Application of Derivatives Test - 10

Let, small charge in \(x\) be \(\Delta x\) and the corresponding change in \(y\) is \(\Delta y\).

Therefore, \(\Delta y=\frac{d y}{d x} \Delta x\)

We have to find the value of \(\sqrt{24.99}\)

Let, \(x+\Delta x=24.99=25-0.01\)

Therefore, \(x=25\) and \(\Delta x=-0.01\)

Assume, \(y=x^{\frac{1}{2}}\)

Differentiating with respect to \(x\), we get:

\(\frac{ dy }{ dx }=\frac{1}{2} x ^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}}\)

At \(x=25\)

\(\left[\frac{ dy }{ dx }\right]_{ x =25}=\frac{1}{10}\) and \(y =(25)^{\frac{1}{2}}=5\)

As we know \(\Delta y =\frac{ dy }{ dx } \Delta x\)

So, \(\Delta y =\frac{1}{10} \times(-0.01)=-0.001\)

Therefore, approximate value of \(\sqrt{24.99}=(24.99)^{\frac{1}{2}}= y +\Delta y =5-0.001=4.999\)

Hence, the correct option is (A).

Let, small change in \(x\) be \(\Delta x\) and the corresponding change in \(y\) is \(\Delta y\).

Therefore, \(\Delta y =\frac{ dy }{ dx } \Delta x\)

We have to find the value of \(\sqrt{36.01}\)

Let, \(x+\Delta x=36.01=36+0.01\)

Therefore, \(x=36\) and \(\Delta x=-0.01\)

Assume, \(y=x^{\frac{1}{2}}\)

Differentiating with respect to \(x\), we get:

\(\frac{ dy }{ dx }=\frac{1}{2} x ^{-\frac{1}{2}}=\frac{1}{2 \sqrt{ x }}\)

At \(x=36\)

\(\left[\frac{ dy }{ dx }\right]_{ x =36}=\frac{1}{12}\) and \(y =(36)^{\frac{1}{2}}=6\)

As we know \(\Delta y=\frac{d y}{d x} \Delta x\)

So, \(\Delta y =\frac{1}{12} \times(0.01)=0.000833\)

Therefore, approximate value of \(\sqrt{36.01}=(36.01)^{\frac{1}{2}}= y +\Delta y =6+0.00083=6.00083\)

Hence, the correct option is (C).

Given: Let \(f(x)=\frac{\operatorname{ln} x}{x}\)

Differentiating both sides, we get:

\(f ^{\prime}( x )=\frac{ d }{ dx }\left(\frac{\operatorname{ln} x }{ x }\right)\)

\(=\frac{\frac{1}{x}( x )-1(\ln x )}{ x ^{2}}\)

\(\Rightarrow f ^{\prime}( x )=\frac{1-\ln x }{ x ^{2}}\)

\(f ^{\prime \prime}( x )=\frac{ d }{ dx }\left( f ^{\prime}( x )\right)\)

\(=\frac{ d }{ dx }\left(\frac{1-\ln x }{ x ^{2}}\right)\)

\(=\frac{ x ^{2} \frac{d}{d x}(1-\ln x )-(1-\ln x ) \frac{ d }{d x}\left( x ^{2}\right)}{\left( x ^{2}\right)^{2}}\)

\(=\frac{ x ^{2}\left(-\frac{1}{x}\right)-(1-\ln x )(2 x )}{ x ^{4}}\)

\(=\frac{- x -2 x +2 x (\ln x )}{ x ^{4}}\)

\(=\frac{ x (-3+2(\ln x ))}{ x ^{4}}\)

\(\Rightarrow f ^{\prime \prime}( x )=\frac{-(3-2 \ln x )}{ x ^{3}}\)

To find the value of \(x\),

\(f ^{\prime}( x )=0\)

\(\Rightarrow \frac{1-ln x }{ x ^{2}}=0\)

\(\Rightarrow 1-\ln x =0\)

\(\Rightarrow \ln x =1\)

\(\Rightarrow x = e ^{1}= e \quad\left(\ln _{ a } b = c \Rightarrow b = a ^{ c }\right)\)

Now, at \(x=e\),

\(f ^{\prime \prime}( e )=\frac{-(3-2 \ln e )}{ e ^{3}}\)

\(=\frac{-3+2}{ e ^{3}}\)

\(=\frac{-1}{ e ^{3}}<0\)

At \(x=e\), maximum value of \(f(x)\) obtain

\(\therefore f ( x = e )=\frac{\operatorname{ln} x }{ x }\)

\(=\frac{\ln e }{ e }\)

\(=\frac{1}{ e } \quad(\because \ln e =1)\)

Hence, the correct option is (B).

We know that the area of a circle of radius units is given by \(A =\pi r ^{2}\) sq.units.

\(\therefore \frac{ dA }{ dt }=\frac{ d }{ dt }\left(\pi r ^{2}\right)=2 \pi r \left(\frac{ dr }{ dt }\right)\)

Now, \(\left[\frac{ dA }{ dt }\right]_{ r =4}=8 \pi(0.01)\) m\(^{2}/\) sec \(=0.08\) mm\(^{2}/\)sec

Hence, the correct option is (C).

Let, the radius of a sphere is r meters.

The surface area of Sphere \((A)=4 \pi r^{2}\)

Given: Radius of a sphere \(=6 m\) and \(\Delta r =0.02 m\)

Now, the volume of the sphere is given by,

\(A=4 \pi r^{2}\)

Differentiating with respect to \(r\), we get:

\(\frac{ d S }{ dr }=8 \pi r\)

\(\Rightarrow dS =8 mr \cdot dr\)

\(\Rightarrow dS =8 \pi \cdot 6 \cdot(0.02)\)

\(\Rightarrow dS =0.96\) square meters

Hence, the correct option is (B).

Let, \(f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+1\) be a function defined on the interval \([1,4]\)

First we have to find \(f^{\prime}(x)\)

\(f^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48\)

Now let's find the roots of the equation \(f^{\prime}(x)=0\)

\(12 x^{3}-24 x^{2}+24 x-48=0\)

\(\Rightarrow 12 \cdot\left(x^{3}-2 x^{2}+2 x-4\right)=0\)

\(\Rightarrow 12 \cdot\left(x^{2}+2\right) \cdot(x-2)=0\)

\(\Rightarrow x-2=0 \quad \quad \quad\left[\because\left(x^{2}+2\right) \neq 0\right.\) as we are dealing with real valued functions]

\(\Rightarrow x =2\)

Now let's find out \(f^{\prime \prime}(x)\) i.e., \(\frac{d^{2}(f(x))}{d x^{2}}\)

\( f^{\prime \prime}(x)=36 x^{2}-48 x+24\)

Now evaluate the value of \(f^{\prime \prime}(x)\) at \(x=2\), we get:

\(f^{\prime \prime}(2)=72>0\)

As we know that according to second derivative test if \(f^{\prime \prime}(c)>0\) then \(x=c\) is a point of local minima

So, \(x=2\) is a point of local minima

So \(f(2)=-63\)

Let's calculate \(f (0)\) and \(f (3)\)

\(f (0)=1 \text { and } f (3)=-8\)

As we know that if a function is defined in the closed interval \([a, b]\) then the minimum value of \(f ( x )\) on \([ a , b ]\) is the smallest of \(m , f ( a )\) and \(f ( b )\).

So, the minimum value of the given function on the interval \([0,3]\) is \(-63\).

Hence, the correct option is (D).

The slope of the tangent to the curve:

The slope of curve \(y = f ( x )\) at some point \(P \left( x _{1}, y _{1}\right)\) means the slope of the tangent to the curve at \(P \left( x _{1}, y _{1}\right)\).

The slope is given as \(m\), where \(m =\frac{ dy }{ dx }\) at \(\left( x _{1}, y _{1}\right)\).

Given: \(2 x^{3}+3 y=2 y^{3}+3 x\)

Differentiating with respect to \(x\)

\(2\left(3 x^{2}\right)+3 \frac{d y}{d x}=2\left(3 y^{2} \frac{d y}{d x}\right)+3(1)\)

\(6 x^{2}+3 \frac{d y}{d x}=6 y^{2} \frac{d y}{d x}+3\)

Taking \(\frac{ dy }{ dx }\) on one side, we get:

\(6 x^{2}-3=\left(6 y^{2}-3\right) \frac{d y}{d x}\)

\(\frac{d y}{d x}=\frac{6 x^{2}-3}{6 y^{2}-3}\)

Hence, the correct option is (B).

Let, small charge in \(x\) be \(\Delta x\) and the corresponding change in \(y\) is \(\Delta y\).

\(\Delta y =\frac{ dy }{ dx } \Delta x = f ^{\prime}( x ) \Delta x\)

Now that \(\Delta y = f ( x +\Delta x )- f ( x )\)

Therefore, \(f(x+\Delta x)=f(x)+\Delta y\)

Given: \(f(x)=3 x^{2}+3\)

Let, \(x+\Delta x=3.01=3+0.01\)

Therefore, \(x=3\) and \(\Delta x=0.01\)

\(f (x +\Delta x)= f (x)+\Delta y\)

\(\Rightarrow f (x +\Delta x)= f ( x )+ f ^{\prime}( x ) \Delta x\)

\(\Rightarrow f (3.01)=3 x ^{2}+3+(6 x ) \Delta x\)

\(\Rightarrow f (3.01)=3(3)^{2}+3+(6 \cdot 3)(0.01)\)

\(\Rightarrow f (3.01)=30+0.18\)

\(\Rightarrow f (3.01)=30.18\)

Hence, the correct option is (A).

Given diameter of spherical balloon \(D =\frac{3}{2}(4 x +3)\)

Radius \(R =\frac{ D }{2}=\frac{3}{4}(4 x +3)\)

Rate of change of radius \(R\) wrt \(x =\frac{ d R }{ dx }=\frac{ d }{ dx }\left[\frac{3}{4}(4 x +3)\right]\)

\(\frac{ dR }{ dx }=\frac{3}{4} \times \frac{ d }{ dx }(4 x +3)\)

\(\frac{ d R }{ dx }=\frac{3}{4} \times 4=3\)

Volume of the spherical balloon \(V=\frac{4 \pi}{3} \times R^{3}\)

Rate of change of radius \(V\) wrt \(x =\frac{ d V }{ dx }=\frac{ dV }{ dR } \times \frac{ dR }{ dx }\)

\(\frac{ d V }{ dx }=\frac{ d }{ dR }\left[\frac{4 \pi}{3} \times R ^{3}\right] \times \frac{ dR }{ dx }\)

\(\frac{ d V }{ dx }=\frac{4 \pi}{3} \times 3 R ^{2} \times 3=12 \pi R ^{2}\)

Hence, the correct option is (A).

Given: Equation of curve is \(y=\sqrt{5 x-3}-2\) and the tangent to the curve \(y=\sqrt{5 x-3}-2\) is parallel to the line \(4 x-2 y+3=0\)

The given line \(4 x -2 y +3=0\) can be re-written as:

\(y =2 x +\left(\frac{3}{2}\right)=0\)

Now by comparing the above equation of line with \(y = m x + c\) we get,

\(m =2\) and \(c =\frac{3}{2}\)

\(\because\) The line \(4 x-2 y+3=0\) is parallel to the tangent to the curve \(y=\sqrt{5 x-3}-2\)

As we know that if two lines are parallel then their slope is same.

So, the slope of the tangent to the curve \(y=\sqrt{5 x-3}-2\) is \(m =2\)

Let, the point of contact be \(\left( x _{1}, y _{1}\right)\)

As we know that slope of the tangent at any point say \(\left(x_{1}, y_{1}\right)\) to a curve is given by:

\(m=\left[\frac{d y}{d x}\right]_{\left(x_{1}, y_{1}\right)}\)

\(\Rightarrow \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{5 x-3}} \cdot 5-0=\frac{5}{2 \sqrt{5 x-3}}\)

\(\Rightarrow\left[\frac{d y}{d x}\right]_{\left(x_{1}, y_{1}\right)}=\frac{5}{2 \sqrt{5 x_{1}-3}}\)

\(\because\) Slope of tangent to the curve \(y=\sqrt{5 x-3}-2\) is \(m =2\)

\(\Rightarrow 2=\frac{5}{2 \sqrt{5 x_{1}-3}}\)

By squaring both the sides of the above equation we get:

\(4=\frac{25}{4 \cdot\left(5 x_{1}-3\right)}\)

\(\Rightarrow x_{1}=\frac{73}{80}\)

\(\because\left( x _{1}, y _{1}\right)\) is point of conctact i.e., \(\left( x _{1}, y _{1}\right)\) will satisfy the equation of curve:

\(y=\sqrt{5 x-3}-2\)

\(\Rightarrow y_{1}=\sqrt{5 x_{1}-3}-2\)

By substituting \(x_{1}=\frac{73}{80}\) in the above equation we get:

\(y _{1}=-\frac{3}{4}\)

So, the point of contact is: \(\left(\frac{73}{80},-\frac{3}{4}\right)\)

As we know that equation of tangent at any point say \(\left(x_{1}, y_{1}\right)\) is given by:

\(y-y_{1}=\left[\frac{d y}{d x}\right]_{\left(x_{1}, y_{1}\right)} \cdot\left(x-x_{1}\right)\)

\(\Rightarrow y+\frac{3}{4}=2 \cdot\left(x-\frac{73}{80}\right)\)

\(\Rightarrow 80 x-40 y-103=0\)

So, the equation of tangent to the given curve at the point \(\left(\frac{73}{80},-\frac{3}{4}\right)\) is \(80 x -40 y-103=0\)

Hence, the correct option is (D).

Given: \(f(x)\) is an increasing function and \(g(x)\) is a decreasing function

If \(f^{\prime}(x)>0\) then the function is said to be increasing.

If \(f ^{\prime}( x )<0\) then the function is said to be decreasing.

\(\therefore f ^{\prime}( x )>0\) and \(g ^{\prime}( x )<0\)

Let, \(h(x)=\operatorname{gof}(x)=g(f(x))\)

Differentiating with respect to \(x\), we get

\(h^{\prime}(x)=g^{\prime}(f(x)) \times f^{\prime}(x)\)

We know, \(f^{\prime}(x)>0\) and \(g^{\prime}(x)<0\)

Therefore, \(h^{\prime}(x)=(\)Negative\()\times(\)Positive\()=\) Negative

\(\therefore h ^{\prime}( x )<0\)

So, \(\operatorname{gof}( x )\) is decreasing function.

Hence, the correct option is (B).

Given: \(y = x ^{2}\) and \(x=1\)

As we know that equation of tangent is \(\frac{y-f(a)}{x-a}=f^{\prime}(a)\) for the function \(y = f ( x )\) at \(x\) \(=a\)

First find \(f ( a )\) and \(f ^{\prime}( a )\)

\(f(1)=1\)

\(\Rightarrow f^{\prime}(x)=2 x \text { and } f^{\prime}(1)=2\)

As we know that equation of tangent is \(\frac{y-f(a)}{x-a}=f^{\prime}(a)\) for the function \(y = f ( x )\) at \(x\) \(=a\)

\(\frac{y-1}{x-1}=2\)

\(\Rightarrow y=2 x-1\)

So, the equation of tangent is \(y =2 x -1\).

Hence, the correct option is (A).

If \(f^{\prime}(x) \geq 0\) at each point in an interval, then the function is said to be increasing.

Given, \(f(x)=2 x^{3}-15 x^{2}+36 x+12\)

Differentiating, we get:

\(f^{\prime}(x)=6 x^{2}-30 x+36\)

\(f' ( x )\) is increasing function:

\(f^{\prime}(x) \geq 0\)

\(=6 x^{2}-30 x+36 \geq 0\)

\(=x^{2}-5 x+6 \geq 0\)

\(=(x-2)(x-3) \geq 0\)

Thus, \(x \in(-\infty, 2] \cup[3, \infty)\)

The interval in which the function \(f (x)=2 x^{3}-15 x^{2}+36 x+10\) is increasing in \((-\infty, 2] \cup\) \([3, \infty)\)

Hence, the correct option is (C).

Let, \(f(x)=\sin 2 x \cdot \cos 2 x\)

\(\Rightarrow f^{\prime}(x)=2 \cos ^{2} 2 x-2 \sin ^{2} 2 x\)

\(\Rightarrow f^{\prime}(x)=2 \cdot\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)\)

As we know that, \(\cos 2 x=\cos ^{2} x-\sin ^{2} x\)

\(f^{\prime}(x)=2 \cos 4 x\)

If \(f^{\prime}(x)=0\) then \(2 \cos 4 x=0\)

\(\Rightarrow x=\frac{\pi}{8}\)

\(\Rightarrow f^{\prime \prime}(x)=-8 \sin 4 x\)

\(\Rightarrow f^{\prime \prime}(x)=-8<0\)

So, \(x=\frac{\pi}{8}\) is the point of maxima.

So, the maximum value of \(f(x)=\sin 2 x \cdot \cos 2 x\) is given by \(f\left(\frac{\pi}{8}\right)=\sin\left(\frac{\pi}{4}\right) \cdot \cos \left(\frac{\pi}{4}\right)\) \(=\frac{1}{2}\)

Hence, the correct option is (A).

Given that the edge length \(( L )\) of the square changes at the rate \(=\frac{ dL }{ dt }=2\) cm/s

Area of the square \(A=L^{2}\)

Rate of change of area \(A =\frac{ dA }{ dt }=\frac{ dA }{ d L } \times \frac{ dL }{ dt }\)

\(\frac{ dA }{ dt }=\frac{ d }{ dL }\left[ L ^{2}\right] \times \frac{ d L }{ dt }\)

\(\frac{ dA }{ dt }=2 L \times 2=4 L\)

As, L \(=12\) cm

\(\frac{ dA }{ dt }=4 \times 12\)

\(\frac{ d A }{ dt }=48\) cm\(^{2} /\)s

Hence, the correct option is (D).

Given: \(f(x)=\log (\sin x)\)

Let's examine the function \(f ( x )\) on the interval \(\left(0, \frac{\pi}{2}\right)\)

So, first let's calculate \(f ^{\prime}( x )\)

\(f^{\prime}(x)=\frac{1}{\sin x} \cdot \frac{d(\sin x)}{d x}=\cot x\)

As we know that \(\cot x>0 \forall x \in \left(0, \frac{\pi}{2}\right)\)

\(f^{\prime}(x)>0 \forall x \in\left(0, \frac{\pi}{2}\right)\)

As we know that for a strictly increasing function \(f^{\prime}(x)>0\) for all \(x \in(a, b)\) So, the given function \(f ( x )\) is a strictly increasing function on \(\left(0, \frac{\pi}{2}\right)\)

\(\therefore\) Option (A) is true

Now let's the function \(f ( x )\) on the interval \(\left(0, \frac{\pi}{2}\right)\)

As we know that, \(f^{\prime}(x)=\cot x\) and \(\cot x<0 \forall x \in\left(0, \frac{\pi}{2}\right)\)

\(f^{\prime}(x)<0 \forall x \in\left(0, \frac{\pi}{2}\right)\)

As we know that for a strictly decreasing function \(f^{\prime}(x)<0\) for all \(x \in(a, b)\) So, the given function \(f ( x )\) is a strictly decreasing function on \(\left(\frac{\pi}{2}, \pi\right)\)

\(\therefore\) Option (B) is true.

Hence, the correct option is (D).

If \(f^{\prime}(x)>0\) at each point in an interval then the function is said to be increasing.

If \(f ^{\prime}( x )<0\) at each point in an interval I, then the function is said to be decreasing.

Here, derivative of \(e^{x}\) is \(e^{x}\).

and it is greater than zero at any interval

\(\therefore e^{ x }\) is an increasing function.

Hence, the correct option is (A).

The slope of the tangent to the curve \(=\frac{d y}{d x}\)

The slope of normal to the curve \(=\frac{-1}{\left(\frac{ dy }{ dx }\right)}\)

Point-slope is the general form: \(y-y_{1}=m\left(x-x_{1}\right)\), Where \(m=\) slope

Here, \(y=3 x^{2}+1\)

\(\frac{ dy }{ dx }=6 x\)

\(\left.\frac{ dy }{ dx }\right|_{ x =2}=12\)

Slope of normal to the curve \(=\frac{-1}{\left(\frac{d y}{d x}\right)}=\frac{-1}{12}\)

Equation of normal to curve passing through \((2,13)\) is:

\(y-13=\frac{-1}{12}(x-2)\)

\(\Rightarrow 12 y-156=-x+2\)

\(\Rightarrow x+12 y-158=0\)

Hence, the correct option is (D).

Given: Equation of line is \(y=4 x+c\) and the line is tangent to the circle \(x^{2}+y^{2}=9\) Here, we have to find the value of \(c\).

As we know that, if the line \(y=m x+c\) is a tangent to the circle \(x^{2}+y^{2}=a^{2}\) then \(c^{2}=\) \(a^{2}\left(1+m^{2}\right)\)

By comparing the given equation of line and circle with \(y = m x + c\) and \(x ^{2}+ y ^{2}=\) \(a ^{2}\) respectively, we get:

\(m =4, a ^{2}=9\)

\(\because m =4\)

\(\Rightarrow m ^{2}=16\)

\(\because c ^{2}= a ^{2}\left(1+ m ^{2}\right)\)

By substituting \(m ^{2}=16\) and \(a ^{2}=9\) in the above equation we get,

\(\Rightarrow c^{2}=9 \times(1+16)=9 \times 17\)

\(\Rightarrow c=\pm 3 \sqrt{17}\)

Hence, the correct option is (A).

Let, small charge in \(x\) be \(\Delta x\) and the corresponding change in \(y\) is \(\Delta y\).

Therefore, \(\Delta y =\frac{ dy }{ dx } \Delta x\)

We have to find an approximate value of \((1.04)^{\frac{1}{4}}\)

Let, \(x+\Delta x=1.04=1+0.04\)

Therefore, \(x=1\) and \(\Delta x=0.04\)

Assume, \(y = x ^{\frac{1}{4}}\)

Differentiating with respect to \(x\), we get

\(\frac{ dy }{ dx }=\frac{1}{4} x ^{-\frac{3}{4}}\)

At \(x=1\)

\(\left[\frac{ dy }{ dx }\right]_{ x =1}=\frac{1}{4}\) and \(y =1^{\frac{1}{4}}=1\)

As we know \(\Delta y=\frac{d y}{d x} \Delta x\)

So, \(\Delta y =\frac{1}{4} \times 0.04=0.01\)

Therefore, approximate value of \((1.04)^{\frac{1}{4}}=y+\Delta y=1+0.01=1.01\)

Hence, the correct option is (A).

Let, small charge in \(x\) be \(\Delta x\) and the corresponding change in \(y\) is \(\Delta y\).

Therefore, \(\Delta y =\frac{ dy }{ dx } \Delta x\)

We have to find the value of \((242)^{\frac{1}{5}}\)

Let, \(x+\Delta x=242=243-1\)

Therefore, \(x=243\) and \(\Delta x=-1\)

Assume, \(y = x ^{\frac{1}{5}}\)

Differentiating with respect to \(x\), we get:

\(\frac{ dy }{ dx }=\frac{1}{5} x ^{-\frac{4}{5}}=\frac{1}{5( x )^{\frac{4}{5}}}\)

At \(x=243\)

\(\left[\frac{ dy }{ dx }\right]_{ x =243}=\frac{1}{405}\) and \(y =(243)^{\frac{1}{5}}=3\)

As we know \(\Delta y =\frac{ dy }{ dx } \Delta x\)

So, \(\Delta y=\frac{1}{405} \times(-1)=0.0024\)

Therefore, approximate value of \((242)^{\frac{1}{5}}= y +\Delta y =3-0.0024=2.997\)

Hence, the correct option is (A).

Following steps to finding maxima and minima using derivatives.

Find the derivative of the function.

Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.

Now we have find second derivative.

\(f'(x)\) is less than 0 then the given function is said to be maxima

If \(f'(x)\) Is greater than 0 then the function is said to be minima

 Let, \(f(x)=\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)\)

Differentiate with respect to \(x\), we get:

\(f^{\prime}( x )=\cos \left( x +\frac{\pi}{6}\right)-\sin \left( x +\frac{\pi}{6}\right)\)

Again differentiate with respect to \(x\),

\(f ^{\prime \prime}( x )=-\sin \left( x +\frac{\pi}{6}\right)-\cos \left( x +\frac{\pi}{6}\right)=-\left[\sin \left( x +\frac{\pi}{6}\right)+\cos \left( x +\frac{\pi}{6}\right)\right]\)

For maximum value \(f ^{\prime}( x )=0\)

\(\cos \left( x +\frac{\pi}{6}\right)-\sin \left( x +\frac{\pi}{6}\right)=0\)

\(\Rightarrow \cos \left( x +\frac{\pi}{6}\right)=\sin \left( x +\frac{\pi}{6}\right)\)

\(\Rightarrow \tan \left( x +\frac{\pi}{6}\right)=1\)

\(\therefore x +\left(\frac{\pi}{6}\right)=\left(\frac{\pi}{4}\right)\)

\(\Rightarrow x =\frac{\pi}{12}\)

At \(x=\frac{\pi}{12}, f^{\prime \prime}(x)<0\)

So, function is maximum at \(x =\frac{\pi}{12}\)

Hence, the correct option is (A).

Rate of change of '\(x\)' is given by \(\frac{d x}{d t}\)

Given that, \(y=2 x-x^{2}\) and \(\frac{d x}{d t}=3\) units\(/\)sec

Then, the slope of the curve, \(\frac{d y}{d x}=2-2 x=m\)

\(\frac{ dm }{ dt }=0 - 2 \times \frac{ dx }{ dt }\)

\(=-2(3)\)

\(=-6\) units per second

Thus, the slope of the curve is decreasing at the rate of 6 units per second when \(x\) is increasing at the rate of 3 units per second.

Hence, the correct option is (B).

Let, the pairs of positive numbers with sum 24 be: \(x\) and \(24-x\).

Then, let \(f ( x )\) denotes the product of such pairs

i.e., \(f(x)=x \times(24-x)=24 x-x^{2}\)

Here we have to find that pair of number for which \(f ( x )\) is maximum.

First we have to find \(f ^{\prime}( x )\)

\(f^{\prime}(x)=24-2 x\)

Now let's find the roots of the equation \(f^{\prime}(x)=0\)

\(2 x=24\)

\(\Rightarrow x=12\)

Now let's find out \(f^{\prime \prime}(x)\) i.e., \(\frac{d^{2}(f(x))}{d x^{2}}\)

\(f^{\prime \prime}(x)=-2\)

Now evaluate the value of \(f ^{\prime \prime}( x )\) at \(x =12\)

\(f^{\prime \prime}(12)=-2<0\)

As we know that according to second derivative test if \(f^{\prime \prime}(c)<0\) then \(x=c\) is a point of maxima

So, \(x=12\) is a point of maxima

So, when \(x=12\) then \(24-x=12\)

Thus, the required numbers are 12 and 12

Hence, the correct option is (B).

Given:

\(f(x)=X^{5}-20 X^{3}+240 X\)

Differentiate w.r.t \(x\),

\(f^{\prime}(x)=5 X^{4}-60 X^{2}+240=X^{4}-12 X^{2}+48\)

For checking whether the function is Monotonically increasing or decreasing put \(X_{1}=-1\)

\(f^{\prime}\left(X_{1}\right)=(-1)^{4}-12(-1)^{2}+48=1-12+48=37\)

Now, put \(X_{2}=0\)

\(f^{\prime}\left(X_{2}\right)=0-0+48=48\)

Therefore, \(f^{\prime}\left(X_{1}\right) \leq f^{\prime}\left(X_{2}\right)\)

From the above condition we can assume that given function is Monotonically increasing everywhere.

Hence, the correct option is (D).

Given: \(f(x)=\log (1+x)-\frac{x}{(1+x)}\)

Here we have to find the interval in which \(f ( x )\) is decreasing.

Let's first calculate \(f ^{\prime}( x )\)

As we know that \(\frac{d}{d x}\{f(x) \pm g(x)\}=\frac{d\{f(x)\}}{d x} \pm \frac{d\{g(x)\}}{d x}\),

\(\frac{d(\log x)}{d x}=\frac{1}{x}\), for \(x>0\) and

\(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x) \cdot \frac{d(f(x))}{d x}-f(x) \cdot \frac{d g(g))}{d x}}{[g(x)]^{2}}\)

\(\Rightarrow f^{\prime}(x)=\left\{\frac{1}{1+x}-\frac{(1+x) \cdot 1-x \cdot 1}{\left(1+x^{2}\right)}\right\}=\frac{x}{(1+x)^{2}}\)

As we know that for a decreasing function say \(f(x)\) we have \(f^{\prime}(x) \leq 0\)

\( \frac{x}{(1+x)^{2}} \leq 0\)

\(\Rightarrow x \leq 0 \ldots \ldots\left(\because(1+x)^{2} \geq 0\right)\)

So, the given function is decreasing for the interval \((-\infty, 0]\)

Hence, the correct option is (B).

Following steps to finding maxima and minima using derivatives.

Find the derivative of the function. Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points. Now we have to find the second derivative.

\(f'(x)\) is less than 0 then the given function is said to be maxima

If \(f'(x)\) Is greater than 0 then the function is said to be minima

Let, \(f(x)=x^{3}-12 x^{2}+k x-8\)

Differentiating with respect to \(x\), we get:

\(f^{\prime}(x)=3 x^{2}-24 x+k\)

It is given that function attains its maximum value of the interval \([0,3]\) at \(x=2\)

\(\therefore f^{\prime}(2)=0\)

\(3 \times 2^{2}-(24 \times 2)+k=0\)

\(\therefore k=36\)

Hence, the correct option is (C).

Given, at any instant \(t\), for a sphere, \(r\) denotes the radius, S denotes the surface area. Surface area \(S=4 \pi r^{2}\)

Differentiating w.r.to \(t\), we get:

\(\frac{ dS}{ dt }=4 \pi \cdot 2 r \frac{ dr }{ dt }\)

\(\Rightarrow \frac{ dS }{ dt }=8 \pi r \frac{ dr }{ dt }\)

\(\Rightarrow \frac{ dr }{ dt }=\frac{1}{8 \pi r } \frac{ ds }{ dt } \quad \quad \ldots..\) (1)

Volume of sphere \(V =\frac{4}{3} \pi r ^{3}\)

Differentiating w.r.to \(t\), we get:

\(\frac{ dV }{ dt }=\frac{4}{3} \pi \cdot 3 r ^{2} \frac{ dr }{ dt }\)

\(\Rightarrow \frac{ dV }{ dt }=4 \pi r ^{2} \frac{ dr }{ dt }\)

From equation (1), we have

\(\frac{ dV }{ dt }=4 \pi r ^{2} \frac{1}{8 \pi r } \frac{ dS }{ dt }\)

\(\Rightarrow \frac{ dV }{ dt }=\frac{ r }{2} \frac{ dS }{ dt }\)

Thus, if at any instant \(t\), for a sphere, \(r\) denotes the radius, \(S\) denotes the surface area and \(V\) denotes the volume, then \(\frac{ dV }{ dt }=\frac{1}{2} r \frac{ d S }{ d t }\)

Hence, the correct option is (B).

Slope of the curve \(=\frac{dy}{dx}\)

Given: Equation of the curve \(y=x^{2} \quad \quad\ldots\) (1)

Let's find Slope of tangent at any point on curve \((x, y)\)

\(y=x^{2}\)

Differentiating with respect to \(x\), we get:

\(\frac{ dy }{ dx }=2 x\)

According to question, Slope of the tangent \(=y\)-coordinate of the point

\(2 x=y\)

\(\Rightarrow 2 x=x^{2}\)

\(\Rightarrow x^{2}-2 x=0\)

\(\Rightarrow x(x-2)=0\)

\(\therefore x=0 \text { or } 2\)

Put the value of \(x\) in equation \(1^{\text{st}}\), we get:

\(y =0 \text { or } 4\)

Therefore, the required points are \((0,0)\) and \((2,4)\)

Hence, the correct option is (C).

Minimum and maximum value of cos x is -1 and 1

-1 ≤ cos x ≤ 1

So, the minimum and maximum value of cos 2x is -1 and 1

-1 ≤ cos 2x ≤ 1

Maximum value of  |cos 2x + 7|

Here, cos 2x = 1

|1+7|=|8|=8" role="presentation">|1+7| = |8| = 8

Minimum value of  |cos 2x + 7|

Here, cos 2x = -1

|(−1)+7|=|6|=6" role="presentation">|(−1)+7| = |6| = 6

Maximum value = 8, Minimum value = 6

Hence, the correct option is (C).