85279
Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are
1 \((12,8)\)
2 \((15,5)\)
3 \((10,10)\)
4 \((2,18)\)
Explanation:
(B) : Let the first part \(=\mathrm{y}\) Then other part \(=(20-y)\) \(f(y)=(20-y) y^{3}\) or \(f(y)=20 y^{3}-y^{4}\) For maximum or minimum put \(\mathrm{f}^{\prime}(\mathrm{y})=0\) \(\mathrm{f}^{\prime}(\mathrm{y})=60 \mathrm{y}^{2}-4 \mathrm{y}^{3}=0\) \(4 \mathrm{y}^{2}(15-\mathrm{y})=0\) \(\mathrm{y}=0\) or \(\mathrm{y}=15\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2}\) \(\therefore \mathrm{f}(\mathrm{y})\) is maximum at \(\mathrm{y}=15\) Hence, 20 can be divided in 15 and 5
COMEDK-2012
Application of Derivatives
85280
The function \(f(x)=\cot ^{-1} x+x\) increases in the interval
1 \((1, \infty)\)
2 \((-1, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
(D) : Given function, \(f(x)=\cot ^{-1} x+x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^{2}}+1\) \(f^{\prime}(x)=\frac{x^{2}}{1+x^{2}} \geq 0\). for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing on \((-\infty, \infty)\).
COMEDK-2016
Application of Derivatives
85282
The function \(f(x)=x^{3}-6 x^{2}+12 x-16, x \in \mathrm{R}\) is
1 increasing for all \(x \in \mathrm{R}\)
2 decreasing for all \(x \in \mathrm{R}\)
3 increasing for all \(x \in(-1, \infty)\)
4 decreasing for all \(\mathrm{x} \in(2, \infty)\)
Explanation:
(A) : Given, \(f(x)=x^{3}-6 x^{2}+12 x-16\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-12 \mathrm{x}+12=3(\mathrm{x}-2)^{2}>0\) for all \(x \in R, x \neq 2\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing for all \(\mathrm{x} \in \mathrm{R}\).
85279
Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are
1 \((12,8)\)
2 \((15,5)\)
3 \((10,10)\)
4 \((2,18)\)
Explanation:
(B) : Let the first part \(=\mathrm{y}\) Then other part \(=(20-y)\) \(f(y)=(20-y) y^{3}\) or \(f(y)=20 y^{3}-y^{4}\) For maximum or minimum put \(\mathrm{f}^{\prime}(\mathrm{y})=0\) \(\mathrm{f}^{\prime}(\mathrm{y})=60 \mathrm{y}^{2}-4 \mathrm{y}^{3}=0\) \(4 \mathrm{y}^{2}(15-\mathrm{y})=0\) \(\mathrm{y}=0\) or \(\mathrm{y}=15\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2}\) \(\therefore \mathrm{f}(\mathrm{y})\) is maximum at \(\mathrm{y}=15\) Hence, 20 can be divided in 15 and 5
COMEDK-2012
Application of Derivatives
85280
The function \(f(x)=\cot ^{-1} x+x\) increases in the interval
1 \((1, \infty)\)
2 \((-1, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
(D) : Given function, \(f(x)=\cot ^{-1} x+x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^{2}}+1\) \(f^{\prime}(x)=\frac{x^{2}}{1+x^{2}} \geq 0\). for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing on \((-\infty, \infty)\).
COMEDK-2016
Application of Derivatives
85282
The function \(f(x)=x^{3}-6 x^{2}+12 x-16, x \in \mathrm{R}\) is
1 increasing for all \(x \in \mathrm{R}\)
2 decreasing for all \(x \in \mathrm{R}\)
3 increasing for all \(x \in(-1, \infty)\)
4 decreasing for all \(\mathrm{x} \in(2, \infty)\)
Explanation:
(A) : Given, \(f(x)=x^{3}-6 x^{2}+12 x-16\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-12 \mathrm{x}+12=3(\mathrm{x}-2)^{2}>0\) for all \(x \in R, x \neq 2\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing for all \(\mathrm{x} \in \mathrm{R}\).
85279
Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are
1 \((12,8)\)
2 \((15,5)\)
3 \((10,10)\)
4 \((2,18)\)
Explanation:
(B) : Let the first part \(=\mathrm{y}\) Then other part \(=(20-y)\) \(f(y)=(20-y) y^{3}\) or \(f(y)=20 y^{3}-y^{4}\) For maximum or minimum put \(\mathrm{f}^{\prime}(\mathrm{y})=0\) \(\mathrm{f}^{\prime}(\mathrm{y})=60 \mathrm{y}^{2}-4 \mathrm{y}^{3}=0\) \(4 \mathrm{y}^{2}(15-\mathrm{y})=0\) \(\mathrm{y}=0\) or \(\mathrm{y}=15\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2}\) \(\therefore \mathrm{f}(\mathrm{y})\) is maximum at \(\mathrm{y}=15\) Hence, 20 can be divided in 15 and 5
COMEDK-2012
Application of Derivatives
85280
The function \(f(x)=\cot ^{-1} x+x\) increases in the interval
1 \((1, \infty)\)
2 \((-1, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
(D) : Given function, \(f(x)=\cot ^{-1} x+x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^{2}}+1\) \(f^{\prime}(x)=\frac{x^{2}}{1+x^{2}} \geq 0\). for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing on \((-\infty, \infty)\).
COMEDK-2016
Application of Derivatives
85282
The function \(f(x)=x^{3}-6 x^{2}+12 x-16, x \in \mathrm{R}\) is
1 increasing for all \(x \in \mathrm{R}\)
2 decreasing for all \(x \in \mathrm{R}\)
3 increasing for all \(x \in(-1, \infty)\)
4 decreasing for all \(\mathrm{x} \in(2, \infty)\)
Explanation:
(A) : Given, \(f(x)=x^{3}-6 x^{2}+12 x-16\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-12 \mathrm{x}+12=3(\mathrm{x}-2)^{2}>0\) for all \(x \in R, x \neq 2\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing for all \(\mathrm{x} \in \mathrm{R}\).
85279
Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are
1 \((12,8)\)
2 \((15,5)\)
3 \((10,10)\)
4 \((2,18)\)
Explanation:
(B) : Let the first part \(=\mathrm{y}\) Then other part \(=(20-y)\) \(f(y)=(20-y) y^{3}\) or \(f(y)=20 y^{3}-y^{4}\) For maximum or minimum put \(\mathrm{f}^{\prime}(\mathrm{y})=0\) \(\mathrm{f}^{\prime}(\mathrm{y})=60 \mathrm{y}^{2}-4 \mathrm{y}^{3}=0\) \(4 \mathrm{y}^{2}(15-\mathrm{y})=0\) \(\mathrm{y}=0\) or \(\mathrm{y}=15\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{y})=120 \mathrm{y}-12 \mathrm{y}^{2}\) \(\therefore \mathrm{f}(\mathrm{y})\) is maximum at \(\mathrm{y}=15\) Hence, 20 can be divided in 15 and 5
COMEDK-2012
Application of Derivatives
85280
The function \(f(x)=\cot ^{-1} x+x\) increases in the interval
1 \((1, \infty)\)
2 \((-1, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
(D) : Given function, \(f(x)=\cot ^{-1} x+x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^{2}}+1\) \(f^{\prime}(x)=\frac{x^{2}}{1+x^{2}} \geq 0\). for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing on \((-\infty, \infty)\).
COMEDK-2016
Application of Derivatives
85282
The function \(f(x)=x^{3}-6 x^{2}+12 x-16, x \in \mathrm{R}\) is
1 increasing for all \(x \in \mathrm{R}\)
2 decreasing for all \(x \in \mathrm{R}\)
3 increasing for all \(x \in(-1, \infty)\)
4 decreasing for all \(\mathrm{x} \in(2, \infty)\)
Explanation:
(A) : Given, \(f(x)=x^{3}-6 x^{2}+12 x-16\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-12 \mathrm{x}+12=3(\mathrm{x}-2)^{2}>0\) for all \(x \in R, x \neq 2\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) for all \(\mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing for all \(\mathrm{x} \in \mathrm{R}\).
