If f(x)=sin left(cos ^{-1}left(frac{1-2^{2 x} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \sin \left(\cos ^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)$.Let $2^x = t$. Then $f(x) = \sin(\cos^{-1}(\frac{1-t^2}{1+t^2}))

Vedclass · Accepted Answer

$481$

Vedclass · Answer

(B) Given $f(x) = \sin \left(\cos ^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}ight)ight)$.Let $2^x = t$. Then $f(x) = \sin(\cos^{-1}(\frac{1-t^2}{1+t^2}))$.We know that $\cos^{-1}(\frac{1-t^2}{1+t^2}) = 2	an^{-1}(t)$ for $t \ge 0$.So,$f(x) = \sin(2	an^{-1}(2^x))$.Using the identity $\sin(2	heta) = \frac{2	an	heta}{1+	an^2	heta}$,where $	heta = 	an^{-1}(2^x)$,we get $	an	heta = 2^x$.Thus,$f(x) = \frac{2(2^x)}{1+(2^x)^2} = \frac{2 \cdot 2^x}{1+2^{2x}}$.Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:$f'(x) = \frac{(1+2^{2x}) \cdot \frac{d}{dx}(2 \cdot 2^x) - (2 \cdot 2^x) \cdot \frac{d}{dx}(1+2^{2x})}{(1+2^{2x})^2}$.$f'(x) = \frac{(1+2^{2x})(2 \cdot 2^x \ln 2) - (2 \cdot 2^x)(2^{2x} \ln 2 \cdot 2)}{(1+2^{2x})^2}$.At $x=1$,$2^x = 2$ and $2^{2x} = 4$.$f'(1) = \frac{(1+4)(2 \cdot 2 \ln 2) - (2 \cdot 2)(4 \ln 2 \cdot 2)}{(1+4)^2} = \frac{5(4 \ln 2) - 4(8 \ln 2)}{25} = \frac{20 \ln 2 - 32 \ln 2}{25} = -\frac{12}{25} \ln 2$.Comparing with $-\frac{b}{a} \ln 2$,we get $b=12$ and $a=25$.Therefore,$|a^2 - b^2| = |25^2 - 12^2| = |625 - 144| = 481$.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

$(a)$ $x\|x\|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{\|x\|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $\|x-1\|+\|x+1\|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

If $f(x)=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x}}{1+2^{2 x}}\right)\right)$ and its first derivative with respect to $x$ is $-\frac{ b }{ a } \log _{ e } 2$ when $x =1,$ where $a$ and $b$ are integers,then the minimum value of $\left| a ^{2}- b ^{2}\right|$ is.........

Similar Questions

Let $f$ and $g$ be twice differentiable functions on $R$ such that
$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$
$f^{\prime}(1)=4, g^{\prime}(1)=3$
$f(2)=12, g(2)=4$
Then which of the following is $NOT$ true?

If $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$,where $f^{\prime \prime}(x) = -f(x)$ and $g(x) = f^{\prime}(x)$,and given $F(5) = 5$,then $F(10)$ is equal to

Let $f: ( -\infty, \infty ) \to ( -\infty, \infty )$ be defined by $f(x) = x^3 + 1$.
Statement $1$: The function $f$ has a local extremum at $x = 0$.
Statement $2$: The function $f$ is continuous and differentiable on $( -\infty, \infty )$ and $f'(0) = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module