Limits, Continuity and Differentiability

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.

$f(x) = x\cos (\pi (x + [x]))$

At x = 0

$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} x\cos (\pi (x + [x]) = 0$

and f(x) = 0

$ \therefore $ It is continuous at x = 0 and clearly discontinuous at other integer points.

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

$ = \mathop {\lim }\limits_{x \to 0} {{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\left( {x + {{{x^3}} \over 3} + {{2{x^5}} \over {15}} + ....} \right) - x\left( {2x + {{{2^3}{x^3}} \over 3} + 2.{{{2^5}{x^5}} \over {15}}+...} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}^4}} \over {{x^4}\left( {{2 \over 3} - {8 \over 3}} \right) + {x^6}\left( {{4 \over {15}} - {{64} \over {15}}} \right) + ....}}$

$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {1 - {{{x^2}} \over {3!}} + {{{3^4}} \over {5!}} - ....} \right)}^4}} \over { - 2 + {x^2}\left( { - {{60} \over {15}}} \right) + ....}}$

(By dividing numerator and denominator by x⁴)

$ = {{4{{\left( {1 - 0} \right)}^4}} \over { - 2 + 0}}$

$ = {4 \over { - 2}}$

$=$ $-$ 2

f(x) is continuous at x = 1

$ \therefore $    ${{2{{\left( 1 \right)}^2}} \over a} = a$

$ \Rightarrow $   a² = 2

$ \Rightarrow $   a = $ \pm $ $\sqrt 2 $

Also f(x) is continuous at x = $\sqrt 2 $

$ \therefore $   a = ${{2{b^2} - 4b} \over {2\sqrt 2 }}$

Now, when a = $\sqrt 2 ,$ then

4 = 2b² $-$ 4b

$ \Rightarrow $   b² $-$ 2b = 2

$ \Rightarrow $   b² $-$ 2b $-$ 2 = 0

$ \Rightarrow $   b = ${{2 \mp \sqrt {4 + 4.2} } \over 2}$

= 1 $ \pm $ $\sqrt 3 $

$ \therefore $   (a, b) = $\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$

When a = $-$ $\sqrt 2 $, then

$-$ $\sqrt 2 $ = ${{2{b^2} - 4b} \over {2\sqrt 2 }}$

$ \Rightarrow $   $-$ 4 = 2b² $-$ 4b

$ \Rightarrow $   b² $-$ 2b + 2 = 0

$ \therefore $    b = ${{2 \pm \sqrt {4 - 8} } \over 2}$ = 1 $ \pm $ i

As  b $ \in $ Real number so,

b = 1 $ \pm $ i  is not accepted.

$ \therefore $   (a, b) = $\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$   or  $\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$

As   f(x) is differentiable at x = 1

$ \therefore $   $\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a + {{\cos }^{ - 1}}\left( {x + b} \right)} \right) = f(1)$

$ \Rightarrow $   $-$1 $=$ a + cos^$-$1(1 + b)

$ \Rightarrow $   cos^$-$1 (1 + b) $=$ $-$1 $-$ a . . . . . .(1)

As   f(x) is differentiable, so,

2 . H . D $=$ R . H . D

Here, L . H . D $=$ $\mathop {\lim }\limits_{h \to 0} $ ${{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$

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

$ = \mathop {\lim }\limits_{x \to 0} {{ - 1 + h + 1} \over { - h}}$

$=$ $ = \mathop {\lim }\limits_{x \to 0} {h \over { - h}}$

$=$ $-$ 1

R. H. D $ = \mathop {\lim }\limits_{x \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.$

$ = \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }}$ [ Using L' Hospital Rule]

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

$ \therefore $   $ - 1 = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$

$ \Rightarrow 1 - {\left( {1 + b} \right)^2} = 1$

$ \Rightarrow $  ${\left( {1 + b} \right)^2} = 0$

$ \Rightarrow $   $b = - 1$

putting value of b in equation (1), we get,

${\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a$

$ \Rightarrow $   ${\pi \over 2} = - 1 - a$

$ \Rightarrow $   $a = - 1 - {\pi \over 2}$

$ \therefore $   ${a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}$

Given,

$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}$

So,   $\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$

$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left[ {\left( {1 + {a \over x} - {4 \over {{x^2}}} - 1} \right)2x} \right]}}$

$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}$

$ = {e^{2a}}$

$ \therefore $   e^2a = e³

$ \therefore $    2a = 3

$ \Rightarrow $   a $=$ ${3 \over 2}$

$\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)$

$\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)$

Applying $L$ Hospital's rule :

$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}$

$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}$

$ = {1 \over 2}$

$g\left( x \right) = f\left( {f\left( x \right)} \right)$

In the neighbourhood of $x=0,$

$f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)$

$\therefore$ $g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.$

$ = \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)$

$\therefore$ $g(x)$ is differentiable

and $g'\left( x \right) = - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)$

$ \Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)$

$f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$

If ${x^3} - x \ge 0$

$ \Rightarrow \cos \left| {{x^3} - x} \right| = \cos ({x^3} - x)$

and if ${x^3} - x \le 0$

$ \Rightarrow \cos |{x^3} - x| = \cos ({x^3} - x)$

$\therefore$ $\cos (|{x^3} - x|) = \cos ({x^3} - x),\,\forall x \in R$ ..... (i)

Again, if ${x^3} + x \ge 0$

$ \Rightarrow |x|\sin (|{x^3} + x|) = x\sin ({x^3} + x)$

and if ${x^3} + x \le 0$

$ \Rightarrow |x|\sin (|{x^3} + x|) = - x\sin \{ - ({x^3} + x)\} $

$\therefore$ $|x|\sin (|{x^3} + x|) = x\sin ({x^3} + x),\,\forall x \in R$ ...... (ii)

$ \Rightarrow f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$

$\therefore$ $f(x) = a\cos ({x^3} - x) + bx\sin ({x^3} + x)$ ..... (iii)

For a function to be differentiable at x = 0, the function must be continuous.

$f(0) = a\cos (0 - 0) + b(0)\sin (0) = a$

Therefore,

$f({0^ + }) = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

$f({0^ - }) = \mathop {\lim }\limits_{h \to \infty } [a(\cos ( - {h^3} + h) + b( - h)\sin ( - {h^3} - h)]$

$ = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

which is continuous at x = 0; hence, f(x) is differentiable for all values of a and b.

Therefore,

$f(1) = a\cos (1 - 1) + 1\sin (1 + 1) = a + b\sin 2$

$f({1^ + }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 + h)^3} - (1 + h) + b(1 + h)\sin {(1 + h)^3} + (1 + h)$

$ = f(1)$

$f({1^ - }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 - h)^3} - (1 - h) + b(1 - h)\sin {(1 - h)^3} + (1 - h)$

$ = f(1)$

Thus, f(x) is continuous and we can also see that f is differentiable at x = 0 and x = 1.

$f(x) = [{x^2} - 3] = [{x^2}] - 3$

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

and $g(x) = |x|f(x) + |4x - 7|f(x)$

$ = (|x| + |4x - 7|)f(x)$

$ = (|x| + |4x - 7|)[{x^2} - 3]$

$ = \left\{ {\matrix{ {( - x - 4x - 7)( - 3),} & { - 1/2 \le x < 0} \cr {(x - 4x + 7)( - 3),} & {0 \le x < 1} \cr {(x - 4x + 7)( - 2),} & {1 \le x < \sqrt 2 } \cr {(x - 4x + 7)( - 1),} & {\sqrt 2 \le x < \sqrt 3 } \cr {(x - 4x + 7)(0),} & {\sqrt 3 \le x < 7/4} \cr {(x + 4x - 7)(0),} & {7/4 \le x < 2} \cr {(x + 4x - 7)(1),} & {x = 2} \cr } } \right.$

$\therefore$ $g(x) = \left\{ {\matrix{ {15x + 21,} & { - 1/2 \le x < 0} \cr {9x - 21,} & {0 \le x < 1} \cr {6x - 14,} & {1 \le x < \sqrt 2 } \cr {3x - 7,} & {\sqrt 2 \le x\sqrt 3 } \cr {0,} & {\sqrt 3 \le x < 2} \cr {5x - 7,} & {x = 2} \cr } } \right.$

Now, the graphs of f(x) and g(x) are shown below.

Graph for f(x)

Clearly, f(x) is discontinuous at 4 points.

$\therefore$ Option (b) is correct.

Graph for g(x)

Clearly, g(x) is not differentiable at 4 points, when

x $\in$ ($-$1 / 2, 2).

$\therefore$ Option (c) is correct.

Multiply and divide by $x$ in the given expression, we get

$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$

$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}$

$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$

$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$

Since $g(x)$ is differentiable, -

it will be continuous at $x=3$

$\therefore$ $\mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right)$

$2k = 3m + 2\,\,\,\,\,...\left( 1 \right)$

Also $g(x)$ is differentiable at $x=0$

$\therefore$ $\mathop {\lim }\limits_{x \to {3^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g'\left( x \right)$

${K \over {2\sqrt {3 + 1} }} = m$

$k=4m$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Solving $(1)$ and $(2)$, we get

$m = {2 \over 5},\,\,k = {8 \over 5}$

$\therefore$ $k+m=2$

Rewrite f as

$f(x) = \left\{ {\matrix{ {g(x),} & {x > 0} \cr 0 & {x = 0} \cr { - g(x),} & {x < 0} \cr } } \right.$

We have, $f'(x) = \left\{ {\matrix{ {g'(x),} & {x > 0} \cr { - g'(x),} & {x < 0} \cr } } \right.$

At x = 0

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

$ = \mathop {\lim }\limits_{h \to 0} {{g(h) - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{g(h) - g(0)} \over h} = g'(0)$

$f'(x) = \left\{ {\matrix{ {g'(x),} & {x \ge 0} \cr { - g'(x),} & {x < 0} \cr } } \right.$

f is differentiable at x = 0.

$\therefore$ Option (a) is correct.

(b) $h(x) = {e^{|x|}} = \left\{ {\matrix{ {{e^x},} & {x \ge 0} \cr {{e^{ - x}},} & {x < 0} \cr } } \right.$

$ \Rightarrow h'(x) = \left\{ {\matrix{ {{e^x},} & {x \ge 0} \cr { - {e^{ - x}},} & {x < 0} \cr } } \right.$

$ \Rightarrow h'({0^ + }) = 1$

and $h'({0^ - }) = - 1$

So, h(x) is not differentiable at x = 0.

$\therefore$ Option (b) is not correct.

(c) $(foh)(x) = f\{ h(x)\} $, as $h(x) > 0$

$ = \left\{ {\matrix{ {g({e^x}),} & {x \ge 0} \cr {g({e^{ - x}}),} & {x < 0} \cr } } \right.$

$ \Rightarrow (foh)'(x) = \left\{ {\matrix{ {{e^x}g'({e^x}),} & {x \ge 0} \cr { - {e^x}g'({e^{ - x}}),} & {x < 0} \cr } } \right.$

$ \Rightarrow (foh)'({0^ + }) = g'(1),(foh)'({0^ - })$

$ = - g'(1)$

So, $(hof)(x)$ is not differentiable at

$x = 0$.

$\therefore$ Option (c) is not correct.

(d) $(hof)(x) = {e^{\left| {f(x)} \right|}} = \left\{ {\matrix{ {{e^{\left| {g(x)} \right|}},} & {x \ne 0} \cr {{e^0} = 1,} & {x = 0} \cr } } \right.$

Now, $(hof)'(0) = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over x}$

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

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

$ = 1\,.\,g'(0)\,.\,\mathop {\lim }\limits_{h \to 0} {{|x|} \over x}$

= 0, as g'(0) = 0

$\therefore$ Option (d) is correct.

Consider $\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$

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

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

$\left[ \, \right.$ As $\sin \left( {\pi - \theta } \right) = \sin \theta $ $\left. \, \right]$

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

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

Given that $f:(a,b) \to [1,\infty )$

and $g(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {\int_a^x {f(t)dt} } & , & {a \le x \le b} \cr {\int_a^b {f(t)dt} } & , & {x > b} \cr } } \right.$

Now, $g({a^ - }) = 0 = g({a^ + }) = g(a)$

[as $g({a^ + }) = \mathop {\lim }\limits_{x \to {a^ + }} \int_a^x {f(t)dt = 0} $ and $g(a) = \int_a^a {f(t)dt = 0} $]

$g({b^ - }) = g({b^ + }) = g(b) = \int_a^b {f(t)dt} $

$\Rightarrow$ g is continuous, $\forall$x$\in$R.

Now, $g'(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {f(x)} & , & {a < x < b} \cr 0 & , & {x > b} \cr } } \right.$

g'(a^$-$) = 0 but g'(a⁺) = f(a) $\ge$ 1

[$\because$ Range of f(x) is [1, $\infty$), $\forall$ x $\in$[a, b]]

$\Rightarrow$ g is non-differentiable at x = a

and g'(b⁺) = 0

but g'(b^$-$) = f(b) $\ge$ 1

$\Rightarrow$ g is not differentiable at x = b.

Multiply and divide by $x$ in the given expression, we get

$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x_2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$

$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan 4x}}$

$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$

$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$

$\mathop {\lim }\limits_{n \to \infty } {{({1^a} + {2^a} + ... + {n^a})} \over {{{(n + 1)}^{a - 1}}[(na + 1) + (na + 2) + ... + (na + n)]}}$

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

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

$ = {{\int\limits_a^1 {{x^a}dx} } \over {a + {1 \over 2}}} = {1 \over {(a + 1)\left( {a + {1 \over 2}} \right)}} = {1 \over {60}}$ given.

$ \Rightarrow (2a + 1)(a + 1) = 120 \Rightarrow 2{a^2} + 3a - 119 = 0$

$ \Rightarrow 2{a^2} - 14a + 17a - 119 = 0$

$ \Rightarrow 2a(a - 7) + 17(a - 7) = 0 \Rightarrow (2a + 17)(a - 7) = 0$.

$\therefore$ $a = 7,\, - {{17} \over 2}$

But $a = - {{17} \over 2}$ has to be discarded because integral $\int\limits_a^1 {{x^a}dx} $ converges if a > $-$1

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

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {\matrix{ {x - 2\,\,\,,} & {x \ge 2} \cr {2 - x\,\,\,,} & {x \le 2} \cr } } \right.$

Similarly, $f\left( x \right) = \left| {x - 5} \right| = \left\{ {\matrix{ {x - 5\,\,\,,} & {x \ge 5} \cr {5 - x\,\,\,,} & {x \le 5} \cr } } \right.$

$\therefore$ $f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|$

$ = \left\{ {x - 2 + 5 - x = 3,2 \le x \le 5} \right\}$

Thus $\,\,\,\,\,\,\,\,\,f\left( x \right) = 3,2 \le x \le 5$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( x \right) = 0,2 < x < 5$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( 4 \right) = 0$

$\therefore$ Statement $1$ is true.



As $f\left( 2 \right) = 0 + \left| {2 - 5} \right| = 3$

and $f\left( 5 \right) = \left| {5 - 2} \right| + 0 = 3$

$\therefore$ Statement - $2$ is also true but not a correct explanation for statement $1.$