Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blanks:
The equation of normal to the curve y = tan x at (0, 0) is ________.

Answer

The equation of normal to the curve y = tan x at (0, 0) is x + y = 0.
Solution:
We have, y = tan x
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\sec^2\text{x}$
The slope of the curve y = tan x at (0, 0) is given by
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\sec^20=1$
Now, we know that, the product of the slopes of two perpendicular lines is -1
$\therefore$ 1 × Slope of normal = -1
⇒ Slope of normal = -1
Therefore, the equation of normal to the curve y = tan x at (0, 0) is
y - 0 = -1(x - 0)
⇒ x + y = 0
Therefore, the equation of normal to the curve y = tan x at (0, 0) is x + y = 0.

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Question 21 Mark

Fill in the blanks:
The values of a for which the function f(x) = sinx - ax + b increases on R are ______.

Answer

The values of a for which the function f(x) = sinx - ax + b increases on R are $(-\infty-1).$
Solution:
The values of a for which the function f(x) = sinx - ax + b increases on R are $(-\infty-1).$
$\because\ \text{f}'(\text{x})=\cos\text{x}-\text{a}$
and $\text{f}'(\text{x})>0\Rightarrow\cos\text{x}>\text{a}$
Since, $\cos\text{x}\in[-1,1]$
$\Rightarrow\ \text{a}<-1\Rightarrow\text{a}\in(-\infty,-1)$

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Question 31 Mark

Fill in the blanks:
The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval _______.

Answer

The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval $(1,\infty).$
Solution:
We have, $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4}$
$\Rightarrow\ \text{f}'(\text{x})=\frac{\text{x}^4\cdot4\text{x}-(2\text{x}^2-1)\cdot4\text{x}^3}{\text{x}^8}$ $\bigg[\because\Big(\frac{\text{f}}{\text{g}}\Big)'=\frac{\text{gf}'-\text{fg}'}{\text{g}^2}\bigg]$
$=\frac{4\text{x}^5-8\text{x}^5+4\text{x}^3}{\text{x}^8}$
$=\frac{-4\text{x}^5+4\text{x}^3}{\text{x}^8}$
$=\frac{4\text{x}^3(-\text{x}^2+1)}{\text{x}^8}$
For decreasing, f'(x) < 0
$\Rightarrow\ \frac{4\text{x}^3(1-\text{x}^2)}{\text{x}^8}<0$
$\Rightarrow\ 1-\text{x}^2<0$
$\Rightarrow\ \text{x}^2>1$
$\Rightarrow\ \text{x}>\pm1$
Since, x > 0
Hence, $\text{x}\in(1,\infty)$
Therefore, the function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},$ where x > 0, decreases in the interval $(1,\infty).$

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Question 41 Mark

Fill in the blanks:
The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is ______.

Answer

The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is $2\sqrt{\text{ab}}$
Solution:
$\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{x}^2}$
$\text{f(x)}=0$
$\Rightarrow\ \text{a}=\frac{\text{b}}{\text{x}^2}$ or $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}\ \ (\text{as x}>0)$
Now, $\text{f}''(\text{x})=\frac{2\text{b}}{\text{x}^3}>0$ for $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$
Thus least value of f(x) is $\text{f}\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\text{a}\cdot\sqrt{\frac{\text{b}}{\text{a}}}+\frac{\text{b}}{\sqrt{\frac{\text{b}}{\text{a}}}}$ $=\sqrt{\text{ab}}+\sqrt{\text{ab}}=2\sqrt{\text{ab}}$

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Question 51 Mark

Fill in the blanks:
The curves $y = 4x^2+ 2x - 8$ and $y = x^3- x + 13$ touch each other at the point _____.

Answer

The curves $y = 4x^2+ 2x - 8$ and $y = x^3- x + 13$ touch each other at the point (3, 34).
Solution:
Given, equation of curves are $y = 4x^2+ 2x - 5$ and $y - x^3- x + 13$
For first curve $\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=8\text{x}+2$
For the secqnd curve $\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=3\text{x}^2-1$
When both curve touch, the slope of both curves should be same
$\therefore\ 8\text{x}+2=3\text{x}^2-1$
$\Rightarrow\ 3\text{x}^2-8\text{x}-3=0$
$\Rightarrow\ (3\text{x}+1)(\text{x}-3)=0$
$\therefore\ \text{x}=-\frac{1}{2}$ and $ \text{x} = 3$
Now consider $x = 3$
For first curve $y(3) = 4(3)^2+ 2(3) - 5 = 37$
For second curve $y(3) = (3)^3- 3 + 13 = 37$
Thus at (3, 37), both curves touch.
Now, $\text{x}=-\frac{1}{3},$
For first curve $\text{y}\Big(-\frac{1}{3}\Big)=4\Big(-\frac{1}{3}\Big)^2+2\Big(\frac{-1}{3}\Big)-5=-\frac{47}{9}$
For first curve $\text{y}\Big(-\frac{1}{3}\Big)=\Big(-\frac{1}{3}\Big)^3-\Big(\frac{-1}{3}\Big)+13=\frac{8}{27}+13$
Thus at $\text{x}=-\frac{1}{3},$ both curve do not meet but their tangent are parallel.
So, the only point where both curve touch is (3, 34)

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Question 61 Mark

The equation of normal to the curve y = tan x at (0, 0) is ________.

Answer

The equation of normal to the curve y = tan x at (0, 0) is x + y = 0.
Solution:
We have, y = tan x
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\sec^2\text{x}$
The slope of the curve y = tan x at (0, 0) is given by
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\sec^20=1$
Now, we know that, the product of the slopes of two perpendicular lines is -1
$\therefore$ 1 × Slope of normal = -1
⇒ Slope of normal = -1
Therefore, the equation of normal to the curve y = tan x at (0, 0) is
y - 0 = -1(x - 0)
⇒ x + y = 0
Therefore, the equation of normal to the curve y = tan x at (0, 0) is x + y = 0.

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Question 71 Mark

The values of a for which the function f(x) = sinx - ax + b increases on R are ______.

Answer

The values of a for which the function f(x) = sinx - ax + b increases on R are $(-\infty-1).$
Solution:
The values of a for which the function f(x) = sinx - ax + b increases on R are $(-\infty-1).$
$\because\ \text{f}'(\text{x})=\cos\text{x}-\text{a}$
and $\text{f}'(\text{x})>0\Rightarrow\cos\text{x}>\text{a}$
Since, $\cos\text{x}\in[-1,1]$
$\Rightarrow\ \text{a}<-1\Rightarrow\text{a}\in(-\infty,-1)$

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Question 81 Mark

The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval _______.

Answer

The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval $(1,\infty).$
Solution:
We have, $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4}$
$\Rightarrow\ \text{f}'(\text{x})=\frac{\text{x}^4\cdot4\text{x}-(2\text{x}^2-1)\cdot4\text{x}^3}{\text{x}^8}$ $\bigg[\because\Big(\frac{\text{f}}{\text{g}}\Big)'=\frac{\text{gf}'-\text{fg}'}{\text{g}^2}\bigg]$
$=\frac{4\text{x}^5-8\text{x}^5+4\text{x}^3}{\text{x}^8}$
$=\frac{-4\text{x}^5+4\text{x}^3}{\text{x}^8}$
$=\frac{4\text{x}^3(-\text{x}^2+1)}{\text{x}^8}$
For decreasing, f'(x) < 0
$\Rightarrow\ \frac{4\text{x}^3(1-\text{x}^2)}{\text{x}^8}<0$
$\Rightarrow\ 1-\text{x}^2<0$
$\Rightarrow\ \text{x}^2>1$
$\Rightarrow\ \text{x}>\pm1$
Since, x > 0
Hence, $\text{x}\in(1,\infty)$
Therefore, the function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},$ where x > 0, decreases in the interval $(1,\infty).$

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Question 91 Mark

The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is ______.

Answer

The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is $2\sqrt{\text{ab}}$
Solution:
$\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{x}^2}$
$\text{f(x)}=0$
$\Rightarrow\ \text{a}=\frac{\text{b}}{\text{x}^2}$ or $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}\ \ (\text{as x}>0)$
Now, $\text{f}''(\text{x})=\frac{2\text{b}}{\text{x}^3}>0$ for $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$
Thus least value of f(x) is $\text{f}\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\text{a}\cdot\sqrt{\frac{\text{b}}{\text{a}}}+\frac{\text{b}}{\sqrt{\frac{\text{b}}{\text{a}}}}$ $=\sqrt{\text{ab}}+\sqrt{\text{ab}}=2\sqrt{\text{ab}}$

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Question 101 Mark

The curves $y = 4x^2 + 2x - 8$ and $y = x^3 - x + 13$ touch each other at the point $.......$

Answer

Given$,$ equation of curves are $y = 4x^2 + 2x - 5$ and $y - x^3 - x + 13$
For first curve $\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=8\text{x}+2$
For the secqnd curve $\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=3\text{x}^2-1$
When both curve touch$,$ the slope of both curves should be same
$\therefore\ 8\text{x}+2=3\text{x}^2-1$
$\Rightarrow\ 3\text{x}^2-8\text{x}-3=0$
$\Rightarrow\ (3\text{x}+1)(\text{x}-3)=0$
$\therefore\ \text{x}=-\frac{1}{2}$ and $ \text{x} = 3$
Now consider $x = 3$
For first curve $y(3) = 4(3)^2 + 2(3) - 5 = 37$
For second curve $y(3) = (3)^3 - 3 + 13 = 37$
Thus at $(3, 37),$ both curves touch.
Now, $\text{x}=-\frac{1}{3},$
For first curve $\text{y}\Big(-\frac{1}{3}\Big)=4\Big(-\frac{1}{3}\Big)^2+2\Big(\frac{-1}{3}\Big)-5=-\frac{47}{9}$
For first curve $\text{y}\Big(-\frac{1}{3}\Big)=\Big(-\frac{1}{3}\Big)^3-\Big(\frac{-1}{3}\Big)+13=\frac{8}{27}+13$
Thus at $\text{x}=-\frac{1}{3},$ both curve do not meet but their tangent are parallel.
So$,$ the only point where both curve touch is $(3, 34)$

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MATHS Fill In The Blanks[1 Marks ]

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