

(iv) Given, ${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right),\,if\,x \ne 0 \hfill \cr 0,\,if\,x = 0 \hfill \cr} \right.$

Now, $\mathop {\lim }\limits_{x \to 0} {f_4}(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \left( {{1 \over x}} \right) = 0$

${f_4}'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$

For $x = 0,\,{f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$

$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{{h^2}\sin \left( {{1 \over h}} \right) - 0} \over h}$

$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} h\sin \left( {{1 \over h}} \right) = 0$

Thus,

${f_4}'(x) = \left\{ \matrix{ 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right),\,x \ne 0 \hfill \cr 0,\,x = 0 \hfill \cr} \right.$

Again, $\mathop {\lim }\limits_{x \to 0} $

$f'(x) = \mathop {\lim }\limits_{x \to 0} \left( {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right)$

does not exists.

Since, $\mathop {\lim }\limits_{x \to 0} $ cos${\left( {{1 \over x}} \right)}$ does not exists.

Hence, f'(x) is not continuous at x = 0.

$ \therefore $ Option (3) for S.

Given,

$\mathop {\lim }\limits_{t \to x} {{f(x)\sin t - f(t)\sin x} \over {t - x}} = {\sin ^2}x$

Using L' Hospital rules

$\mathop {\lim }\limits_{t \to x} {{f(x)\cos t - f'(t)\sin x} \over 1} = {\sin ^2}x$

$ \Rightarrow f(x)\cos x - f'(x)\sin x = {\sin ^2}x$

$ \Rightarrow f'(x)\sin x - f(x)\cos x = - {\sin ^2}x$

$ \Rightarrow {{f'(x)\sin x - f(x)\cos x} \over {{{\sin }^2}x}} = - 1$

$ \Rightarrow d\left( {{{f(x)} \over {\sin x}}} \right) = - 1$

On integrating, we get

${{f(x)} \over {\sin x}} = - x + C$

$ \Rightarrow f(x) = - x\sin x + C\sin x$

It is given that $x = {\pi \over 6},\,f\left( {{\pi \over 6}} \right) = - {\pi \over {12}}$

$ \therefore $ $f\left( {{\pi \over 6}} \right) = - {\pi \over 6}\sin {\pi \over 6} + C\sin {\pi \over 6}$

$ = - {\pi \over {12}} = - {\pi \over {12}} + {1 \over 2}C$

$ \Rightarrow C = 0$

$ \therefore $ $f(x) = - x\sin x$

(a) $f(x) = - x\sin x$

$f\left( {{\pi \over 4}} \right) = - {\pi \over 4}\sin {\pi \over 4} = - {\pi \over {4\sqrt 2 }}$ false

(b) f(x) = $-$ x sin x

$\sin x > x - {{{x^3}} \over 6},\,\forall x \in (0,\pi )$

$ \Rightarrow - x\sin x < - {x^2} + {{{x^4}} \over 6}$

$ \Rightarrow f(x) < {{{x^4}} \over 6} - {x^2},\,\forall x \in (0,\,\pi )$

It is true.

(c) f(x) = $-$x sin x

f'(x) = $-$ sin x $-$ x cos x

f'(x) = 0

$ \Rightarrow $ $-$ sin x $-$ x cos x = 0

tan x = $-$ x



$ \Rightarrow $ Their exists $\alpha $$ \in $(0, $\pi $) for which f'($\alpha $) = 0

It is true.

(d) $f(x) = - x\sin x$

$f'(x) = - \sin x - x\cos x$

$f''(x) = - 2\cos x + x\sin x$

$f''\left( {{\pi \over 2}} \right) = {\pi \over 2},\,f\left( {{\pi \over 2}} \right) = - {\pi \over 2}$

$ \therefore $ $f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = 0$

It is true.

We have,

${(f(0))^2} + {(f'(0))^2} = 85$

and $f:R \to [ - 2,2]$

(a) Since, f is twice differentiable function, so f is continuous function.

$ \therefore $ This is true for every continuous function.

Hence, we can always find x $ \in $ (r, s), where f(x) is one-one.

$ \therefore $ This statement is true.

(b) By L.M.V.T.

$f'(c) = {{f(b) - f(a)} \over {b - a}}$

$ \Rightarrow |f'(c)| = \left| {{{f(b) - f(a)} \over {b - a}}} \right|$

$ \Rightarrow |f'({x_0})| = \left| {{{f(0) - f( - 4)} \over {0 + 4}}} \right|$

$ = \left| {{{f(0) - f( - 4)} \over 4}} \right|$

Range of f is [$-$2, 2]

$ \therefore $ $ - 4 \le f(0) - f( - 4) \le 4$

$ \Rightarrow 0 \le \left| {{{f(0) - f( - 4)} \over 4}} \right| \le 1$

Hence, |f'(x₀)| = 1.

Hence, statement is true.

(c) As no function is given, then we assume

$f(x) = 2\sin \left( {{{\sqrt {85} x} \over 2}} \right)$

$ \therefore $ $f'(x) = \sqrt {85} \cos \left( {{{\sqrt {85} x} \over 2}} \right)$

Now, ${(f(0))^2} + {(f'(0))^2} = {(2\sin 0)^2} + {(\sqrt {85} \cos 0)^2}$

${(f(0))^2} + {(f'(0))^2} = 85$

and $\mathop {\lim }\limits_{x \to \infty } f(x)$ does not exists.

Hence, statement is false.

(d) From option b, $|f'({x_0})|\, \le 1$

${x_0} \in ( - 4,0)$

$ \therefore $ ${(f'({x_0}))^2}\, \le 1$

Hence, $g({x_0}) = {(f({x_0}))^2} + {(f'({x_0}))^2}\, \le 4 + 1$

[$ \because $ f(x₀) $ \in $ [$-$2, 2]

$ \Rightarrow g({x_0})\, \le 5$

Now, let p $ \in $ ($-$4, 0) for which g(p) = 5

Similarly, let q be smallest positive number q $ \in $ (0, 4)

such that g(q) = 5

Hence, by Rolle's theorem is (p, q)

g'(c) = 0 for $\alpha $ $ \in $ ($-$4, 4) and since g(x) is greater than 5 as we move from x = p to x = q

and f(x))² $ \le $ 4

$ \Rightarrow $ (f'(x))² $ \ge $ 1 in (p, q)

Thus, g'(c) = 0

$ \Rightarrow $ f'f + f'f" = 0

So, f($\alpha $) + f"($\alpha $) = 0 and f'($\alpha $) $ \ne $ 0

Hence, statement is true.

We have,

$f'(x) = {e^{(f(x) - g(x))}}g'(x)\forall x \in R$

$ \Rightarrow f'(x) = {{{e^{f(x)}}} \over {{e^{g(x)}}}}g'(x)$

$ \Rightarrow {{f'(x)} \over {{e^{f(x)}}}} = {{g'(x)} \over {{e^{g(x)}}}}$

$ \Rightarrow {e^{ - f(x)}}f'(x) = {e^{ - g(x)}}g'(x)$

On integrating both side, we get

${e^{ - f(x)}}$ = ${e^{ - g(x)}}$ + C

At x = 1

${e^{ - f(1)}}$ = ${e^{ - g(1)}}$ + C

${e^{ - 1}} = {e^{ - g(1)}} + C$ [$ \because $ f(1) = 1] ...(i)

At x = 2

${e^{ - f(2)}}$ = ${e^{ - g(2)}}$ + C

$ \Rightarrow $ ${e^{ - f(2)}}$ = ${e^{ - 1}}$ + C [$ \because $ g(2) = 1] .... (ii)

From Eqs. (i) and (ii)

${e^{ - f(2)}}$ = $2{e^{ - 1}}$ $-$ ${{e^{ - g(1)}}}$ ...(iii)

$ \Rightarrow $ ${e^{ - f(2)}}$ > $2{e^{ - 1}}$

We know that, e^$-$x is decreasing

$ \therefore $ $-$f(2) < log_e 2$-$1

f(2) > 1 $-$ log_e 2

$ \Rightarrow $ ${{e^{ - g(1)}}}$ + ${e^{ - f(2)}}$ = $2{e^{ - 1}}$ [from Eq. (iii)]

$ \Rightarrow $ ${{e^{ - g(1)}}}$ < $2{e^{ - 1}}$ $-$g(1) < log_e 2 $-$1

$ \Rightarrow $ g(1) >$-$log_e 2

$f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$

$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$

$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$

$\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}$

$\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}$

$\Rightarrow\left(\frac{4}{5}\right)^0=k+\frac{2}{5}$

$\Rightarrow k+\frac{2}{5}=1 \Rightarrow k=1-\frac{2}{5} \Rightarrow k=\frac{3}{5}$

Given,

      $\mathop {\lim }\limits_{x \to 3} $ ${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$

Here if you put x = 3 in ${{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}$

you will get ${0 \over 0}$ form.

So, we can apply L' Hospital rule

$\therefore\,\,\,$ $\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$

= $\mathop {\lim }\limits_{x \to 3} $ ${{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}$ (applying L' Hospital rule)

= ${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$

= ${1 \over 2}$ $ \times $ $\sqrt 2 $

= ${1 \over {\sqrt 2 }}$

$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$

Put ${{\pi \over 2} - x}$ = t

$ \Rightarrow $ x $ \to $ ${{\pi \over 2}}$

t $ \to $ 0

= $\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}$

= $\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}$

= $\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$

= $\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$

= $\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}$

= ${1 \over 4} \times {1 \over 4}$

= ${1 \over {16}}$

f'(x) is increasing

For some x in $\left( {{1 \over 2},\,1} \right)$

f'(x) = 1 [LMVT]

$ \therefore $ f'(1) > 1

$f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)$

Now, $\mathop {\lim }\limits_{x \to {1^ - }} f(x)$

$ = \mathop {\lim }\limits_{x \to {1^ - }} {{1 - x(1 + 1 - x)} \over {1 - x}}\cos \left( {{1 \over {1 - x}}} \right)$

$ = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)\cos \left( {{1 \over {1 - x}}} \right) = 0$

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $

${{1 - x(1 + 1 - x)} \over {x - 1}}\cos \left( {{1 \over {1 - x}}} \right)$

$ = \mathop {\lim }\limits_{x \to {1^ + }} - (x + 1).cos\left( {{1 \over {x + 1}}} \right)$, which does not exist.

(a) $ \because $ ${e^x} \in (1,e)$ in (0, 1) and $\int_0^x {f(t)\sin t\,dt \in } $ (0, 1) in (0, 1)

$ \therefore $ ${e^x} - \int_0^x {f(t)\sin t\,dt} $ cannot be zero.

So, option (a) is incorrect.

(b) $f(x) + \int_0^{{\pi \over 2}} {f(t)\sin t\,dt} $ always positive

$ \therefore $ Option (b) is incorrect.

(c) Let $h(x) = x - \int_0^{{\pi \over 2} - x} {f(t)\cos t\,dt} $,

$h(0) = - \int_0^{{\pi \over 2}} {f(t)\cos t\,dt} < 0$

$h(1) = 1 - \int_0^{{\pi \over 2} - 1} {f(t)\cos t\,dt} > 0$

$ \therefore $ Option (c) is correct.

(d) Let $g(x) = {x^9} - f(x)$

$g(0) = - f(0) < 0$, $g(1) = 1 - f(1) > 0$

$ \therefore $ Option (d) is correct.