\displaystyle \textbf{Question 1. }\text{In which one of the following intervals is the function} \ \
\displaystyle f(x)=x^{3}-12x\text{ increasing? \hspace{6.2cm} ISC 2024} \ \
\displaystyle \text{(a) }(-2,2) \ \           \displaystyle \text{(b) }(-\infty,-2)\cup(2,\infty) \ \           \displaystyle \text{(c) }(-2,\infty) \ \           \displaystyle \text{(d) }(-\infty,2) \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) We have, } f(x)=x^{3}-12x
\displaystyle f'(x)=3x^{2}-12
\displaystyle \text{For increasing, } f'(x)>0
\displaystyle \Rightarrow 3x^{2}-12>0
\displaystyle \Rightarrow 3(x^{2}-4)>0
\displaystyle \Rightarrow 3(x-2)(x+2)>0
\displaystyle \text{Now, sign of } f'(x) \text{ is positive on } (-\infty,-2) \text{ and } (2,\infty)
\displaystyle \text{Hence, } f(x) \text{ is increasing in } (-\infty,-2)\cup(2,\infty)
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\displaystyle \textbf{Question 2. }\text{Given that }\frac{1}{y}+\frac{1}{x}=\frac{1}{12}\text{ and }y\text{ decreases at a rate of } 1\ \mathrm{cm\,s}^{-1}, \ \
\displaystyle \text{ find the rate of change of }x\text{ when }x=5\ \mathrm{cm}\text{ and } y=1\ \mathrm{cm}.\text{ \hspace{3.2cm} ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } \frac{1}{y}+\frac{1}{x}=\frac{1}{12} \qquad \cdots(i)
\displaystyle \text{and } \frac{dy}{dt}=-1 \text{ cm/s}
\displaystyle \text{On differentiating Eq. (i) w.r.t. } t, \text{ we get}
\displaystyle -\frac{1}{y^{2}}\frac{dy}{dt}-\frac{1}{x^{2}}\frac{dx}{dt}=0
\displaystyle \Rightarrow \frac{1}{y^{2}}-\frac{1}{x^{2}}\frac{dx}{dt}=0 \quad \left[\because \frac{dy}{dt}=-1\right]
\displaystyle \Rightarrow \frac{dx}{dt}=\frac{x^{2}}{y^{2}}
\displaystyle \text{When } x=5 \text{ cm and } y=1 \text{ cm, then}
\displaystyle \frac{dx}{dt}=\frac{5^{2}}{1^{2}}=25 \text{ cm/s}
\displaystyle \text{Hence, the rate of change of } x \text{, i.e. } \frac{dx}{dt}, \text{ is } 25 \text{ cm/s.}
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\displaystyle \textbf{Question 3. }\text{The interval in which the function } f(x)=5+36x-3x^{2} \ \
\displaystyle \text{ increases will be \hspace{0.2cm} ISC 2023} \ \
\displaystyle \text{(a) }(-\infty,6) \ \           \displaystyle \text{(b) }(6,\infty) \ \           \displaystyle \text{(c) }(-6,6) \ \            \displaystyle \text{(d) }(0,-6) \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } f(x)=5+36x-3x^{2}
\displaystyle f'(x)=36-6x
\displaystyle f(x) \text{ increases.}
\displaystyle \therefore f'(x)>0 \Rightarrow 36-6x>0
\displaystyle \Rightarrow x<6
\displaystyle \Rightarrow x\in(-\infty,6)
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\displaystyle \textbf{Question 4. }\text{The edge of a variable cube is increasing at the rate of } 10\ \mathrm{cm/s}.
\displaystyle \text{ How fast is the volume of the cube increasing } \text{when the edge is }5\ \mathrm{cm}\text{ long? \hspace{0.2cm} ISC 2020} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ be the edge of cube and } V \text{ be its volume.}
\displaystyle \text{Given, } \frac{dx}{dt}=10 \text{ cm/s}
\displaystyle \text{Now, } V=x^{3}
\displaystyle \Rightarrow \frac{dV}{dt}=3x^{2}\frac{dx}{dt}
\displaystyle \text{To find } \frac{dV}{dt}, \text{ when } x=5 \text{ is given by}
\displaystyle \left(\frac{dV}{dt}\right)_{x=5}=3(5)^{2}(10)
\displaystyle \Rightarrow \frac{dV}{dt}=750 \text{ cm}^{3}\text{/s}
\displaystyle \text{Thus, the volume of cube is increasing at the rate of } 750 \text{ cm}^{3}\text{/s.}
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\displaystyle \textbf{Question 5. }\text{Prove that the function }f(x)=x^{3}-6x^{2}+12x+5\text{ is } \text{increasing on }R.
\displaystyle \text{ \hspace{0.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=x^{3}-6x^{2}+12x+5
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle f'(x)=3x^{2}-12x+12
\displaystyle =3(x^{2}-4x+4)
\displaystyle =3(x-2)^{2}\geq 0
\displaystyle \therefore f'(x)\geq 0,\ \forall x\in R
\displaystyle \text{Hence, } f(x) \text{ is an increasing function.}
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\displaystyle \textbf{Question 6. }\text{A }13\ \mathrm{m}\text{ long ladder is leaning against a wall, touching the wall at a certain} 
\displaystyle \text{height from the ground level. The bottom of the ladder is pulled away from the wall,} \ \
\displaystyle \text{along the ground, at the rate of }2\ \mathrm{m/s}.\text{ How fast is the height on the wall decreasing }
\displaystyle \text{when the foot of the }  \text{ladder is }5\ \mathrm{m}\text{ away from the wall? \hspace{3.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } AC=13\text{ m be the length of ladder and } AB \text{ be the distance from wall at time } t,\ AB=x,\ BC=y.
\displaystyle \text{Given, } x=5\text{ m and } \frac{dx}{dt}=2\text{ m/s}
\displaystyle \text{In right angled } \triangle ABC,
\displaystyle AB^{2}+BC^{2}=AC^{2}
\displaystyle \Rightarrow x^{2}+y^{2}=13^{2}=169 \qquad \cdots(i)
\displaystyle \text{When } x=5,
\displaystyle 5^{2}+y^{2}=169
\displaystyle \Rightarrow y^{2}=169-25=144 \Rightarrow y=12
\displaystyle \text{Differentiating Eq. (i) w.r.t. } t,
\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
\displaystyle \Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}=0
\displaystyle \Rightarrow 5(2)+12\frac{dy}{dt}=0
\displaystyle \Rightarrow 10+12\frac{dy}{dt}=0
\displaystyle \Rightarrow \frac{dy}{dt}=-\frac{10}{12}=-\frac{5}{6}\text{ m/s}
\displaystyle \text{Hence, height of ladder is decreasing at rate } \frac{5}{6}\text{ m/s.}
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\displaystyle \textbf{Question 7. }\text{Water is dripping out from a conical funnel of semi-vertical angle } \frac{\pi}{4} \\ \text{ at the uniform rate of }2\ \mathrm{cm^3/s} \text{ in the surface, through a tiny hole at the vertex of the }
\displaystyle \text{bottom. When the slant height of the water level is }4\ \mathrm{cm}, \text{ find the rate of decrease} \\ \text{of the slant height of the water. \hspace{6.2cm} ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be the radius, } l \text{ be the slant height, } \alpha \text{ be the semi-vertical angle of the cone and } \\ S \text{ be the curved surface area of the cone.}
\displaystyle \text{Given, } \frac{dS}{dt}=-2 \text{ cm}^{2}\text{/s},\ l=4 \text{ cm and } \alpha=\frac{\pi}{4}
\displaystyle \text{In } \triangle OAB,
\displaystyle \sin\alpha=\frac{r}{l} \Rightarrow \sin\frac{\pi}{4}=\frac{r}{l} \Rightarrow r=\frac{l}{\sqrt{2}}
\displaystyle \therefore \text{Curved surface area of cone }=\pi rl
\displaystyle \text{i.e. } S=\pi rl=\frac{\pi l^{2}}{\sqrt{2}} \quad \left[\because r=\frac{l}{\sqrt{2}}\right]
\displaystyle \text{On differentiating w.r.t. } t, \text{ we get}
\displaystyle \frac{dS}{dt}=\frac{2\pi l}{\sqrt{2}}\frac{dl}{dt}
\displaystyle \Rightarrow -2=\frac{2\pi}{\sqrt{2}}(4)\frac{dl}{dt}
\displaystyle \Rightarrow \frac{dl}{dt}=-\frac{\sqrt{2}}{4\pi} \text{ cm/s}
\displaystyle \text{Hence, the water level is decreasing at the rate of } \frac{\sqrt{2}}{4\pi} \text{ cm/s.}
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\displaystyle \textbf{Question 8. }\text{A circular disc of radius }3\ \mathrm{cm} \text{ is heated. Due to expansion its radius} \\ \text{increases at the rate of }0.05\ \mathrm{cm/s}. \ \text{Find the rate at which its area increases when the} \\ \text{radius is }3.2\ \mathrm{cm}. \text{ \hspace{6.2cm} ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be the radius of circular disc and } A \text{ be the area of circular disc.}
\displaystyle \text{Also, rate of change of radius with respect to time is } 0.05 \text{ cm/s.}
\displaystyle \therefore \frac{dr}{dt}=0.05
\displaystyle \text{Now, } A=\pi r^{2}
\displaystyle \text{On differentiating w.r.t. } t, \text{ both sides, we get}
\displaystyle \frac{dA}{dt}=\pi(2r)\frac{dr}{dt}
\displaystyle \Rightarrow \frac{dA}{dt}=2\pi r(0.05)=0.1\pi r
\displaystyle \text{When radius } r=3.2 \text{ cm}
\displaystyle \frac{dA}{dt}=0.1\pi(3.2)=0.32\pi \text{ cm}^{2}\text{/s}
\displaystyle \text{Hence, the required rate of increase of the area is } 0.32\pi \text{ cm}^{2}\text{/s.}
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