Limits, Continuity and Differentiability

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

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 $1.$

Let $f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$

Doubtful points are $x = n,n \in I$

$L.H.L$ $ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $

$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$

( As $\left[ x \right]$ is the greatest integer function )

$R.H.L. = \mathop {\lim }\limits_{x \to {n^ + }} \,\left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi = n\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$

Now, value of the function at $x = n$ is $f(n)=0$

Since, $L.H.L.$ $ = R.H.L. = f\left( n \right)$

$\therefore$ $f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$ is continuous for every real $x.$

$\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + x + 1} \over {x + 1}} - ax - b} \right) = 4$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{{x^2} + x + 1 - a{x^2} - ax - bx - b} \over {x + 1}} = 4$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{{x^2}(1 - a) + x(1 - a - b) + (1 - b)} \over {x + 1}} = 4$

Here, we make degree of N^r = degree of D^r

$\therefore$ $1 - a = 0$

and $\mathop {\lim }\limits_{x \to \infty } {{x(1 - a - b) + (1 - b)} \over {x + 1}} = 4$

$ \Rightarrow 1 - a - b = 4$

$ \Rightarrow b = - 4$

To check differentiable at x = 0

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

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos {\pi \over h}} \right| - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} h.\,\left| {\cos {\pi \over h}} \right| = 0$

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

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos \left( { - {\pi \over h}} \right)} \right| - 0} \over { - h}}$

$ = 0$

So, f(x) is differentiable at x = 0

To check differentiability at x = 2

$R\{ f'(2)\} = \mathop {\lim }\limits_{h \to 0} {{f(2 + h) - f(2)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}\left| {\cos \left( {{\pi \over {2 + h}}} \right)} \right| - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\cos \left( {{\pi \over {2 + h}}} \right)} \over h}$

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

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

$ \Rightarrow R\left( {f'(2)} \right) = \pi $

$L\left( {f'(2)} \right) = \mathop {\lim }\limits_{h \to 0} {{f(2 - h) - f(2)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\left| {\cos {\pi \over {2 - h}}} \right| - {2^2}.\left| {\cos {\pi \over 2}} \right|} \over { - h}}$

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

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

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

$ \Rightarrow L(f'(2)) = - \pi $

Thus, f(x) is differentiable at x = 0 but not at x = 2.

We have at the points x = 2n

f(2n) = a_n + sin2n$\pi$ = a_n

Also for the L.H.L., we have

L.H.L. = $\mathop {\lim }\limits_{h \to 0} ({b_n} + \cos \pi (2n - h)) = {b_{n + 1}}$

R.H.L. = $\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin \pi (2n + h)) = {a_n}$

For continuity ${b_{n + 1}} = {a_n}$

Again at $x = 2n + 1$

L.H.L. = $\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin (\pi (2n + 1 - h))) = {a_n}$

R.H.L. = $\mathop {\lim }\limits_{h \to 0} ({b_{n + 1}} + \cos (\pi (2n + 1) - h)) = {b_{n + 1}} - 1$

Also, $f(2n + 1) = {a_n}$

For continuity we require ${b_n} + 1 = {a_n}$ $\therefore$ ${a_n} - {b_n} = 1$

Also, ${a_n} = {b_{n + 1}} - 1$

$\therefore$ ${a_{n - 1}} - {b_n} = - 1$

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

$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \left( { - h} \right)} \over { - h}} = p + 1 + 1 = p + 2$

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

$ = \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}$

and $f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}$

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

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

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

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

Thus $L.H.L. \ne R.H.L.$

Hence, $\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}\,\,$ does not exist.

Here, $\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}}$ [1^$\infty$ from]

$ \Rightarrow {e^{\mathop {\lim }\limits_{x \to 0} \{ x\log (1 + {b^2})\} \,.\,{1 \over x}}}$

$ \Rightarrow {e^{\log (1 + {b^2})}} = {(1 + b)^2}$ ..... (i)

Given,

$\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}} = 2b{\sin ^2}\theta $

$ \Rightarrow (1 + {b^2}) = 2b{\sin ^2}\theta $

$\therefore$ ${\sin ^2}\theta = {{1 + {b^2}} \over {2{b^2}}}$ ..... (ii)

By $AM \ge GM$

${{b + {1 \over b}} \over 2} \ge {\left( {b\,.\,{1 \over b}} \right)^{1/2}} \Rightarrow {{{b^2} + 1} \over {2b}} \ge 1$ ...... (iii)

From Eqs. (ii) and (iii), we get ${\sin ^2}\theta = 1$

$ \Rightarrow \theta = \pm {\pi \over 2}$ as $\theta \in ( - \pi ,\pi ]$

Set x = 0 in the functional equation to obtain

$f(0) = f(0) + f(0)$ $\therefore$ $f(0) = 0$

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

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

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

Thus, $f'(x) = \lambda $ (say). Also $f(x) = \lambda x + \mu $

As $f(x) = 0$ we have $\mu = 0$ $\therefore$ $f(x) = \lambda x$.

$\mathop {\lim }\limits_{x \to {{{\pi ^ - }} \over 2}} f(x) = 0 = f( - \pi /2)$

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

$f'(x) = \left\{ {\matrix{ { - 1,} & {x \le - \pi /2} \cr {\sin x,} & { - \pi /2 < x \le 0} \cr {1,} & {0 < x \le 1} \cr {1/x,} & {x > 1} \cr } } \right.$

Clearly, f(x) is not differentiable at x = 0 as f'(0^$-$) = 0 and f'(0⁺) = 1.

f(x) is differentiable at x = 1 as $f'({1^{ - 1}}) = f'({1^ + }) = 1$.

$f(x)$ is a positive increasing function.

$\therefore$ $0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right)$

$ \Rightarrow 0 < 1 < {{f\left( {2x} \right)} \over {f\left( x \right)}} < {{f\left( {3x} \right)} \over {f\left( x \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } 1 \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {3x} \right)} \over {f\left( x \right)}}$

By Sandwich Theorem.

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

Given that $f\left( x \right) = x\left| x \right|\,\,$ and $\,\,g\left( x \right) = \sin x$

So that go

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

$ = g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$

$ = \left\{ {\matrix{ {\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \cr {\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \cr } } \right.$

$ = \left\{ {\matrix{ { - \sin \,{x^2},} & {if\,\,\,x < 0} \cr {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \cr } } \right.$

$\therefore$ $\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ {\matrix{ { - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \cr {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \cr } } \right.$

Here we observe

$L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)$

$ \Rightarrow $ go $f$ is differentiable at $x=0$

and $\left( {go\,f} \right)'$ is continuous at $x=0$

Now $\left( {go\,f} \right)''\left( x \right) = \left\{ {\matrix{ { - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \cr {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \cr } } \right.$

Here $L\left( {gof} \right)''\left( 0 \right) = - 2$ and $R\left( {go\,f} \right)''\left( 0 \right) = 2$

As $L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)$

$ \Rightarrow go\,f\left( x \right)$ is not twice differentiable at $x=0.$

$\therefore$ Statement - $1$ is true but statement $-2$ is false.

The given limit is

$L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}}$

$ = \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}(a + \sqrt {{a^2} - {x^2}} )}} - {1 \over {4{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} {{(4 - a) - \sqrt {{a^2} - {x^2}} } \over {4{x^2}(a + \sqrt {{a^2} - {x^2}} )}}$

The numerator $\to$ 0 if $a = 2$; therefore, $L = {1 \over {64}}$.

We have $f\left( x \right) = \left\{ {\matrix{ {\left( {x - 1} \right)\sin \left( {{1 \over {x - 1}}} \right),} & {if\,\,x \ne 1} \cr {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {if\,\,x = 1} \cr } } \right.$

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

$ = \mathop {\lim }\limits_{h \to 0} {{h\,\sin {1 \over h} - 0} \over h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin {1 \over h} = a$ finite number

Let this finite number be $l$

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

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

$ = - \mathop {\lim }\limits_{h \to 0} \sin \left( {{1 \over h}} \right) = - (\,$ a finite number $\,)$ $=-l$

Thus $Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$

$\therefore$ $f$ is not differentiable at $x=1$

Also, $f'\left( 0 \right) = \sin {1 \over {\left( {x - 1} \right)}} - {{x - 1} \over {{{\left( {x - 1} \right)}^2}}}\cos {\left. {\left( {{1 \over {x - 1}}} \right)} \right]_{x = 0}}$

$ = - \sin 1 + \cos \,1$

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

As when $x \in ( - 1,1),f'(x) < 0$

So, f(x) is decreasing on ($-1,1$) at x = 1

$f''(1) = {{4a} \over {{{(a + 2)}^2}}} > 0$

So, local minima at x = 1

Given that,

$g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$ for $u \in ( - \infty ,\infty )$

$g( - u) = 2{\tan ^{ - 1}}{e^{ - u}} - {\pi \over 2} = 2{\cot ^{ - 1}}({e^u}) - {\pi \over 2}$

$ = 2\left( {{\pi \over 2} - {{\tan }^{ - 1}}({e^u}) - {\pi \over 2}} \right)$

$ = - 2{\tan ^{ - 1}}({e^u}) + {\pi \over 2}$

$ = - g(u)$

$\therefore$ $g(u) = - g(u)$

$ \Rightarrow g(u)$ is an odd function

We have, $g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$

$g'(u) = {{2{e^u}} \over {1 + 2{e^{2u}}}}$

$g'(u) > 0,\forall u \in R$ [$\because$ ${e^u} > 0$]

So, $g'(u)$ is increasing function.

Given, $g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}},0 < x < 2,m \ne 0,n > 0,m,n$ are integer.

Left hand derivative (L.H.D.) of $|x - 1|$ at $x = 1$ is P.

As $|x - 1| = \left\{ {\matrix{ {x - 1,} & {x \ge 1} \cr { - (1 - x),} & {x < 1} \cr } } \right.$

$\therefore$ L.H.D. at $x = 1$ will be $-1$.

[$\therefore$ L.H.D. at $x = 1$ is $\mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$]

$p = - 1$

Also, $\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(1 + h - 1)}^n}} \over {\log {{\cos }^m}(1 + h - 1)}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {\log {{\cos }^m}h}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {m\log \cos \,h}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{n\,.\,{h^{n - 1}}} \over {m\log \cos \,h}} = - 1$ [using L. Hospital rule]

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {{{ - n} \over m}} \right)\,.\,{{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = - 1$

$ \Rightarrow \left( {{n \over m}} \right)\mathop {\lim }\limits_{h \to 0} {{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = 1$

$n - 2 = 0$ and ${n \over m} = 1 \Rightarrow m = n = 2$

or

$[f(x) = |x - 1| = - x + 1,x < 1f'(x) = - 1$

So, L.H.D. at $x = - 1$ is $-1$

$\therefore$ $P = - 1$

Now, $\mathop {\lim }\limits_{x \to {1^ + }} {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}} = p$ .... (i)

Let $x - 1 = t$ in L.H.S. of eq. (i)

$\mathop {\lim }\limits_{t \to } {{{t^n}} \over {m\log \cos \,t}},\left( {{0 \over 0}\,\mathrm{form}} \right)$ [using L' Hospital rule]

We get, $\mathop {\lim }\limits_{t \to 0} {{n{t^{n - 1}}} \over {m\tan \,t}} = {{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{{t^{n - 1}}} \over {\tan \,t}}\left( {{0 \over 0}\,\mathrm{form}} \right)$

Again using L' Hospital rule,

${{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = P$

$ \Rightarrow {{ + n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = - 1$

For non-zero answer

$n - 2 = 0 \Rightarrow n = 2$ Also $m = n = 2$.

$f(x)$ is a non constant twice differential function such that

$f(x) = f(1 - x) \Rightarrow f'(x) = - f'(1 - x)$ .... (i)

For $x = {1 \over 2}$

We get $f'\left( {{1 \over 2}} \right) = - f'\left( {1 - {1 \over 2}} \right)$

$ = f'\left( {{1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) = 0 \Rightarrow f'\left( {{1 \over 2}} \right) = 0$

For $x = {1 \over 4}$, we get $f'\left( {{1 \over 4}} \right) = - f'\left( {{3 \over 4}} \right)$

But given that $f'\left( {{1 \over 4}} \right) = 0:f'\left( {{1 \over 4}} \right) = f'\left( {{3 \over 4}} \right) = 0$

Hence $f'(x)$ satisfies all conditions of roller's theorem for $x \in \left[ {{1 \over 4},{1 \over 2}} \right]$ and $\left[ {{1 \over 2},{3 \over 4}} \right]$, so there exist at least one point ${C_1} \in \left( {{1 \over 4},{1 \over 2}} \right)$ and at least one point ${C_2} \in \left( {{1 \over 2},{3 \over 4}} \right)$ such that

$f''(G) = 0$ and $f''{(C)_2} = 0$

$\therefore$ $f''(x)$ vanishes at least twice on [0, 1]

Also using $f(x) = f(1 - x)$

$f\left( {x + {1 \over 2}} \right) = f\left( {1 - x - {1 \over 2}} \right) = f\left( { - x + {1 \over 2}} \right)$

$f'\left( {x + {1 \over 2}} \right)$ is an odd function

$ \Rightarrow \sin x\,f\left( {x + {1 \over x}} \right)$ is an even function

$\int\limits_{{{ - 1} \over 2}}^{{1 \over 2}} {f\left( {x + {1 \over x}} \right)\sin x\,dx = 0} $

$f\left( x \right) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}$

$ \Rightarrow f\left( x \right) = x + 1\,\forall \,x \in R$



Hence, $f(x)$ is differentiable everywhere for all $x \in R.$

Given, $f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$

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

$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}$ $\left[ \, \right.$ ${0 \over 0}$ form $\left. \, \right]$

$\therefore$ using, $L$'Hospital rule

$f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {2\left( {x{e^{2x}}2 + {e^{2x}}.1} \right) + {e^{2x}}.2}}$

$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4x{e^{2x}} + 2{e^{2x}} + 2{e^{2x}}}}\,\,$ $\left[ {\,\,} \right.$ ${0 \over 0}$ form $\left. {\,\,} \right]$

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

$f(x)=2+\cos x$ is differentiable everywhere and $f'(x)=-\sin x$

$\therefore f'(c)=0 \Rightarrow c=0, \pm \pi, \pm 2 \pi, \pm 3 \pi \ldots \ldots$.

$\therefore$ for each real $t,[t, t+\pi]$ contains a value from $(0, \pm \pi, \pm 2 \pi, \pm 3 \pi \ldots)$

$\therefore$ Statement 1 is true.

$f(t+2 \pi)=2+\cos (t+2 \pi)=2+\cos t=f(t)$

So, statement 2 is true.

Therefore, $f(x)$ is a periodic function of period $2 \pi$.

$f'(x)=k e^{x}-1 < 0$ if $k \leq 0$

$\therefore f(x)$ is decreasing for all $x$.

Now, $\lim_\limits{x \rightarrow+\infty} f(x)=\lim_\limits{x \rightarrow+\infty} k e^{x}-x=-\infty$

Also, $f(k)=k\left(e^{k}-1\right) \geq 0$

$\left[\therefore k \leq 0 \Rightarrow e^{k}-1 \leq 0\right]$

$\therefore f(x)=0$ has exactly one root in $[k, \infty)$.

$\therefore \quad y=x$ meets $y=k e^{x}$ for $k \leq 0$ at one point.

Let’s call the function $ f(x) = k e^{x} - x $.

To find when it has only one root, check where its graph just touches the x-axis. This happens when $ f(x) = 0 $ and the slope is zero at the same point.

Find the derivative: $ f'(x) = k e^{x} - 1 $.

Set $ f'(x) = 0 $ to find the minimum or maximum point: $ k e^{x} - 1 = 0 $. This gives $ x = -\log k $.

Now check the second derivative: $ f''(x) = k e^{x} $.

Plugging in $ x = -\log k $, we get $ f''(-\log k) = k e^{-\log k} $.

The value of $ f''(-\ln k) = 1 $, which is greater than 0, so this is a minimum point.

The function will only touch the x-axis once if the minimum value is zero. So set $ f(-\ln k) = 0 $:

$ f(-\ln k) = k e^{-\ln k} - (-\ln k) = 1 + \ln k $.

Set $ 1 + \log k = 0 $ to get the value of $ k $:

$ k = \frac{1}{e}$.

We want the equation $ k e^{x} - x = 0 $ to have two different (distinct) solutions for $ x $.

Let’s call the function $ f(x) = k e^{x} - x $. For this equation to have two different solutions, the graph of $ f(x) $ must cross the x-axis at two points.

This will happen if the lowest point (minimum) of $ f(x) $ is below zero, or $ f(x)_{\min} < 0 $.

To find the minimum, take the derivative: $ f'(x) = k e^{x} - 1 $. Set this to zero: $ k e^{x} - 1 = 0 \implies e^{x} = \frac{1}{k} \implies x = -\log k $.

Now, plug $ x = -\log k $ back into $ f(x) $:
$ f(-\log k) = k e^{-\log k} - (-\log k) = k \cdot \frac{1}{k} + \log k = 1 + \log k $.

For two distinct roots, this minimum must be less than zero:
$ 1 + \log k < 0 $

This means $ \log k < -1 $, or $ k < \frac{1}{e} $. We also need $ k > 0 $.

So the possible values for $ k $ are $ 0 < k < \frac{1}{e} $.

Therefore, $ k \in \left(0, \frac{1}{e}\right) $.

First, factor the numerator and denominator of $f(x)$:

$f(x)=\frac{x^{2}-6x+5}{x^{2}-5x+6} = \frac{(x-5)(x-1)}{(x-2)(x-3)}$

(A) When $-1 < x < 1$:

Let's check the sign of each factor in this interval:

For $x$ between $-1$ and $1$:

• $x-5 < 0$ (negative)
• $x-1 < 0$ (negative)
• $x-2 < 0$ (negative)
• $x-3 < 0$ (negative)

So, multiplying two negatives in the numerator and two negatives in the denominator: $f(x) = \frac{(negative) \times (negative)}{(negative) \times (negative)} = \frac{positive}{positive} > 0$

Now check if $f(x) < 1$. Let's look at $f(x) - 1$: $f(x)-1 = \frac{-(x+1)}{(x-2)(x-3)}$

For $-1 < x < 1$, $(x+1) > 0$ and $(x-2), (x-3) < 0$. So numerator is negative, denominator is positive (since two negatives). Therefore, $f(x)-1 < 0 \implies f(x) < 1$.

So, for $-1 < x < 1$: $0 < f(x) < 1$. This matches option (P).

(B) When $1 < x < 2$:

• $x-5 < 0$ (negative)
• $x-1 > 0$ (positive)
• $x-2 < 0$ (negative)
• $x-3 < 0$ (negative)

So, numerator: $negative \times positive = negative$. Denominator: negative $\times$ negative = positive.

So $f(x) = \frac{negative}{positive} = negative$, i.e., $f(x) < 0$. This matches (Q).

(C) When $3 < x < 5$:

• $x-5 < 0$ (negative)
• $x-1 > 0$ (positive)
• $x-2 > 0$ (positive)
• $x-3 > 0$ (positive)

Numerator: $negative \times positive = negative$.
Denominator: $positive \times positive = positive$.

So $f(x) = \frac{negative}{positive} = negative$, so $f(x) < 0$. This matches (Q).

(D) When $x > 5$:

• $x-5 > 0$ (positive)
• $x-1 > 0$ (positive)
• $x-2 > 0$ (positive)
• $x-3 > 0$ (positive)

Numerator: $positive \times positive = positive$.
Denominator: $positive \times positive = positive$.

So $f(x) = \frac{positive}{positive} = positive$, i.e., $f(x) > 0$.

Also, let's check $f(x) < 1$ again. Use previous result: $f(x) - 1 = \frac{-(x+1)}{(x-2)(x-3)}$ For $x > 5$, numerator is negative, denominator is positive, so $f(x)-1 < 0$, so $f(x) < 1$.

Therefore, for $x > 5$, $0 < f(x) < 1$, which matches (P).

(A) $y = x|x| = \left\{ {\matrix{ { - {x^2}} & {if\,x < 0} \cr {{x^2}} & {if\,x \ge 0} \cr } } \right.$

From graph $y=x|x|$ is (continuous in ($-1,1$) (p) differentiable in ($-1,1$) (q) Strictly increasing in ($-1,1$) (r)

(B) $y = \sqrt {|x|} $

$ = \left\{ {\matrix{ {\sqrt { - x} } & {if\,x < 0} \cr {\sqrt x } & {if\,x \ge 0} \cr } } \right.$

${y^2} = - x,x < 0$ {where y can take only +ve values}

and ${y^2} = x,x \ge 0$

$\therefore$ Graph is as follows:

From graph $y = \sqrt {|x|} $ is continuous in ($-1,1$) (P) not differentiable at x = 0 (s)

(C) Note this step

$y = x + [x] = \left\{ {\matrix{ {x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {x + 1,} & {x + 1} \cr } } \right.$

$\therefore$ Graph of $y = x + [x]$ is

From graph, $y = x + [x]$ is neither continuous, nor differentiable at x = 0 and hence in ($-1,1$) (s).

(D) $ = |x - 1| + |x + 1|$

$ = \left\{ {\matrix{ { - 2x} & {x < - 1} \cr 2 & { - 1 \le x < 1} \cr {2x} & {x \ge 1} \cr } } \right.$

Graph is as follows:

From graph, $y=f(x)$ is continuous (p) differentiable (q) in ($-1,1$) but not strictly increasing in ($-1,1$).

$f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right.$

$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$

$\therefore$ $f'\left( x \right)$ exist at everywhere.

$ \begin{aligned} & \lim _{x \rightarrow 0}\left[(\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right] \\ = & \lim _{x \rightarrow 0}(\sin x)^{\frac{1}{x}}+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \end{aligned} $

Evaluate the given indeterminate form i.e,

$\quad$ let $y=\lim\limits_{x \rightarrow 0}\left(\frac{1}{x}\right) \sin x$

Taking $\ln$ on both sides

$ \begin{aligned} \ln y & =\lim _{x \rightarrow 0} \sin x \log \left(\frac{1}{x}\right) \\ & =\lim _{x \rightarrow 0} \sin x(-\log x) \\ & =\lim _{x \rightarrow 0}\left(\frac{-\log x}{\operatorname{cosec} x}\right) \end{aligned} $

Applying L-Hopital's rule.

$ \begin{aligned} \ln y & =\lim _{x \rightarrow 0} \frac{\frac{-1}{x}}{-\operatorname{cosec} x \cot x} \\ \Rightarrow \quad \ln y & =\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \tan x \end{aligned} $

$ \begin{aligned} & =1 \times 0 & \therefore & \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ & =0 . & & \lim _{x \rightarrow 0} \tan x=0 \\ \Rightarrow \quad y & =e^0=1 & & \end{aligned} $

In this graph we can observe

$ f(x)=1 $(minimum of the given curve as $x>1$ )

$f(x)=x^3$ (minimum of the curve as $x<1$ )

i.e, $f(x)=\left\{\begin{array}{rr}x^3 & x \leq 1 \\ 1 & x>1\end{array}\right.$

From graph it is clear that

$\Rightarrow f(x)$ is continuous $\forall x \in \mathbf{R}$

$\Rightarrow$ Since, the obtained $f(x)$ has sharp corner in graph at $x=1$ like $f(x)=|x|$ at $x=0$

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

Concept: Function is non differentiable if it has sharp corner because it indicates that function change its definition abruptly results multiple tangent at that pointed part of curve.

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

$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|$

$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0$

$ \Rightarrow f\left( x \right) = $ constant

As $f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0$