APPLICATION
OF DERIVATIVES
(SUM
29 TO 33)
ALL EDUCATION BOARD
Question 29
Without using derivatives show that
`f(x)=7x-3 ` is a strictly
increasing function on `
\mathbf{R}`
Answer
Here `\quad f(x)=7 x-3`.
Let `x_1, x_2 \in \mathbf{R}`
and `x_1<x_2`.
`\therefore \quad f\left(x_1\right)`
=`7 x_1-3`and
`f\left(x_2\right)=7 x_2-3`
Now, `x_1<x_2 `
`Rightarrow 7
x_1<7 x_2 `
`Rightarrow 7
x_1-3<7 x_2-3`
`\Rightarrow \quad f\left(x_1\right)<f\left(x_2\right)`
Thus `x_1<x_2 \Rightarrow f\left(x_1\right)<f\left(x_2\right)`
Hence, `f(x)`is a strictly increasing function on `\mathbb{R}`.
Question 30
Prove that
`f(x)`=`x^{3}-3 x^{2} + 3x-100x` is increasing on `R`.
Here
`f(x)=x^5+10 x-2 `
` f^{\prime}(x)=5 x^4+10=5\left(x^4+2\right)`
Now, for `x \in \mathbf{R}, \quad x^4 \geq 0 \Rightarrow x^4+2>0`
`
`Rightarrow `
`5\left(x^4+2\right)>0 `
`Rightarrow `
`f^{\prime}(x)>0`
`
`therefore` f(x)` is a strictly increasing function on `\mathbf{R}`.
Question31 Deternine the intervals in which the following
functions are
strictly increasing or strictly decresing:
`f(x) =\frac{3}{10} x^4-\frac{4}{5}
x^3-3 x^2+\frac{36}{5} x+11 `
Answer Here
`f(x) =\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x+11 `
`\therefore \quad f^{\prime}(x)`
=`\frac{3}{10}\left(4
x^3\right)-\frac{4}{5} \cdot\left(3 x^2\right)-3(2 x)+\frac{36}{5} `
` =\frac{6}{5} x^3-\frac{12}{5} x^2-6 x+\frac{36}{5}`=`\frac{6}{5}\left(x^3-2
x^2-5 x+6\right)`
For` x=1, f^{\prime}(x)=0 \Rightarrow x=1` is a root of `f^{\prime}(x)`
Dividing (1) by `(x-1)`, we get
`f^{\prime}(x)=\frac{6}{5}(x-1)(x+2)(x-3)`
Now `f^{\prime}(x)=0 \Rightarrow x=-2,1`and 3 , which are the critical values.
These values of `x` divide the real line into four disjoint intervals
namely `(-\infty,-2),(-2,1),(1,3)`and `(3,
\infty)`.
When `x<-2`,
` \quad(x-1)<0,(x+2)<0,(x-3)<0`
` \therefore \quad f^{\prime}(x)=(-)(-)(-)=-{ve} `
`\Rightarrow \quad f(x) \text { is strictly decreasing for } x<-2 . `
`\text { When }-2<x<1, `
`\therefore \quad(x-1)<0,(x+2)>0,(x-3)<0 `
`\therefore \quad f^{\prime}(x)=(-)(+)(-)=+ \text { ve. }`
`\Rightarrow f(x)`is strictly decreasing for `x<-2.`
When `-2<x<1`,
when
`(x-1) <0,(x+2)>0,(x-3)<0 `
`\therefore \quad f^{\prime}(x) =(-)(+)(-)=+ \text { -ve. }`
`\Rightarrow f(x)` is strictly increasing for `-2<x<1`.
When `1<x<3`,
`
`(x-1) >0,(x+2)>0,(x-3)<0 `
`therefore \quad f^{\prime}(x) ` =(+)(+)(-)= -v e
`Rightarrow f(x)`is strictly decreasing for `1<x<3`.
When `x>3`,
`(x-1) >0,(x+2)>0,(x-3)>0 `
`therefore \quad f^{\prime}(x) =(+)(+)(+)=+ { -ve }`
`Rightarrow f(x)` is strictly increasing for `x>3`,
`\cos x=\sin x`
Hence, `f(x)` is strictly increasing for `-2<x<1, x>3`
ie., on `(-2,1)`, `(3, \infty)` and
strictly decreasing for `x<-2,1<x<3` ie.,
on `(-\infty,-2),(1,3)`.
Question 32. Prove that the function `x^2-x+1` is neither
increasing nor decreasing strictly on
`(-1,1)`.
Solution. Let `f(x)=x^2-x+1`
``
`\therefore \quad f^{\prime}(x)=2 x-1=2\left(x-\frac{1}{2}\right)`
``
Now, `f(x)` is strictly increasing if `f^{\prime}(x)>0`
i.e., `\quad 2\left(x-\frac{1}{2}\right)>0 \Rightarrow x-\frac{1}{2}>0
\Rightarrow x>\frac{1}{2}`
Again, `f(x)` is strictly decreasing if `f^{\prime}(x)<0`
i.e., `\quad 2\left(x-\frac{1}{2}\right)<0 \Rightarrow x-\frac{1}{2}<0
\Rightarrow x<\frac{1}{2}`
Thus `f(x)` is strictly increasing on the interval`\left(\frac{1}{2}, 1\right)`
and strictly decreasing on the interval `\left(-1, \frac{1}{2}\right)`.
Hence, the given function is neither increasing nor
decreasing strictly on the
entire interval `(-1,1)`.
Question 33.
Find the least value of `a` such that the function `x^2+a x+1`
is increasing on `(1,2)`.
Solution. Let `f(x)=x^2+a x+1`
`\therefore \quad f^{\prime}(x)=2 x+a`
Now, `\quad 1<x<2 \Rightarrow 2<2 x<4 \Rightarrow 2+a<2
x+a<4+a`
`\Rightarrow \quad 2+a<f^{\prime}(x)<4+a`
For `f(x)` to be increasing, we have `f^{\prime}(x) \geq 0`
Now, `\quad f^{\prime}(x) \geq 0 \Rightarrow 2+a \geq 0 \Rightarrow a \geq-2`
Hence, the least value of `a` such that the function `x^2+a x+1`
is increasing on `(1,2)` is - 2 .
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