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

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

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

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

Q: Assertion(A): Let \${ f }({ x })\$ be twice differentiable function such that \$f^{ '' }(x)=-{ f }({ x })\$ and \$f^{ ' }(x)={ g }({ x })\$. lf \${ h }({ x })=[{ f }({ x })]^{ 2 }+[{ g }({ x })]^{ 2 }\$ and \${ h }(1)=8\$, then \${ h }(2)=8\$Reason (R): Derivative of a constant function is zero.

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

Q: Assertion (A): lf \$f(x)=\cos^{2}x+\cos^{2}\left(x+\dfrac{\pi}3\right)- \cos x \cos \left(x+\dfrac{\pi}3\right)\$ then \$f'(x)=0\$Reason(R): Derivative of constant function is zero

\$f(x)=\cos ^{2}x+\cos ^{2}(x+\cfrac{\pi }{3})-2\cos x \cos (x+\cfrac{\pi }{3})+\cos x \cos (x+\cfrac{\pi }{3})\$ \$=(\cos x - \cos (x+\cfrac{\pi }{3}))^{2} + \cos x \cos (x+\cfrac{\pi }{3}) \$ \$=(\cos x - (\cfrac{1}{2}\cos x - \cfrac{\sqrt{3}}{2} \sin x ))^{2} + \cos x \cos (x+\cfrac{\pi }{3})\$ \$=(\cfrac{\cos x }{2} + \cfrac{\sqrt{3}}{2} \sin x )^{2} + \cos x \cos (x+\cfrac{\pi }{3})\$ \$=\cfrac{\cos^2x}4 + \cfrac{3\sin^2x}4 +\cfrac{\sqrt3}2 \sin x\cos x + \cos x\left(\cfrac{\cos x}2 - \cfrac{\sqrt3}2\sin x\right)\$\$f(x)=\cfrac{3}{4}\$\$f{}'(x)=0\$

Q: lf \$ \displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}.\sin \frac{1}{x}\$ for \$x \neq 0,\ \mathrm{f}(\mathrm{0})=0\$ then?

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

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

Since, \$f\left( x \right)\$ is continuous at \$x=0\$, therefore, \$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =0\$\$\displaystyle\Rightarrow \lim _{ x\rightarrow 0 }{ { x }^{ n } } \sin { \left( \frac { 1 }{ x } \right) } =0\Rightarrow n>0\$\$f(x)\$ is differential at \$x=0\$ if \$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { f\left( x \right)-f\left( 0 \right) }{ x-0 } } \$ exists finitely\$\displaystyle \Rightarrow \lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }\sin { \left( \frac { 1 }{ x } \right) } -0 }{ x } } \$ exists finitely\$\displaystyle \Rightarrow { x }^{ n-1 }\sin { \left( \frac { 1 }{ x } \right) } \$ exists finitely.\$\Rightarrow n-1>0 \Rightarrow n>1\$If \$n\le1\$, then \$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ n-1 }\sin { \left( \frac { 1 }{ x } \right) } } \$ does not exists and hence, \$f(x)\$ is not differentiable at \$x=0\$Hence, \$f(x)\$ is continuous but not differentiable at \$x=0\$ for \$0<n\le1\$i.e., \$n\in (0,1]\$

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

Given \$y=f(x)\$ has a unique tangent at the point \$(e^a,0)\$. So, as \$x\rightarrow e^{a}\$, \$f(x)\rightarrow 0\$Now, \$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}\$\$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) } \$\$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) } \$\$=\dfrac{7-1}{3}=2\$

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

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

Q: lf \$\mathrm{f}(\mathrm{x})\$ is a quadratic expression which is positive for all real vaues of \$\mathrm{x}\$ and \$\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}'(\mathrm{x})+\mathrm{f}''(\mathrm{x})\$ then for any real value of \$\mathrm{x}\$

Let \$f(x) =ax^2+bx+c\$According to the given condition \$a>0, b^2-4ac 0\$ for any \$x \in R\$

Q: Let \$f\$ be a twice differentiable function such that \$f''\left( x \right) =-f\left( x \right) \$ and \$f'(x)=g(x).\$.If \$h'\left( x \right) ={ \left[ f\left( x \right) \right] }^{ 2 }+{ \left[ g\left( x \right) \right] }^{ 2 },h\left( 1 \right) =6\$ and \$h(0)=4\$ then \$h(4)\$ is equal to?

Given \$h'\left( x \right) ={ \left[ f\left( x \right) \right] }^{ 2 }+{ \left[ g\left( x \right) \right] }^{ 2 }\$Differentiating both side w.r.t \$x\$, we get\$h''\left( x \right) =2f\left( x \right) f'\left( x \right) +2g\left( x \right) g'\left( x \right) \\ =2f\left( x \right) g\left( x \right) +2g\left( x \right) g''\left( x \right) \quad \quad \quad \left[ \because f'\left( x \right) =g\left( x \right) \right] \\ =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0\quad \quad \quad \quad \left[ \because f''\left( x \right) =-f\left( x \right) \right] \$Thus \$h'(x)=k,\$ a constant for all \$x\in R\$.Hence \$h(x)=ax+b,\$ so that form \$h(0)=4,\$ we get \$b=4\$and from \$h(1)=6\$ we get \$a=2\$Therefore \$h(4)=12\$

Question 1

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

Accepted Answer

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

Question 2

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

Accepted Answer

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

Question 3

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

Accepted Answer

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

Question 4

Assertion (A): lf $f(x)=\cos^{2}x+\cos^{2}\left(x+\dfrac{\pi}3\right)- \cos x \cos \left(x+\dfrac{\pi}3\right)$ then $f'(x)=0$Reason(R): Derivative of constant function is zero

Accepted Answer

$f(x)=\cos ^{2}x+\cos ^{2}(x+\cfrac{\pi }{3})-2\cos   x  \cos (x+\cfrac{\pi }{3})+\cos   x  \cos (x+\cfrac{\pi }{3})$     $=(\cos  x - \cos (x+\cfrac{\pi }{3}))^{2} + \cos  x \cos (x+\cfrac{\pi }{3}) $     $=(\cos  x - (\cfrac{1}{2}\cos  x - \cfrac{\sqrt{3}}{2} \sin  x ))^{2} + \cos  x \cos  (x+\cfrac{\pi }{3})$     $=(\cfrac{\cos  x }{2} + \cfrac{\sqrt{3}}{2} \sin  x )^{2} + \cos  x \cos  (x+\cfrac{\pi }{3})$     $=\cfrac{\cos^2x}4 + \cfrac{3\sin^2x}4 +\cfrac{\sqrt3}2 \sin x\cos x + \cos x\left(\cfrac{\cos x}2 - \cfrac{\sqrt3}2\sin x\right)$$f(x)=\cfrac{3}{4}$$f{}'(x)=0$

Question 5

lf $ \displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}.\sin \frac{1}{x}$ for $x \neq 0,\ \mathrm{f}(\mathrm{0})=0$ then?

Accepted Answer

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

Question 6

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

Accepted Answer

Since, $f\left( x \right)$ is continuous at $x=0$, therefore, $\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =0$$\displaystyle\Rightarrow \lim _{ x\rightarrow 0 }{ { x }^{ n } } \sin { \left( \frac { 1 }{ x }  \right)  } =0\Rightarrow n>0$$f(x)$ is differential at $x=0$ if $\displaystyle \lim _{ x\rightarrow 0 }{ \frac { f\left( x \right)-f\left( 0 \right)  }{ x-0 }  } $ exists finitely$\displaystyle \Rightarrow \lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }\sin { \left( \frac { 1 }{ x }  \right)  } -0 }{ x }  } $ exists finitely$\displaystyle \Rightarrow { x }^{ n-1 }\sin { \left( \frac { 1 }{ x }  \right)  } $ exists finitely.$\Rightarrow n-1>0 \Rightarrow  n>1$If $n\le1$, then $\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ n-1 }\sin { \left( \frac { 1 }{ x }  \right)  }  } $ does not exists and hence, $f(x)$ is not differentiable at $x=0$Hence, $f(x)$ is continuous but not differentiable at $x=0$ for $0<n\le1$i.e., $n\in (0,1]$

Question 7

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

Accepted Answer

Given $y=f(x)$ has a unique tangent at the point $(e^a,0)$. So, as $x\rightarrow e^{a}$, $f(x)\rightarrow 0$Now, $\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}$$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) } $$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) } $$=\dfrac{7-1}{3}=2$

Question 8

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

Accepted Answer

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

Question 9

lf $\mathrm{f}(\mathrm{x})$ is a quadratic expression which is positive for all real vaues of $\mathrm{x}$ and $\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}'(\mathrm{x})+\mathrm{f}''(\mathrm{x})$ then for any real value of $\mathrm{x}$

Accepted Answer

Let $f(x) =ax^2+bx+c$According to the given condition $a>0, b^2-4ac <0$.........(i)$\therefore g(x) = ax^2+bx+c+2ax+b+2a=ax^2+(b+2a)x+(b+c+2a)$Now discriminant of $g(x)$ is $D=(b+2a)^2-4a(2a+c+b)=b^2+4a^2+4ab-4ab-4ac-8a^2=(b^2-4ac)-4a^2< 0 $ using (i)Hence $g(x)> 0$ for any $x \in R$

Question 10

Let $f$ be a twice differentiable function such that $f''\left( x \right) =-f\left( x \right) $ and $f'(x)=g(x).$.If $h'\left( x \right) ={ \left[ f\left( x \right) \right]  }^{ 2 }+{ \left[ g\left( x \right) \right]  }^{ 2 },h\left( 1 \right) =6$ and $h(0)=4$ then $h(4)$ is equal to?

Accepted Answer

Given $h'\left( x \right) ={ \left[ f\left( x \right)  \right]  }^{ 2 }+{ \left[ g\left( x \right)  \right]  }^{ 2 }$Differentiating both side w.r.t $x$, we get$h''\left( x \right) =2f\left( x \right) f'\left( x \right) +2g\left( x \right) g'\left( x \right) \ =2f\left( x \right) g\left( x \right) +2g\left( x \right) g''\left( x \right) \quad \quad \quad \left[ \because f'\left( x \right) =g\left( x \right)  \right] \ =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0\quad \quad \quad \quad \left[ \because f''\left( x \right) =-f\left( x \right)  \right] $Thus $h'(x)=k,$ a constant for all $x\in R$.Hence $h(x)=ax+b,$ so that form $h(0)=4,$ we get $b=4$and from $h(1)=6$ we get $a=2$Therefore $h(4)=12$

Limits And Derivatives

Maths
4789 Views

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

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

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

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

Assertion (A): lf $f (x) = cos^{2} x + cos^{2} (x + \frac{π}{3}) - cos x cos (x + \frac{π}{3})$ then $f^{'} (x) = 0$

Reason(R): Derivative of constant function is zero

lf $f (x) = x . sin \frac{1}{x}$ for $x \neq = 0, f (0) = 0$ then?

Let $f (x) = {x^{n} sin \frac{1}{x}, x \neq = 0 0, x = 0$ , then f(x) is continuous but not differentiable at x=0 if

The graph of the function $y = f (x)$ has a unique tangent at the point $(e^{a}, 0)$ through which the graph passes then $x \to e^{a} lim \frac{l o g _{e} { 1 + 7 f ( x ) } - s i n f ( x )}{3 f ( x )}$ is

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

lf $f (x)$ is a quadratic expression which is positive for all real vaues of $x$ and $g (x) = f (x) + f^{'} (x) + f^{''} (x)$ then for any real value of $x$

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ and $f^{'} (x) = g (x) .$ .
If $h^{'} (x) = [f (x)]^{2} + [g (x)]^{2}, h (1) = 6$ and $h (0) = 4$ then $h (4)$ is equal to?

Limits And Derivatives

Maths 4789 Views

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

lf f′(x)=g(x) and g′(x)=−f(x) for all x and f(2)=4=g(2), then f2(24)+g2(24) is

Assertion(A): Let f(x) be twice differentiable function such that f′′(x)=−f(x) and f′(x)=g(x). lf h(x)=[f(x)]2+[g(x)]2 and h(1)=8, then h(2)=8 Reason (R): Derivative of a constant function is zero.

Assertion (A): lf f(x)=cos2x+cos2(x+3π​)−cosxcos(x+3π​) then f′(x)=0 Reason(R): Derivative of constant function is zero

lf f(x)=x.sinx1​ for x=0, f(0)=0 then?

Let f(x)={xnsinx1​,x=00,x=0​ , then f(x) is continuous but not differentiable at x=0 if

The graph of the function y=f(x) has a unique tangent at the point (ea,0) through which the graph passes then x→ealim​3f(x)loge​{1+7f(x)}−sinf(x)​ is

f(x)=∣x−1∣+∣x−3∣ then f′(2)=

lf f(x) is a quadratic expression which is positive for all real vaues of x and g(x)=f(x)+f′(x)+f′′(x) then for any real value of x

Let f be a twice differentiable function such that f′′(x)=−f(x) and f′(x)=g(x)..If h′(x)=[f(x)]2+[g(x)]2,h(1)=6 and h(0)=4 then h(4) is equal to?