Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a thrice differentiable odd function satisfying $f^{\prime}(x) \geq 0, f^{\prime\prime}(x)=f(x), f(0)=0, f^{\prime}(0)=3$. Then $9 f\left(\log _e 3\right)$ is equal to __________ .
Explanation:
$\begin{aligned} &f^{\prime}(x) \geq 0, f^{\prime \prime}(x)=f(x)\\ &\text { Second order differential equation } \end{aligned}$
$\begin{aligned} & f(x)=A e^x+B e^{-x} \\ & f(0)=0 \Rightarrow A=-B \\ & \Rightarrow f(x)=A\left(e^x-e^{-x}\right) \\ & f^{\prime}(x)=A e^x+A e^{-x}=A\left(e^x+e^{-x}\right) \\ & f^{\prime}(0)=3=A\left(e^0+e^{-0}\right)=2 A \Rightarrow A=\frac{3}{2} \\ & f(x)=\frac{3}{2}\left(e^x-e^{-x}\right) \\ & \text { If }(\ln 3)=\frac{27}{2}\left(e^{\ln 3}-e^{-\ln 3}\right)=\frac{27}{2}\left(3-\frac{1}{3}\right)=\frac{27}{2} \cdot \frac{8}{3} \\ & =36 \end{aligned}$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is __________.
Explanation:
$\because f: R \rightarrow R \text { and } f(0)=0, f(1)=1, f(2)=-1 \text {, }$
$f(3)=2$ and $f(4)=-2$ then
$f(x)$ has atleast 4 real roots.
Then $f(x)$ has atleast 3 real roots and $f^{\prime}(x)$ has atleast 2 real roots.
Now we know that
$\begin{aligned} \frac{d}{d x}\left(f^3 \cdot f^{\prime \prime}\right) & =3 f^2 \cdot f^{\prime} \cdot f^{\prime \prime}+f^3 \cdot f^{\prime \prime \prime} \\ & =f^2\left(3 f^{\prime} \cdot f^{\prime}+f \cdot f^{\prime \prime}\right) \end{aligned}$
Here $f^3 \cdot f'$ has atleast 6 roots.
Then its differentiation has atleast 5 distinct roots.
Explanation:
$\begin{aligned} & y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x \\\\ & y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\\\ & =\frac{32-9+12}{32}=\frac{35}{32} \\\\ & \therefore 96 y^{\prime}\left(\frac{\pi}{6}\right)=105\end{aligned}$
Explanation:
$\begin{aligned} & f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\ & f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\ & f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\ & f^{\prime \prime \prime}(x)=6 \\ & f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(3)=6 \\ & f(x)=x^3+x^2 \cdot(-5)+x \cdot(2)+6 \\ & f^{\prime}(x)=3 x^2-10 x+2 \\ & f^{\prime}(10)=300-100+2=202 \end{aligned}$
Let $f(x)=\sum_\limits{k=1}^{10} k x^{k}, x \in \mathbb{R}$. If $2 f(2)+f^{\prime}(2)=119(2)^{\mathrm{n}}+1$ then $\mathrm{n}$ is equal to ___________
Explanation:
$ \begin{aligned} & f(x)=x+2 x^2+\ldots \ldots \ldots+10 x^{10} \\\\ & f(x) . x=x^2+2 x^3+\ldots \ldots \ldots+9 x^{10}+10 x^{11} \\\\ & f(x)(1-x)=x+x^2+x^3+\ldots \ldots \ldots+x^{10}-10 x^{11} \\\\ & \therefore f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{10 x^{11}}{(1-x)} \end{aligned} $
$ \Rightarrow $ $ f(x)=\frac{x-x^{11}-10 x^{11}+10 x^{12}}{(1-x)^2} = \frac{10 x^{12}-11 x^{11}+x}{(1-x)^2} $
So,
$ \begin{aligned} & f(2)=2\left(1-2^{10}\right)+10 \cdot 2^{11} \\\\ & =2+18 \cdot 2^{10} \end{aligned} $
$f'\left( x \right) = {{{{\left( {1 - x} \right)}^2}\left( {120{x^{11}} - 121{x^{10}} + 1} \right) + 2\left( {1 - x} \right)\left( {10{x^{12}} - {{11.x}^{11}} + 2} \right)} \over {{{\left( {1 - x} \right)}^4}}}$
So,
$f'\left( x \right) = {{1\left( {{{120.2}^{11}} - {{121.2}^{10}} + 1} \right) + 2\left( { - 1} \right)\left( {{{10.2}^{12}} - {{11.2}^{11}} + 2} \right)} \over {{{\left( { - 1} \right)}^4}}}$
= ${2^{10}}\left( {83} \right) - 3$
Hence $2 f(2)+f^{\prime}(2)=119.2^{10}+1$
$ \Rightarrow \text { So, } \mathrm{n}=10 $
If $f(x)=x^{2}+g^{\prime}(1) x+g^{\prime \prime}(2)$ and $g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$, then the value of $f(4)-g(4)$ is equal to ____________.
Explanation:
$\Rightarrow f(x)=x^{2}+a x+b$
Now, $f(1)=1+a+b ; f^{\prime}(x)=2 x+a ; f^{\prime \prime}(x)=2$
$g(x)=(1+a+b) x^{2}+x(2 x+a)+2$
$\Rightarrow g(x)=(a+b+3) x^{2}+a x+2$
$\Rightarrow g^{\prime}(x)=2 x(a+b+3)+a \Rightarrow g^{\prime}(1)=2(a+b+3)$+a=a$
$\Rightarrow a+b+3=0$ .......(1)
$g^{\prime \prime}(x)=2(a+b+3)=b$
$\Rightarrow 2 a+b+6=0$ ........(2)
Solving (i) and (ii), we get
$a=-3$ and $b=0$
$f(x)=x^{2}-3 x$ and $g(x)=-3 x+2$
$f(4)=4$ and $g(4)=-12+2=-10$
$\Rightarrow f(4)-g(4)=16-2=14$
Let $f^{1}(x)=\frac{3 x+2}{2 x+3}, x \in \mathbf{R}-\left\{\frac{-3}{2}\right\}$ For $\mathrm{n} \geq 2$, define $f^{\mathrm{n}}(x)=f^{1} \mathrm{o} f^{\mathrm{n}-1}(x)$. If $f^{5}(x)=\frac{\mathrm{a} x+\mathrm{b}}{\mathrm{b} x+\mathrm{a}}, \operatorname{gcd}(\mathrm{a}, \mathrm{b})=1$, then $\mathrm{a}+\mathrm{b}$ is equal to ____________.
Explanation:
$f'(x) = {{3x + 2} \over {2x + 3}}x \in R - \left\{ { - {3 \over 2}} \right\}$
${f^5}(x) = {f_o}{f_o}{f_o}{f_o}f(x)$
${f_o}f(x) = {{13x + 12} \over {12x + 13}}$
${f_o}{f_o}{f_o}{f_o}f(x) = {{1563x + 1562} \over {1562x + 1563}}$
$ \equiv {{ax + b} \over {bx + a}}$
$\therefore$ $a = 1563,b = 1562$
$ = 3125$
Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function that satisfies the relation $f(x+y)=f(x)+f(y)-1,\forall x,y\in\mathbb{R}$. If $f'(0)=2$, then $|f(-2)|$ is equal to ___________.
Explanation:
$ \begin{aligned} & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\\\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\\\ & f^{\prime}(x)=2 \Rightarrow d y=2 d x \\\\ & y=2 x+C \\\\ & \mathrm{x}=0, \mathrm{y}=1, \mathrm{c}=1 \\\\ & \mathrm{y}=2 \mathrm{x}+1 \\\\ & |f(-2)|=|-4+1|=|-3|=3 \end{aligned} $
For the curve $C:\left(x^{2}+y^{2}-3\right)+\left(x^{2}-y^{2}-1\right)^{5}=0$, the value of $3 y^{\prime}-y^{3} y^{\prime \prime}$, at the point $(\alpha, \alpha)$, $\alpha>0$, on C, is equal to ____________.
Explanation:
$\because$ $C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$ for point ($\alpha$, $\alpha$)
${\alpha ^2} + {\alpha ^2} - 3 + {({\alpha ^2} - {\alpha ^2} - 1)^5} = 0$
$\therefore$ $\alpha = \sqrt 2 $
On differentiating $({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$ we get
$x + yy' + 5{({x^2} - {y^2} - 1)^4}(x - yy') = 0$ ...... (i)
When $x = y = \sqrt 2 $ then $y' = {3 \over 2}$
Again on differentiating eq. (i) we get :
$1 + {(y')^2} + yy'' + 20({x^2} - {y^2} - 1)(2x - 2yy')(x - y'y) + 5{({x^2} - {y^2} - 1)^4}(1 - y{'^2} - yy'') = 0$
For $x = y = \sqrt 2 $ and $y' = {3 \over 2}$ we get $y'' = - {{23} \over {4\sqrt 2 }}$
$\therefore$ $3y' - {y^3}y'' = 3\,.\,{3 \over 2} - {\left( {\sqrt 2 } \right)^3}\,.\,\left( { - {{23} \over {4\sqrt 2 }}} \right) = 16$
Let f and g be twice differentiable even functions on ($-$2, 2) such that $f\left( {{1 \over 4}} \right) = 0$, $f\left( {{1 \over 2}} \right) = 0$, $f(1) = 1$ and $g\left( {{3 \over 4}} \right) = 0$, $g(1) = 2$. Then, the minimum number of solutions of $f(x)g''(x) + f'(x)g'(x) = 0$ in $( - 2,2)$ is equal to ________.
Explanation:
As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd
and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 .
So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes
$\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$,
has minimum 4 zeroes
If $y(x) = {\left( {{x^x}} \right)^x},\,x > 0$, then ${{{d^2}x} \over {d{y^2}}} + 20$ at x = 1 is equal to ____________.
Explanation:
$\because$ $y(x) = {\left( {{x^x}} \right)^x}$
$\therefore$ $y = {x^{{x^2}}}$
$\therefore$ ${{dy} \over {dx}} = {x^2}\,.\,{x^{{x^2} - 1}} + {x^{{x^2}}}\ln x\,.\,2x$
$\therefore$ ${{dx} \over {dy}} = {1 \over {{x^{{x^2} + 1}}(1 + 2\ln x)}}$ ..... (i)
Now, ${{{d^2}x} \over {dx^2}} = {d \over {dx}}\left( {{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 1}}} \right)\,.\,{{dx} \over {dy}}$
$ = {{ - x{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 2}}\,.\,{x^{{x^2}}}(1 + 2\ln x)({x^2} + 2{x^2}\ln x + 3)} \over {{x^{{x^2}}}(1 + 2\ln x)}}$
$ = {{ - {x^{{x^2}}}(1 + 2\ln x)({x^3} + 3 + 2{x^2}\ln x)} \over {{{\left( {{x^{{x^2}}}(1 + 2\ln x)} \right)}^3}}}$
${{{d^2}x} \over {d{y^2}(at\,x = 1)}} = - 4$
$\therefore$ ${{{d^2}x} \over {d{y^2}(at\,x = 1)}} + 20 = 16$
Let f : R $\to$ R satisfy $f(x + y) = {2^x}f(y) + {4^y}f(x)$, $\forall$x, y $\in$ R. If f(2) = 3, then $14.\,{{f'(4)} \over {f'(2)}}$ is equal to ____________.
Explanation:
$\because$ $f(x + y) = {2^x}f(y) + {4^y}f(x)$ ....... (1)
Now, $f(y + x){2^y}f(x) + {4^x}f(y)$ ...... (2)
$\therefore$ ${2^x}f(y) + {4^y}f(x) = {2^y}f(x) + {4^x}f(y)$
$({4^y} - {2^y})f(x) = ({4^x} - {2^x})f(y)$
${{f(x)} \over {{4^x} - {2^x}}} = {{f(y)} \over {{4^y} - {2^y}}} = k$
$\therefore$ $f(x) = k({4^x} - {2^x})$
$\because$ $f(2) = 3$ then $k = {1 \over 4}$
$\therefore$ $f(x) = {{{4^x} - {2^x}} \over 4}$
$\therefore$ $f'(x) = {{{4^x}\ln 4 - {2^x}\ln 2} \over 4}$
$f'(x) = {{({{2.4}^x} - {2^x})ln2} \over 4}$
$\therefore$ ${{f'(4)} \over {f'(2)}} = {{2.256 - 16} \over {2.16 - 4}}$
$\therefore$ $14{{f'(4)} \over {f'(2)}} = 248$
Explanation:
x + y = e4xy
$ \Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$
At x = 0
${{dy} \over {dx}} = 3$
${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$
At x = 0, ${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$
$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$
Explanation:
$ = \sin (2{\tan ^{ - 1}}{2^x})$
$f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$
$ \therefore $ $f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$
$ \Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$
$ = - {{12} \over {25}}{\log _e}2$
$ \Rightarrow a = 25,b = 12$
$|{a^2} - {b^2}| = |625 - 144| = 481$
then ${{dy} \over {dx}}$ at x = 0 is _______.
Explanation:
$ \therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx$
$ = \cos \alpha .\cos kx - \sin \alpha .\sin kx$
$ = \cos \left( {\alpha + kx} \right)$
So, $y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\alpha + kx} \right)} \right)} $
$ = \sum\limits_{k = 1}^6 {\left( {{k^2}x + kx} \right)} $
$ \Rightarrow {{dy} \over {dx}} = \sum\limits_{k = 1}^6 {{k^2}} $
$ = {{6 \times 7 \times 13} \over 6} = 91$
Then the value of ${d \over {d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)$ is
Explanation:
$\sin \left( {{{\tan }^{ - 1}}\left( {{{\sin \theta } \over {\sqrt {\cos 2\theta } }}} \right)} \right)$, where $\theta \in \left( { - {\pi \over 4},{\pi \over 4}} \right)$
$\sin \left( {{{\tan }^{ - 1}}\left( {{{\sin \theta } \over {\sqrt {2{{\cos }^2}\theta - 1} }}} \right)} \right) = \sin ({\sin ^{ - 1}}(\tan \theta )) = \tan \theta $
${{d(\tan \theta )} \over {d(\tan \theta )}} = 1$
function such that $\left| {f\left( x \right)} \right| \le 1$ and $f'(x)=g(x).$
If ${f^2}\left( 0 \right) + {g^2}\left( 0 \right) = 9.$ Prove that there exists some $c \in \left( { - 3,3} \right)$
such that $g(c).g''(c)<0.$
If $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)| \leq 1$ and $f'(x)=g(x)$, where,$f^{2}(0)+g^{2}(0)=9$ then prove that there exists some $c \in(-3,3)$ such that $g(c) \circ g^{n}(c) < 0$.
Explanation:
$f(x)$ is differentiable function and $g(x)$ is double differentiable function.
let $f(x) = {{{x^3}} \over 3} + 9,g(x) = {x^2}$
$f'(x) = {x^2}$
So, the function satisfies the criteria $f'(x) = g(x)$
Also, the two function
${f^2}(0) = 0 + 9 = 9$
${g^2}(0) = 0$
$ \Rightarrow {f^2}(0) + {g^2}(0) = 9$
$|f(x)| \le 1$
$ \Rightarrow \left| {{{{x^3}} \over 3} + 9} \right| \le 1$
$ \Rightarrow - 1 \le {{{x^3}} \over 3} + 9 \le 1$
$ \Rightarrow - 10 \le {{{x^3}} \over 3} \le - 8$
$ \Rightarrow - 30 \le {x^3} \le - 24$
$ \Rightarrow {( - 30)^{{1 \over 3}}} \le x \le {( - 24)^{{1 \over 3}}}$
$ \Rightarrow ( - 3.10) \le x \le - 2.88$
Now, $g(c) = {c^2}$
$g'(x) = 2x$
$g''(x) = 2$
It is given $g(c).~g''(c) < 0$
$\Rightarrow 2c^2 < 0$
$\Rightarrow c < 0$
As $x$ lies between $-2.88$ to $-3.10$
So, $c$ lies between $[-3,3]$
prove that ${{y'} \over y} = {1 \over x}\left( {{a \over {a - x}} + {b \over {b - x}} + {c \over {c - x}}} \right)$.
${\left( {\sin y} \right)^{\sin \left( {{\pi \over 2}x} \right)}} + {{\sqrt 3 } \over 2}{\sec ^{ - 1}}\left( {2x} \right) + {2^x}\tan \left( {In\left( {x + 2} \right)} \right) = 0$
that $\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$
then show that $\left| {\matrix{ {A\left( x \right)} & {B\left( x \right)} & {C\left( x \right)} \cr {A\left( \alpha \right)} & {B\left( \alpha \right)} & {C\left( \alpha \right)} \cr {A'\left( \alpha \right)} & {B'\left( \alpha \right)} & {C'\left( \alpha \right)} \cr } } \right|$ is
divisible by $f(x)$, where prime denotes the derivatives.
$f''\left( x \right) = - f\left( x \right),$ and $f'\left( x \right) = g\left( x \right),h\left( x \right) = {\left[ {f\left( x \right)} \right]^2} + {\left[ {g\left( x \right)} \right]^2}$
Find $h\left( {10} \right)$ if $h(5)=11$
at $x=1$
and $F\left( x \right) = \left| {\matrix{ {{f_1}\left( x \right)} & {{f_2}\left( x \right)} & {{f_3}\left( x \right)} \cr {{g_1}\left( x \right)} & {{g_2}\left( x \right)} & {{g_3}\left( x \right)} \cr {{h_1}\left( x \right)} & {{h_2}\left( x \right)} & {{h_3}\left( x \right)} \cr } } \right|$ then $F'\left( x \right)$ at $x = a$ is ...........