3 Marks Question

Question 13 Marks

he sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.

Answer

Let the first term of the A.P. be a and the common difference be d.
$\therefore a=a, b=a+d$ and $c=a+2 d$
a + b + c = 18
$\begin{array}{l}\Rightarrow 3 a+3 d=18 \\ \Rightarrow a+d=6 \ldots(i)\end{array}$
Now, according to the question, a + 4, a + d + 4 and a + 2d + 36 are in G.P.
$\therefore(a+d+4)^2=(a+4)(a+2 d+36)$
$\begin{array}{l}\Rightarrow(6- d + d +4)^2=(6- d +4)(6- d +2 d+36)[\text { using }( i )] \\
\Rightarrow(10)^2=(10- d )(42+ d ) \\ \Rightarrow 100=420+10 d-42 d- d ^2 \\ \Rightarrow
d^2+32 d-320=0 \\ \Rightarrow(d+40)( d -8)=0 \\ \Rightarrow d=8,-40\end{array}$
Now, substituting d = 8, -40 in equation (i), we obtain, a = -2, 46, respectively.
For a = -2 and d = 8, we obtain
a = -2, b = 6, c = 14
And for a = 46 and d = -40, we obtain
a = 46, b = 6, c = -34

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Question 23 Marks

Evaluate the following limits: $\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}$.

Answer

Given: $\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}$
Rationalizing the given equation,
$=\lim _{x \rightarrow 0} \frac{2 x}{(\sqrt{a+x}-\sqrt{a-x})} \frac{(\sqrt{a+x}+\sqrt{a-x})}{(\sqrt{a+x}+\sqrt{a-x})}$
Formula: $( a + b )( a - b )= a ^2- b ^2$
$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{2 x(\sqrt{a+x}+\sqrt{a-x})}{a+x-a+x} \\ =\lim _{x \rightarrow 0} \frac{2 x(\sqrt{a+x}+\sqrt{a-x})}{2 x} \\ =\lim _{x \rightarrow 0} \frac{(\sqrt{a+x}+\sqrt{a-x})}{1}\end{array}$
Now we can see that the indeterminant form is removed, so substituting x as 0
Therefore,$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}=\sqrt{a}+\sqrt{a}=2 \sqrt{a}$

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Question 33 Marks

Are the $E=\left\{x: x \in Z, x^2 \leq 4\right\}$ and $F=\left\{x: x \in Z, x^2=4\right\}$ pairs of equal set?

Answer

We know two sets A and B are said to be equal if they have exactly the same elements & we write A = B
We have, $E=\left\{x: x \in Z, x^2 \leq 4\right\}$
Here, $x \in Z$ and $x^2 \leq 4$
If $x=-2$, then $x^2=(-2)^2=4=4$
If $x=-1$, then $x^2=(-1)^2=1<4$
If $x=0$, then $x^2=(0)^2=0<4$
If $x=1$, then $x^2=(1)^2=1<4$
If $x=2$, then $x^2=(2)^2=4=4$
Therefore, E = {-2, -1, 0, 1, 2}
and $F=\left\{x: x \in Z, x^2=4\right\}$
Here, $x \in Z$ and $x^2=4$
If $x=-2$, then $x^2=(-2)^2=4=4$
If $x=2$, then $x^2=(2)^2=4=4$
Therefore, $F =\{-2,2\}$
$\therefore E \neq F$ because the elements in the both the sets are not equal.

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Question 43 Marks

If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively then obtain the quadratic equation.

Answer

Let a and b be the roots of required quadratic equation.
Then A.M. $=\frac{a+b}{2}=8$
a + b = 16
And G.M. $=\sqrt{a b}=5$
$\Rightarrow a b=25$
Now, Quadratic equation $x^2-$ (Sum of roots) $x+$ (Product of roots) $=0$
$\begin{array}{l}\Rightarrow x^2-(a+b) x+a b=0 \\ \Rightarrow x^2-16 x+25=0\end{array}$
Therefore, required equation is $x^2-16 x+25=0$

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Question 53 Marks

Differentiate $e ^{ ax + b }$ from first principle.

Answer

We need to find derivative of $f(x)=e^{a x+b}$
Derivative of a function f(x) is given by
$f ^{\prime}( x )=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\{$ where h is a very small positive number $\}$
$\therefore$ derivative of $f(x)=e^{a x+b}$ is given as
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\begin{array}{l}\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a(z+h)+b}-e^{a z+b}}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a z+b} e^{a k}-e^{a z+b}}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a z+b}\left(e^{a h}-1\right)}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} e^{a x+b} \times \lim _{h \rightarrow 0} \frac{e^{a h}-1}{h}\end{array}$
As one of the limits $\times \lim _{h \rightarrow 0} \frac{e^{h h}-1}{h}$ can't be evaluated by directly putting the value of $h$ as it will take $\frac{0}{0}$ form.
So we need to take steps to find its value.
$\Rightarrow f ^{\prime}( x )=\lim _{ h \rightarrow 0} e ^{ ax + b } \times \lim _{h \rightarrow 0} \frac{e^{h h}-1}{a h} \times a$
Use the formula: $\lim _{x \rightarrow 0} \frac{e^2-1}{x}=\log _e e =1$
$\begin{array}{l}\Rightarrow f^{\prime}(x)=e^{a x+b} \times(a) \\ \Rightarrow f^{\prime}(x)=a e^{a x+b}\end{array}$
Hence,
Derivative of $f(x)=e^{a x+b}=a e^{a x+b}$

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Question 63 Marks

Using $g$ binomial theorem, expand $\left\{(x+y)^5+(x-y)^5\right\}$ and hence find the value of $\left\{(\sqrt{2}+1)^5+(\sqrt{2}-1)^5\right\}$

Answer

We have
$\begin{array}{l}(x+y)^5+(x-y)^5=2\left[{ }^5 C_0 x^5+{ }^5 C_2 x^3 y^2+{ }^5 C_4 x^1 y^4\right] \\ \quad=2\left(x^5+10 x^3 y^2+5 x y^4\right.\end{array}$
Putting $x=\sqrt{2}$ and $y=1$, we get
$\begin{array}{l}(\sqrt{2}+1)^5+(\sqrt{2}-1)^5=2\left[(\sqrt{2})^5+10(\sqrt{2})^3+5 \sqrt{2}\right] \\ \quad=2[4 \sqrt{2}+20 \sqrt{2}+5 \sqrt{2}] \\ \quad=58 \sqrt{2}\end{array}$

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Question 73 Marks

Find $(x+1)^6+(x-1)^6$. Hence or otherwise evaluate $(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$

Answer

$\begin{array}{l}( x +1)^6+( x -1)^6=\left[{ }^6 C_0 x^6+{ }^6 C_1 x^5+{ }^6 C_2 x^4+{ }^6 C_3 x^3+{ }^6 C_4 x^2+{ }^6 C_5 x+{ }^6 C_6\right] \\ +\left[{ }^6 C_0 x^6+{ }^6 C_1 x^5(-1)+{ }^6 C_2 x^4(-1)^2+{ }^6 C_3 x^3(-1)^3+{ }^6 C_4 x^2(-1)^4+{ }^6 C_5 x(-1)^5+{ }^6 C_6(-1)^6\right] \\ =\left[x^6+6 x^5+15 x^4+20 x^3+15 x^2+6 x +1\right]+\left[x^6-6 x^5+15 x^4-20 x^3+15 x^2-6 x+1\right] \\ =2 x^6+30 x^4+30 x^2+2 \\ =2\left(x^6+15 x^4+15 x^2+1\right)\end{array}$
Putting $x=\sqrt{2}$
$\begin{array}{l}(\sqrt{2}+1)^6+(\sqrt{2}-1)^6=2\left[(\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1\right] \\ =2[8+15 \times 4+15 \times 2+1] \\ =2[8+60+30+1] \\ =2 \times 99=198\end{array}$

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Question 83 Marks

Find the point in yz-plane which is equidistant from the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).

Answer

The general point on yz plane is D(0, y, z).
Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).
$\therefore A D=B D$
$\sqrt{(0-3)^2+(y-2)^2+(z+1)^2}=\sqrt{(0-1)^2+(y+1)^2+(z-0)^2}$
Squaring both sides,
$(0-3)^2+(y-2)^2+(z+1)^2=(0-1)^2+(y+1)^2+(z-0)^2$
$9+y^2-4 y+4+z^2+2 z+1=1+y^2+2 y+1+z^2$
-6y + 2z + 12 = 0 ….(1)
Also, AD = CD
$\sqrt{(0-3)^2+(y-2)^2+(z+1)^2}=\sqrt{(0-2)^2+(y-1)^2+(z-2)^2}$
Squaring both sides,
$(0-3)^2+(y-2)^2+(z+1)^2=(0-2)^2+(y-1)^2+(z-2)^2$
$9+y^2-4 y+4+z^2+2 z+1=4+y^2-2 y+1+z^2-4 z+4$
-2y + 6z + 5 = 0 ….(2)
By solving equation (1) and (2) we get
$y=\frac{31}{16} z=\frac{-3}{16}$
The point which is equidistant to the points $A (3,2,-1), B (1,-1,0)$ and $C (2,1,2)$ is $\left(\frac{31}{16}, \frac{-3}{16}\right)$.

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Question 93 Marks

In how many ways can six persons be seated in a row?

Answer

Given: Six persons are to be arranged in a row.
Assume six seats, now in the first seat, any one of six members can be seated, so the total number of possibilities is 6 Similarly, in the second seat, any one of five members can be seated, so the total number of possibilities is ${ }^5 C _1$ In the third seat, any one of four members can be seated, so the total number of possibilities is ${ }^4 C _1$
In the fourth seat, any one of three members can be seated, so the total number of possibilities is ${ }^3 C _1$
In the fifth seat, any one of two members can be seated, so the total number of possibilities is ${ }^2 C _1$
In the sixth seat, only one remaining person can be seated, so the total number of possibilities is ${ }^1 C _1$
Hence the total number of possible outcomes $={ }^6 C _1 \times{ }^5 C _1 \times{ }^4 C _1 \times{ }^3 C _1 \times{ }^2 C _1 \times{ }^1 C _1=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$

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