100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	If the least bounded by the curves `y=x^(2)` and `y=lamdax+12` is equal to `(alpha)/(beta)`, then `[(alpha)/(20beta)]` is equal to ________(where [.] denotes the greatest integer function)
Answer» Correct Answer - 4 For least area `lamda=0` So, `A_("min")=2int_(0)^(4)(12-(x^(2)-4))dx=256/3`

Discussion

2.	Consider the following statements :Statement I : The local minima of f(x) = (2x - 1)2 + 3 is at x = 1/2Statement II : f'(1/2) > 0 and f"(1/2) < 0.Of these statements : (a) Both the statements are true and Statement II is the correct explanation of Statements I. (b) Both the statements are true, but Statement II is not the correct explanation of Statements I. (c) Statement I is true, but Statement II is false. (d) Statement I is false, but Statement II is true.
Answer» Answer is (c) Statement I is true, but Statement II is false.

Discussion

3.	Number of points of inflexion on the cuve `f(x)=(x-1)^(7)(x+2)^(8)` is equal to `l` then `(5l)/(10^(2))` is equal to __________
Answer» Correct Answer - `00000.15` `f^(')(x)=(x-1)^(5)(x+2)^(6)(195x^(2)+153x+30)` `f^(')(x)=0`, if `x=1,-2x_(1),x_(2)` Clearly there are 3 inflexion points

Discussion

4.	The range of real constant `t` for which `(1-tan^2 t)sin theta^2+tan^2 t*tan theta^2 >= theta^2;` always holds `AA theta in (0,pi/2)` is `[alpha,beta)` then `beta/alpha` is equal to
Answer» Correct Answer - 3 `tan^(2)ge (theta^(2)-sintheta^(2))/(tantheta^(2)-sintheta^(2)) AA theta in(0,(pi)/2)` So, `tan^(2)t ge ((theta^(2)-sintheta^(2))/(tantheta^(2)-sintheta^(2)))_("max")AA theta in (0,(pi)/2)` Since, in `(0,(pi)/2):tantheta^(2)gt theta^(2)` and the same is subtracted from `N^(-r)` and `D^(-r)` both So , maximum value occurs at `theta to O^(+)` Therefore `tan^(2)t ge 1/3, t in ((pi)/6, (pi)/2)`

Discussion

5.	Consider the following statements :Statement l : cos-1 (3/7) = π - cos-1 (3/7)Statement II : 0 ≤ cos-1 x ≤ π, where -1 ≤ x ≤ 1.Of these statements : (a) Both the statements are true and Statement II is the correct explanation of Statements I. (b) Both the statements are true, but Statement II is not the correct explanation of Statements I. (c) Statement I is true, but Statement II is false. (d) Statement I is false, but Statement II is true
Answer» Answer is (a) Both the statements are true and Statement II is the correct explanation of Statements I.

Discussion

6.	f(x) = 2x3 - 15x2 + 36x + 4 is (a) increasing in (-∞,2) (b) increasing in [2,3] (c) decreasing in (3,∞] (d) none of these
Answer» Answer is (b) increasing in [2, 3]

Discussion

7.	Let `f(x)=[x]+{x}^(3)` then the area of the figure bounded by `y=f^(-1)(x),y=0` between the ordinates `x=2` and `x=9/2` is `alpha`, then `alpha-3/(2^(10//3))+1/2` is equal to ________(where [.] denotes the greatest integer function)
Answer» Correct Answer - 9 `f^(-1)(x)=[x]+{x}^(1//3)` So, `int_(2)^(9//2)[x]dx+int_(2)^(9//2){x}^(1//3)=17/2+3/(2^(10//3))`

Discussion

8.	Given , `epsilon-L(di)/(dt)=iR`, find the value of `i` at any time t in terms of constant ` epsilon`, L and R. At t=0 , `i=0` .
Answer» We have , `epsilon-L (di)/(dt)=iR` or `(epsilon -iR)=L (di)/(dt)` or `(di)/((epsilon-iR))=(dt)/(L) " " ` ...(i) Integrating both sides of the equation (i), we get `int (di)/((epsilon-iR))= int (dt)/(L)` Here limit of time varies from 0 to t and corresponding limits of i varies from 0 to i. ` :. "" int_(0)^(i)(di)/((epsilon-iR))=int _(0)^(t) (dt)/(L)` For integration of LHS, substitute ` epsilon-iR=z` Also `"" (d)/(di)(epsilon-iR)=(dz)/(di)` or `(0-R)=(dz)/(di)` `:. "" di=(dz)/((-R))` and ` int _(0)^(t)(di)/((epsilon-iR)) = int _(0)^(i) (dzl(-R))/(z)` ` =((1)/(-R)) int_(0)^(i) (dz)/(z)` `=((1)/(-R))\|ln z\|_(0)^(i)` `=((1)/(-R))\|ln (epsilon-iR)\|_(0)^(i)` `=((1)/(-R)){ln(epsilon-iR)-ln(epsilon-0)}` `=((1)/(-R)) ln ((epsilon-iR))/(epsilon)" " ` ...(ii) and RHS ` int_(0)^(t) (dt)/(L)=((1)/L))\|t\|_(0)^(t)=(1)/(L)(t-0)` `=(t)/(L)` From equation (i) and (ii) , we have `(-(1)/(R)) ln ((epsilon-iR)/(epsilon))=(t)/(L)` or ` " " ln((epsilon-iR)/(epsilon))=-(R)/(L)t` or ` ((epsilon-iR)/(epsilon))=e^(-(Rt)/(L))` or `" " i=(epsilon)/(R)(1-e^(-(tR)/(L)))`

Discussion

9.	Equation of plane parallel to the plane x + 2y - 3z + 4 = 0 is (a) x - 2y + 3z + 4 =0 (b) x + 2y - 3z + 1 = 0 (c) x + 2y + 3z + 4 = 0 (d) x + 2y + 3z + 9 = 0
Answer» Answer is (c) x + 2y + 3z + 4 = 0

Discussion

10.	The line (x - 5)/3 = (y + 4)/7 = (z - 6)/2(a) passes through (-5,4,-6) (b) its direction ratios are 3,7,2 (c) is perpendicular to 3i + 7j + 2k (d) none of these
Answer» Answer is (b) its direction ratios are 3,7,2

Discussion

11.	If \(tan\theta=\frac{2x(x+1)}{2x+1}\) find \(sin\theta\) and \(cos\theta\)
Answer» \(\tan \theta =\frac{2\mathrm x(\mathrm x+1)}{2\mathrm x+1}\) \(\because \sec\theta =\sqrt{1+\tan^2\theta}\) \(=\sqrt{1+\left(\frac{2\mathrm x(\mathrm x+1)}{2\mathrm x+1}\right)^2}\) \(=\sqrt{\frac{(2\mathrm x+1)^2+(2\mathrm x(\mathrm x+1))^2}{2\mathrm x+1}}\) \(=\frac{\sqrt{4\mathrm x^2+4\mathrm x+1+4\mathrm x^2(\mathrm x^2+2\mathrm x+1)}}{2\mathrm x+1}\) \(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}{2\mathrm x+1}\) \(\therefore\) \(\cos\theta = \frac{1}{\sec\theta}\) \(=\frac{2\mathrm x+1}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) Now, \(\sin \theta =\sqrt{1-\cos^2\theta}\) \(=\sqrt{1-\frac{(2\mathrm x+1)^2}{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) \(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1-(4\mathrm x^2+4\mathrm x+1)}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) \(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+4\mathrm x^2}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) \(=\frac{2\mathrm x\sqrt{\mathrm x^2+2\mathrm x+1}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) \(=\frac{2\mathrm x(\mathrm x+1)}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) \(\Big(\because (\mathrm x^2+2\mathrm x+1)^{\frac{1}{2}}\)\(=((\mathrm x+1)^2)^{\frac{1}{2}}= \mathrm x+1\Big)\)

11.

If \(tan\theta=\frac{2x(x+1)}{2x+1}\) find \(sin\theta\) and \(cos\theta\)

Answer»

\(\tan \theta =\frac{2\mathrm x(\mathrm x+1)}{2\mathrm x+1}\)

\(\because \sec\theta =\sqrt{1+\tan^2\theta}\)

\(=\sqrt{1+\left(\frac{2\mathrm x(\mathrm x+1)}{2\mathrm x+1}\right)^2}\)

\(=\sqrt{\frac{(2\mathrm x+1)^2+(2\mathrm x(\mathrm x+1))^2}{2\mathrm x+1}}\)

\(=\frac{\sqrt{4\mathrm x^2+4\mathrm x+1+4\mathrm x^2(\mathrm x^2+2\mathrm x+1)}}{2\mathrm x+1}\)

\(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}{2\mathrm x+1}\)

\(\therefore\) \(\cos\theta = \frac{1}{\sec\theta}\) \(=\frac{2\mathrm x+1}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

Now, \(\sin \theta =\sqrt{1-\cos^2\theta}\)

\(=\sqrt{1-\frac{(2\mathrm x+1)^2}{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1-(4\mathrm x^2+4\mathrm x+1)}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+4\mathrm x^2}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{2\mathrm x\sqrt{\mathrm x^2+2\mathrm x+1}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{2\mathrm x(\mathrm x+1)}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\) \(\Big(\because (\mathrm x^2+2\mathrm x+1)^{\frac{1}{2}}\)\(=((\mathrm x+1)^2)^{\frac{1}{2}}= \mathrm x+1\Big)\)

Discussion

12.	Show that 2+5+8+------+(3n-1)=n(3n+1)/2
Answer» Nth term = 3n-1 Now to find sum of n terms Sum = summation tn sum of n natural numbers = n(n+1)/2 Sum=summation tn =3(n(n+1)/2 - n =(3n²+3n-2n)/2 =3n²+n/2 =n(3n+1)/2 hence proved .

12.

Show that 2+5+8+------+(3n-1)=n(3n+1)/2

Answer»

Nth term = 3n-1

Now to find sum of n terms

Sum = summation tn

sum of n natural numbers = n(n+1)/2

Sum=summation tn

=3(n(n+1)/2 - n

=(3n²+3n-2n)/2

=3n²+n/2

=n(3n+1)/2

hence proved .

Discussion

13.	From a circular sheet of radius 4cm, a circle of radius 3cm is removed. The area of the remaining sheet is
Answer» The area of the circular sheet with 4cm radius is, A=πr² =16π The area of the circular sheet with 3cm radius is, A=πr² =9π Since the circle with radius 3cm is removed, then the remaining area is, A=16π−9π =7π =7×3.14 =21.98cm²

13.

From a circular sheet of radius 4cm, a circle of radius 3cm is removed. The area of the remaining sheet is

Answer»

The area of the circular sheet with 4cm radius is,

A=πr²

=16π

The area of the circular sheet with 3cm radius is, A=πr²

=9π

Since the circle with radius 3cm is removed, then the remaining area is,

A=16π−9π

=7π

=7×3.14

=21.98cm²

Discussion

14.	Akshat wants to store a huge amount information about his firm in a database. Which type of table organization would be most suitable for this purpose? a. Relational b. Flat File c. Either Relational or Flat file d. Hierarchical
Answer» Correct answer is c. Either Relational or Flat file

Discussion

15.	Is it important to clearly communicate instructions, ideas and concepts that can help you find success in any career? State True/ False
Answer» True Is it important to clearly communicate instructions, ideas and concepts that can help you find success in any career? is True.

15.

Is it important to clearly communicate instructions, ideas and concepts that can help you find success in any career? State True/ False

Answer»

True

Is it important to clearly communicate instructions, ideas and concepts that can help you find success in any career? is True.

Discussion

16.	Sunita is making a project in spreadsheets. Her friend has told her how to rename a spreadsheet as it will help her identify the worksheet data easily. She was very excited and told the same to her brother. Her brother told her that in spreadsheets(OpenOffice0, a worksheet can be renamed using _____ways. a. 2 b. 3 c. 4 d. 1
Answer» Correct answer is b. 3

Discussion

17.	Mention the features of spreadsheets.
Answer» Spreadsheets are mainly used for making calculations and mathematical works. Microsoft excel is a spread sheet application. In a single sheet, it consists of rows and columns and cells. Every cell has different address. Features of spreadsheet: 1. Tables: Tables are created with different fields eg – name, age, address, roll no so we add a table to fill these values; 2. Clip art: In this we can add images and also audio, video clips can be added here. 3. Charts: With charts, we can clearly shown products evaluation to the clients. For example which product sale is more or less in this month. 4. Functions: MATHEMATICAL: Add, subtract, div, multiply. LOGICAL: average, sum, mod, product. 5. Images and Backgrounds: In this we add images and backgrounds in sheet. 6. Sorting and Filter: In sorting we can sort our data and also filter our data so that repetitions will be removed. 7. Page layout: In this themes, colors, sheets, margins, size, backgrounds, breaks, print, titles, sheets height, width, scaling, gridness, headings, views, bring to front of font or back alignment, etc will be used.

17.

Mention the features of spreadsheets.

Answer»

Spreadsheets are mainly used for making calculations and mathematical works. Microsoft excel is a spread sheet application. In a single sheet, it consists of rows and columns and cells. Every cell has different address.

Features of spreadsheet:

1. Tables:

Tables are created with different fields eg – name, age, address, roll no so we add a table to fill these values;

2. Clip art:

In this we can add images and also audio, video clips can be added here.

3. Charts:

With charts, we can clearly shown products evaluation to the clients.

For example which product sale is more or less in this month.

4. Functions:

MATHEMATICAL: Add, subtract, div, multiply.

LOGICAL: average, sum, mod, product. 5. Images and Backgrounds: In this we add images and backgrounds in sheet.

6. Sorting and Filter:

In sorting we can sort our data and also filter our data so that repetitions will be removed.

7. Page layout:

In this themes, colors, sheets, margins, size, backgrounds, breaks, print, titles, sheets height, width, scaling, gridness, headings, views, bring to front of font or back alignment, etc will be used.

Discussion

18.	bipartite graph
Answer» A bipartite graph, also called a bigraph, is a set of graph vertices decomposed into two disjoint sets such that no two graph vertices within the same set are adjacent.

Discussion

19.	Explain the basic tags in HTML with examples (any five tags).
Answer» 1. Bold Text: You can bold text to emphasize information on your web page. Bold text is useful for introducing new terms and highlighting important phrases on a Web page. Tag: <B> SOME TEXT </B> EXAMPLE: <BODY> <B> THIS IS A Bold TEXT </B> </BODY> 2. Underline Text: Tag: <U> Some text </U> Example: <BODY> <U> You can underline this text </U> </BODY> 3. Italicise Text: Tag: <I> SOME TEXT </I> EXAMPLE: <BODY> <I> THIS IS A Itilicize TEXT </I> </BODY> 4. Big Text: Tag: <Big> SOME TEXT </Big> EXAMPLE: <BODY> <Big> THIS IS A Big TEXT</Big> </BODY> 5. Marquee Text: To scroll text horizontally. Tag: <Marquee> some text </marquee> Example: <marquee> The text is moving < /marquee>

19.

Explain the basic tags in HTML with examples (any five tags).

Answer»

1. Bold Text:

You can bold text to emphasize information on your web page. Bold text is useful for introducing new terms and highlighting important phrases on a Web page.

Tag:

SOME TEXT

EXAMPLE:

2. Underline Text:

Tag:

Some text

Example:

<BODY> You can underline this text </BODY>

3. Italicise Text:

Tag:

SOME TEXT

EXAMPLE:

<BODY> THIS IS A Itilicize TEXT </BODY>

4. Big Text:

Tag:

EXAMPLE:

5. Marquee Text:

To scroll text horizontally.

Tag:

Example:

<marquee> The text is moving < /marquee>

Discussion

20.	Find the missing y-coordinate that makes the two triangles congruent.Triangle ABC: A(8,4), B(2,6), C(5, 0)Triangle MNO: M(7,4), N(1,2), O(4, y)(1 point)−4−404−2
Answer» Option (d) is correct answer. Since, congruent triangle have equal area. Area of triangle ABC = \(\frac{1}{2}\)\(\begin{vmatrix}8 & 4 & 1 \\2 & 6 & 1 \\5 & 0 & 1\end{vmatrix}\) = \(\frac{1}{2}\) \(\big(\)8(6 - 0) - 4(2 - 5) + 1(0 - 30)\(\big)\) = \(\frac{1}{2}\) \(\big(\)48 + 12 - 30\(\big)\) = \(\frac{1}{2}\)(60 - 30) = 15 Area of triangle MNO = \(\frac{1}{2}\)\(\begin{vmatrix}7 & 4 & 1 \\1 & 2 & 1 \\4 & y & 1\end{vmatrix}\) = \(\frac{1}{2}\) \(\big(\)7(2 - y) - 4(1 - 4) + 1(y - 8)\(\big)\) = \(\frac{1}{2}\) (14 - 7y + 12 + y - 8) = \(\frac{1}{2}\)(18 - 6y) = 9 - 3y since given that triangle ABC and triangle MNO are congruent. Therefore, there are must be equal Therefore 9y - 3y = 15 \(\Rightarrow\) 3y = 9 - 15 = -6 \(\Rightarrow\) y = -2

20.

Find the missing y-coordinate that makes the two triangles congruent.Triangle ABC: A(8,4), B(2,6), C(5, 0)Triangle MNO: M(7,4), N(1,2), O(4, y)(1 point)−4−404−2

Answer»

Option (d) is correct answer.

Since, congruent triangle have equal area.

Area of triangle ABC = \(\frac{1}{2}\)\(\begin{vmatrix}8 & 4 & 1 \\2 & 6 & 1 \\5 & 0 & 1\end{vmatrix}\)

= \(\frac{1}{2}\) \(\big(\)8(6 - 0) - 4(2 - 5) + 1(0 - 30)\(\big)\)

= \(\frac{1}{2}\) \(\big(\)48 + 12 - 30\(\big)\) = \(\frac{1}{2}\)(60 - 30) = 15

Area of triangle MNO = \(\frac{1}{2}\)\(\begin{vmatrix}7 & 4 & 1 \\1 & 2 & 1 \\4 & y & 1\end{vmatrix}\)

= \(\frac{1}{2}\) \(\big(\)7(2 - y) - 4(1 - 4) + 1(y - 8)\(\big)\)

= \(\frac{1}{2}\) (14 - 7y + 12 + y - 8)

= \(\frac{1}{2}\)(18 - 6y) = 9 - 3y

since given that triangle ABC and triangle MNO are congruent.

Therefore, there are must be equal

Therefore 9y - 3y = 15

\(\Rightarrow\) 3y = 9 - 15 = -6

\(\Rightarrow\) y = -2

Discussion

21.	Which triangle is congruent to the triangle with vertices (2,−2), (7,10), and (4,6)?(1 point)(−4,2), (8,7), (1,1)(−5,3), (7,−2), (3,1)(9,6), (1,8), (−1,4)(−7,−5), (−3,−2), (3,3)
Answer» Option (b) is correct Area of given triangle = \(\frac{1}{2}\)\(\begin{vmatrix} 2 & -2 & 1 \\ 7 & 10 & 1 \\ 4 & 6 & 1 \end{vmatrix}\) = \(\frac{1}{2}\) \(\big(\)2(10 - 6) + 2(7 - 4) + 1(42 - 40)\(\big)\) = \(\frac{1}{2}\)(8 + 6 + 2) = \(\frac{16}{2}\) = 8 Area of triangle in option A = \(\frac{1}{2}\)\(\begin{vmatrix} -4 & 2 & 1 \\ 8 & 7 & 1 \\ 1 & 1 & 1 \end{vmatrix}\) = \(\frac{1}{2}\) \(\big(\)-4(7 - 1) - 2(8 - 1) + 1(8 - 7)\(\big)\) = \(\frac{1}{2}\)(-24 - 14 + 1) = \(\frac{1}{2}\)(-37) = \(\frac{37}{2}\) \(\ne\) 8 (Area always positive) Hence, not congruent to given triangle Area of triangle given in option B = \(\frac{1}{2}\) \(\begin{vmatrix} -5 & 3 & 1 \\ 7 & -2 & 1 \\ 3 & 1 & 1 \end{vmatrix}\) = \(\frac{1}{2}\)\(\big(\)-5(-2 - 1) - 3(7 - 3) + 1(7 + 6)\(\big)\) = \(\frac{1}{2}\) (15 - 12 + 13) = \(\frac{1}{2}\) x 16 = 8 (It my be congruent to given triangle) Area of triangle given in option (c) = \(\frac{1}{2}\)\(\begin{vmatrix} 9 & 6 & 1 \\ 1 & 8 & 1 \\ -1 & 4 & 1 \end{vmatrix}\) = \(\frac{1}{2}\)\(\big(\)9(8 - 4) - 6(1 + 1) + 1(4 + 8)\(\big)\) = \(\frac{1}{2}\)(36 - 12 +12) = \(\frac{36}{2}\) = 18 \(\ne\)8 Hence not congruent to given triangle Area of triangle given in option (d) = \(\frac{1}{2}\)\(\begin{vmatrix} -7 & -5 & 1 \\ -3 & -2 & 1 \\ 3 & 3 & 1 \end{vmatrix}\) = \(\frac{1}{2}\)\(\big(\)-7(-2 - 3) + 5(-3 - 3) + 1(-9 + 6)\(\big)\) = \(\frac{1}{2}\)(+35 - 45 - 3) = \(\frac{1}{2}\) x 13 \(\ne\) 8 (Area always positive) Hence, not congruent to given triangle Since, only area of triangle given in option (b) is equal to Area of given triangle. Therefore, only possibility for congruence.

21.

Which triangle is congruent to the triangle with vertices (2,−2), (7,10), and (4,6)?(1 point)(−4,2), (8,7), (1,1)(−5,3), (7,−2), (3,1)(9,6), (1,8), (−1,4)(−7,−5), (−3,−2), (3,3)

Answer»

Option (b) is correct

Area of given triangle = \(\frac{1}{2}\)\(\begin{vmatrix} 2 & -2 & 1 \\ 7 & 10 & 1 \\ 4 & 6 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\) \(\big(\)2(10 - 6) + 2(7 - 4) + 1(42 - 40)\(\big)\)

= \(\frac{1}{2}\)(8 + 6 + 2) = \(\frac{16}{2}\) = 8

Area of triangle in option A = \(\frac{1}{2}\)\(\begin{vmatrix} -4 & 2 & 1 \\ 8 & 7 & 1 \\ 1 & 1 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\) \(\big(\)-4(7 - 1) - 2(8 - 1) + 1(8 - 7)\(\big)\)

= \(\frac{1}{2}\)(-24 - 14 + 1)

= \(\frac{1}{2}\)(-37) = \(\frac{37}{2}\) \(\ne\) 8 (Area always positive)

Hence, not congruent to given triangle

Area of triangle given in option B = \(\frac{1}{2}\) \(\begin{vmatrix} -5 & 3 & 1 \\ 7 & -2 & 1 \\ 3 & 1 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\)\(\big(\)-5(-2 - 1) - 3(7 - 3) + 1(7 + 6)\(\big)\)

= \(\frac{1}{2}\) (15 - 12 + 13)

= \(\frac{1}{2}\) x 16 = 8 (It my be congruent to given triangle)

Area of triangle given in option (c) = \(\frac{1}{2}\)\(\begin{vmatrix} 9 & 6 & 1 \\ 1 & 8 & 1 \\ -1 & 4 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\)\(\big(\)9(8 - 4) - 6(1 + 1) + 1(4 + 8)\(\big)\)

= \(\frac{1}{2}\)(36 - 12 +12)

= \(\frac{36}{2}\) = 18 \(\ne\)8

Hence not congruent to given triangle

Area of triangle given in option (d) = \(\frac{1}{2}\)\(\begin{vmatrix} -7 & -5 & 1 \\ -3 & -2 & 1 \\ 3 & 3 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\)\(\big(\)-7(-2 - 3) + 5(-3 - 3) + 1(-9 + 6)\(\big)\)

= \(\frac{1}{2}\)(+35 - 45 - 3)

= \(\frac{1}{2}\) x 13 \(\ne\) 8 (Area always positive)

Hence, not congruent to given triangle

Since, only area of triangle given in option (b) is equal to Area of given triangle.

Therefore, only possibility for congruence.

Discussion

22.	(√2 + √3)2 is a?1. Rational number2. Irrational number3. Complex Number4. Integer
Answer» Correct Answer - Option 2 : Irrational number Given: (√2 + √3)² we have to find the types of the number Formula Used: (a + b)² = a² + 2ab + b² Calculation: (√2 + √3)² ⇒ (√2)² + 2 × √2 × √3 + (√3)² ⇒ 2 + 2√6 + 3 ⇒ 5 + 2√6 √6 is an irrational number, so, (5 + 2√6) is also a irrational number. ∴ The given number is an Irrational number.

22.

(√2 + √3)2 is a?1. Rational number2. Irrational number3. Complex Number4. Integer

Answer» Correct Answer - Option 2 : Irrational number

Given:

(√2 + √3)² we have to find the types of the number

Formula Used:

(a + b)² = a² + 2ab + b²

Calculation:

(√2 + √3)²

⇒ (√2)² + 2 × √2 × √3 + (√3)²

⇒ 2 + 2√6 + 3

⇒ 5 + 2√6

√6 is an irrational number, so, (5 + 2√6) is also a irrational number.

∴ The given number is an Irrational number.

Discussion

23.	Match the following :Column IColumn IIIf(x) = x2 - 2x + 4, 1 ≤ x ≤ 5 will satisfy mean value theorem, then C =(a) 4IIIf tan(tan-1 (1/2) + tan-1 (2/11)) = tan p, then p = (b) 1IIIThe projection of vector a = 4i + 4j - 6k on 4i - 7j + 4k is (c) 3IVIf [(x2,-1),(2,-3)] + [(-2x,3),(4,5)] = [(-1,2),(6,2)] then x (d) 3/4
Answer» Answer is I. (c) 3 II. (d) 3/4 III. (a) 4 IV. (b) 1

23.

Match the following :Column IColumn IIIf(x) = x2 - 2x + 4, 1 ≤ x ≤ 5 will satisfy mean value theorem, then C =(a) 4IIIf tan(tan-1 (1/2) + tan-1 (2/11)) = tan p, then p = (b) 1IIIThe projection of vector a = 4i + 4j - 6k on 4i - 7j + 4k is (c) 3IVIf [(x2,-1),(2,-3)] + [(-2x,3),(4,5)] = [(-1,2),(6,2)] then x (d) 3/4

Answer»

Answer is

I. (c) 3

II. (d) 3/4

III. (a) 4

IV. (b) 1

Discussion

24.	Total number of distinct `x epsilon [0,1]` for which `int_(0)^(x) (t^(8)+1)/(t^(8)+t^(2)+1) dt=3x-2` is ______
Answer» Correct Answer - 1 Let `f(x)=int_(0)^(x) (t^(8)+1)/(t^(8)+t^(2)+1)dt-3x+2` `f(0)=2, f(1)=` negative So, `f(x)` has one root in`[0,1]`

Discussion

25.	If `I_(1)=int_(0)^(1)(1-(1-x^(3))^(sqrt(2)))^(sqrt(3)) x^(2)dx` and `I_(2)(1-(1-x^(3))^(sqrt(2)))^(sqrt(3)+1).x^(2)dx`. Then `((I_(1))/(I_(2))-(sqrt(3)-1)/(2sqrt(2))+0.2)/10` is equal to _________
Answer» Correct Answer - `00000.12` Let `1-x^(3)=t` `3l_(1)=int_(0)^(1)(1-t^(sqrt(2)))^(sqrt(3)) dt` and `3l_(2)=int_(0)^(1)(1-t^(sqrt(2)))^(sqrt(3)+1).1.dt` `=int_(0)^(1)(sqrt(3)+1)sqrt(2)(1-t^(sqrt(2)))^(sqrt(3))(1-t^(sqrt(2))-1)dt` (Using integration by parts) `3l_(2)=-(sqrt(3)+1)sqrt(2).3l_(2)+(sqrt(3)+1)sqrt(2).3l_(1)` So, `(l_(1))/(l_(2))-(sqrt(3)-1)/(2sqrt(2))+0.2=1+(sqrt(3)-1)/(sqrt(2))-(sqrt(3)-1)/(2sqrt(2))+0.2=1.2`

Discussion

26.	Dividing (x80 + x40 + 5) by (x8 + 1) gives the remainder as ______.A. 5B. 8C. 12D. 131. A2. B3. D4. C
Answer» Correct Answer - Option 1 : A Calculations : (x80 + x40 + 5) = x⁽^{8 × 10)} + x^{(8 × 5)}+ 5 ⇒ (x⁸)¹⁰ + (x⁸)⁵ + 5 Now to get the remainder for (x80 + x40 + 5) when dividing it by x⁸ + 1, we should put x⁸+ 1 = 0 or x⁸ = -1 in given equation ⇒ (x8)10 + (x8)5 + 5 ⇒ (-1)¹⁰ + (-1)⁵ + 5 (x⁸ + 1 = 0 or x⁸ = -1) ⇒ 1 - 1 + 5 ⇒ 5 ∴ Option A will be the correct answer.

26.

Dividing (x80 + x40 + 5) by (x8 + 1) gives the remainder as ______.A. 5B. 8C. 12D. 131. A2. B3. D4. C

Answer» Correct Answer - Option 1 : A

Calculations :

(x80 + x40 + 5) = x⁽^{8 × 10)} + x^{(8 × 5)}+ 5

⇒ (x⁸)¹⁰ + (x⁸)⁵ + 5

Now to get the remainder for (x80 + x40 + 5) when dividing it by x⁸ + 1, we should put x⁸+ 1 = 0 or x⁸ = -1 in given equation

⇒ (x8)10 + (x8)5 + 5

⇒ (-1)¹⁰ + (-1)⁵ + 5 (x⁸ + 1 = 0 or x⁸ = -1)

⇒ 1 - 1 + 5

⇒ 5

∴ Option A will be the correct answer.

Discussion

27.	Find the total numbers between 100 and 200 which are divisible by 12.1. 62. 163. 84. 12
Answer» Correct Answer - Option 3 : 8 Calculation: The total number between 1 and 100 divisible by 12 = 100/12 = 8.33 ⇒ 8 (Taking proper number) The total number between 1 and 200 divisible by 12 = 200/12 = 16.67 ⇒ 16 (Taking proper number) The total number between 100 and 200 which are divisible by 12 = 16 – 8 = 8 ∴ The total number between 100 and 200 which are divisible by 12 is 8. Alternate Solution Numbers divisible by 12 in between 100 and 200 are 108, 120, ... , 192 We know. Last number = First number + (Number of terms - 1) × Common difference ⇒ 192 = 108 + (n - 1) × 12 ⇒ 84 = (n - 1) × 12 ⇒ 7 = n - 1 ⇒ n = 8 ∴ The total number between 100 and 200 which are divisible by 12 is 8.

27.

Find the total numbers between 100 and 200 which are divisible by 12.1. 62. 163. 84. 12

Answer» Correct Answer - Option 3 : 8

Calculation:

The total number between 1 and 100 divisible by 12 = 100/12 = 8.33

⇒ 8 (Taking proper number)

The total number between 1 and 200 divisible by 12 = 200/12 = 16.67

⇒ 16 (Taking proper number)

The total number between 100 and 200 which are divisible by 12 = 16 – 8 = 8

∴ The total number between 100 and 200 which are divisible by 12 is 8.

Alternate Solution

Numbers divisible by 12 in between 100 and 200 are 108, 120, ... , 192

We know.

Last number = First number + (Number of terms - 1) × Common difference

⇒ 192 = 108 + (n - 1) × 12

⇒ 84 = (n - 1) × 12

⇒ 7 = n - 1

⇒ n = 8

∴ The total number between 100 and 200 which are divisible by 12 is 8.

Discussion

28.	If f : R → R is defined by f(x) = x2 - 3x + 2, find f(f(x)).
Answer» f(x) = x² - 3x + 2 f.f(x) = (x²- 3x + 2)² - 3(x²- 3x + 2) + 2 = x⁴ + 9x² + 4 + 6x³ + 12x + 4x² - 3x² - 3x²- 9x - 6 + 2 = x⁴- 6x³ - 10x² + 3x

28.

If f : R → R is defined by f(x) = x2 - 3x + 2, find f(f(x)).

Answer»

f(x) = x² - 3x + 2

f.f(x) = (x²- 3x + 2)² - 3(x²- 3x + 2) + 2

= x⁴ + 9x² + 4 + 6x³ + 12x + 4x² - 3x² - 3x²- 9x - 6 + 2

= x⁴- 6x³ - 10x² + 3x

Discussion

29.	The value of \(\frac{\sqrt{243}}{\sqrt{867}}\) is:1. \(\frac{9}{19}\)2. \(\frac{8}{17}\)3. \(\frac{9}{17}\)4. \(\frac{11}{17}\)
Answer» Correct Answer - Option 3 : \(\frac{9}{17}\) Calculations: \(\frac{√{243}}{√{867}}\) ⇒ √(3 × 3 × 3 × 3 × 3)/√(3 × 17 × 17) ⇒ 9√3/17√3 ⇒ 9/17 ∴ The correct answer is 9/17

29.

The value of \(\frac{\sqrt{243}}{\sqrt{867}}\) is:1. \(\frac{9}{19}\)2. \(\frac{8}{17}\)3. \(\frac{9}{17}\)4. \(\frac{11}{17}\)

Answer» Correct Answer - Option 3 : \(\frac{9}{17}\)

Calculations:

\(\frac{√{243}}{√{867}}\)

⇒ √(3 × 3 × 3 × 3 × 3)/√(3 × 17 × 17)

⇒ 9√3/17√3

⇒ 9/17

∴ The correct answer is 9/17

Discussion

30.	How to count a diameter
Answer» How To Calculate Diameter? Diameter = Circumference ÷ π (when the circumference is given) Diameter = 2 × Radius (when the radius is given) Diameter = 2√[Area/π] (when the area is given)

30.

How to count a diameter

Answer»

How To Calculate Diameter?

Diameter = Circumference ÷ π (when the circumference is given)
Diameter = 2 × Radius (when the radius is given)
Diameter = 2√[Area/π] (when the area is given)

Discussion

31.	The set that contains atomic number of only transition element is - (1) 21, 32, 53, 64 (2) 21, 25, 42, 72 (3) 9, 17, 34, 38 (4) 37, 42, 50, 64
Answer» Correct answer is (2) 21, 25, 42, 72

Discussion

32.	If a number is in the form of 810 × 97 × 78, find the total number of prime factors of the given number.1. 522. 5603. 33604. 25
Answer» Correct Answer - Option 1 : 52 Given: The number is 810 × 97 × 7⁸ Concept used: If a number of the form x^a × y^b × z^c ...... and so on, then total prime factors = a + b + c ..... and so on Where x, y, z, ... are prime numbers Calculation: The number 810 × 97 × 78 can be written as (2³)¹⁰ × (3²)⁷ × 7⁸ The number can ve written as 2³⁰ × 3¹⁴ × 7⁸ Total number of prime factors = 30 + 14 + 8 ∴ The total number of prime factors are 52

32.

If a number is in the form of 810 × 97 × 78, find the total number of prime factors of the given number.1. 522. 5603. 33604. 25

Answer» Correct Answer - Option 1 : 52

Given:

The number is 810 × 97 × 7⁸

Concept used:

If a number of the form x^a × y^b × z^c ...... and so on, then total prime factors = a + b + c ..... and so on

Where x, y, z, ... are prime numbers

Calculation:

The number 810 × 97 × 78 can be written as (2³)¹⁰ × (3²)⁷ × 7⁸

The number can ve written as 2³⁰ × 3¹⁴ × 7⁸

Total number of prime factors = 30 + 14 + 8

∴ The total number of prime factors are 52

Discussion

33.	If y = sin[cos{tan(cot x)}], then find dy/dx.
Answer» y = sin[cos{tan(cot x)}] Differentiating, we get dy/dx = sin[cos{tan(- cosec x.cot x)}] = sin[-sin{sec²x(cot x) - cosec x cot x}] = - sin{sec²x(cot x) - cosec x. cot x}

33.

If y = sin[cos{tan(cot x)}], then find dy/dx.

Answer»

y = sin[cos{tan(cot x)}]

Differentiating, we get

dy/dx = sin[cos{tan(- cosec x.cot x)}]

= sin[-sin{sec²x(cot x) - cosec x cot x}]

= - sin{sec²x(cot x) - cosec x. cot x}

Discussion

34.	Prove that : tan-1 x + cot-1 x = π/2
Answer» To prove, tan^-1 x + cot^-1 x = π/2 Let us take tan^-1x = y, then x = tan y = cot((π/2) - y) Hence, cot^-1 x = ((π/2) - y) = ((π/2) - tan^-1 x) or, tan^-1 x + cot^-1 x = π/2

34.

Prove that : tan-1 x + cot-1 x = π/2

Answer»

To prove,

tan^-1 x + cot^-1 x = π/2

Let us take tan^-1x = y, then

x = tan y = cot((π/2) - y)

Hence, cot^-1 x = ((π/2) - y) = ((π/2) - tan^-1 x)

or, tan^-1 x + cot^-1 x = π/2

Discussion

35.	If f : R → R be a function defined by f(x) = x2, show that the function f is many one into.
Answer» f(x) = x² f(x₁) = f(x₂) x₁²= x₂² x₁² - x₂² (x₁ - x₂) (x₁ + x₂) = 0 Squaring x₁- x₂= 0 x₁= x₂ or, x₁ = x₂ or, x₁ + x₂= 0 x₁= -x₁ Hence, the function is many to one onto function.

35.

If f : R → R be a function defined by f(x) = x2, show that the function f is many one into.

Answer»

f(x) = x²

f(x₁) = f(x₂)

x₁²= x₂²

x₁² - x₂²

(x₁ - x₂) (x₁ + x₂) = 0

Squaring x₁- x₂= 0

x₁= x₂

or, x₁ = x₂

or, x₁ + x₂= 0

x₁= -x₁

Hence, the function is many to one onto function.

Discussion

36.	Consider the following statements : Statement-I : ∫dx/x for x ∈ [2,4] = log 2Statement-II : ∫f(x) dx for x ∈ [a,b] = F(b) - F(a).Of these statements:(a) Both the statements are true and Statement-II is the correct explanation of Statement-I(b) Both the statements are true, but Statement-II is not the correct explanation of Statement-I (c) Statement-I is true, But Statement-II is false.(d) Statement-I is false, but Statement-II is true.
Answer» Answer is (a) Both the statements are true and Statement-II is the correct explanation of Statement-I.

Discussion

37.	\( 2 x y^{2}-4 x^{2} y, 11 x^{2} y-x y^{2},-3 x^{2} y+5 x y^{2} \)
Answer» 2xy² − 4x²y = 2xy (y - 2x) 11x²y − xy² = xy (11x - y) −3x²y + 5xy² = xy (-3x + 5y).

37.

\( 2 x y^{2}-4 x^{2} y, 11 x^{2} y-x y^{2},-3 x^{2} y+5 x y^{2} \)

Answer»

2xy² − 4x²y = 2xy (y - 2x)

11x²y − xy² = xy (11x - y)

−3x²y + 5xy² = xy (-3x + 5y).

Discussion

38.	\( \int \frac{n^{2}+4}{n+3} d n \)
Answer» ya,it is integer

Discussion

39.	Add the following 3x2 - 4x + 5 and 4x2 + 6x + 7a.
Answer» (3x² - 4x + 5) + (4x² + 6x + 7a) = (3x² + 4x²) + (-4x + 6x) + (5 + 7a) = 7x² + 2x + 5 + 7a

39.

Add the following 3x2 - 4x + 5 and 4x2 + 6x + 7a.

Answer»

(3x² - 4x + 5) + (4x² + 6x + 7a)

= (3x² + 4x²) + (-4x + 6x) + (5 + 7a)

= 7x² + 2x + 5 + 7a

Discussion

40.	Use a suitable identity to get each of the following products: (p - 11)(p + 11)
Answer» (p - 11) (p + 11) = p² - 11² = p² - 121

Discussion

41.	Find the product of (4a + 3b/5) (3a - 4b/5)
Answer» Let's multiply it using distributive property. (4a×3a) -(4a×4b/5) +(3b/5×3a) -(3b/5×4b/5) 12a²- 16ab/5 + 9ab/5 - 12b²/25 =12a² - 7ab/5 - 12b²/25

41.

Find the product of (4a + 3b/5) (3a - 4b/5)

Answer»

Let's multiply it using distributive property.

(4a×3a) -(4a×4b/5) +(3b/5×3a) -(3b/5×4b/5)

12a²- 16ab/5 + 9ab/5 - 12b²/25

=12a² - 7ab/5 - 12b²/25

Discussion

42.	Data by itself is use less .explain
Answer» Data is unprocessed and raw facts. It is not useful unless it is processed to obtain information.

Discussion

43.	Write a polynomial p(x) in which p(1) = 0, P(-2) = 0, p(2) = 0.
Answer» If p( 1) = 0, then x – 1 is a factor. If p(–2) = 0, then x + 2 is a factor. If p(2) = 0, then x – 2 is a factor. p(x) = (x – 1) (x + 2) (x – 2) = x³ – x² – 4x + 4

43.

Write a polynomial p(x) in which p(1) = 0, P(-2) = 0, p(2) = 0.

Answer»

If p( 1) = 0, then x – 1 is a factor.

If p(–2) = 0, then x + 2 is a factor.

If p(2) = 0, then x – 2 is a factor.

p(x) = (x – 1) (x + 2) (x – 2) = x³ – x² – 4x + 4

Discussion

44.	In the polynomial p (x) = x2 + ax + b p (3 + √2 )= 0, p (3 – √2 ) = 0, write this polynomial after finding a and b.
Answer» p (x) = x² + ax + b p(3 + √2) = 0, (x – 3 – √2) is a factor (1) p(3 – √2 ) = 0, ( x – 3 + √2) is a factor (1) p(x) = x + ax + b = (x – 3 – √2) (x – 3 + √2) = (x – 3)² – ( √2)² (1) x² + ax + b = x – 6x + 7 (1)

44.

In the polynomial p (x) = x2 + ax + b p (3 + √2 )= 0, p (3 – √2 ) = 0, write this polynomial after finding a and b.

Answer»

p (x) = x² + ax + b

p(3 + √2) = 0, (x – 3 – √2) is a factor (1)

p(3 – √2 ) = 0, ( x – 3 + √2) is a factor (1)

p(x) = x + ax + b = (x – 3 – √2) (x – 3 + √2)

= (x – 3)² – ( √2)² (1)

x² + ax + b = x – 6x + 7 (1)

Discussion

45.	Write a second degree polynomial p(x) in which \(p(\sqrt{2}+1)=p(\sqrt{2}-1)=0\)
Answer» If \(p(\sqrt{2}+1)=0\), then x\(-(\sqrt{2}+1)\) is a factor. If \(p(\sqrt{2}-1)=0\), then x\(-(\sqrt{2}-1)\) is a factor. p(x) = [x-(√2+1)][x-(√2-1)] = [x-(√2-1)][x-(√2+1)] = (x-√2)² - 1 = x² - 2√2 x+2-1 = x²-2√2x+1

45.

Write a second degree polynomial p(x) in which \(p(\sqrt{2}+1)=p(\sqrt{2}-1)=0\)

Answer»

If \(p(\sqrt{2}+1)=0\), then x\(-(\sqrt{2}+1)\) is a factor.

If \(p(\sqrt{2}-1)=0\), then x\(-(\sqrt{2}-1)\) is a factor.

p(x) = [x-(√2+1)][x-(√2-1)]

= [x-(√2-1)][x-(√2+1)]

= (x-√2)² - 1 = x² - 2√2 x+2-1

= x²-2√2x+1

Discussion

46.	Find the remainder and quotient obtained by dividing x3 – 5x2 + 7x + 3by (x + 2).
Answer» Let quotient = x² + ax + b and remainder be c, then x³ – 5x² + 7x + 3 =(x + 2)(x² + ax + b) + c + = x³ – 5x² + 7x + 3 = x³ + ax² + bx + 2x² + 2ax + 2b + c = x³ +( a + 2) x² + (b + 2a)x + 2b + c From this equation, a + 2 = –5, b + 2a = 7, 2b + c = 3 i.e., a = –7 b= 21 c = -39 quotient = x² – 7x + 21, remainder = –39

46.

Find the remainder and quotient obtained by dividing x3 – 5x2 + 7x + 3by (x + 2).

Answer»

Let quotient = x² + ax + b and remainder be c, then

x³ – 5x² + 7x + 3 =(x + 2)(x² + ax + b) + c + = x³ – 5x² + 7x + 3 = x³ + ax² + bx + 2x² + 2ax + 2b + c

= x³ +( a + 2) x² + (b + 2a)x + 2b + c From this equation,

a + 2 = –5,

b + 2a = 7, 2b + c = 3

i.e., a = –7

b= 21

c = -39

quotient = x² – 7x + 21,

remainder = –39

Discussion

47.	Write the product (x – 1) x (x + 1)Find the product of (x – 1),(x + 1),(x + 2) If the product is p(x)find p(1), (-1), p(-2) Write the solution of the equation p(x) = 0.
Answer» (x – 1)x (x + 1) = x² – 1 (x – 1) (x + 1) (x + 2)=(x² – 1) (x + 2) = x³ – x + 2x² – 2 = x³ + 2x² – x – 2 = 0 p(x) = x + 2x – x – 2 = 0 p(1) =1 + 2 – 1 – 2 = 0 p(–1) = – 1 + 2 + 1 – 2 = 0 p(–2) = – 8 + 8 + 2 – 2 = 0 1, – 1, – 2 are the solutions of the equation p(x) = 0.

47.

Write the product (x – 1) x (x + 1)Find the product of (x – 1),(x + 1),(x + 2) If the product is p(x)find p(1), (-1), p(-2) Write the solution of the equation p(x) = 0.

Answer»

(x – 1)x (x + 1) = x² – 1

(x – 1) (x + 1) (x + 2)=(x² – 1) (x + 2)

= x³ – x + 2x² – 2 = x³ + 2x² – x – 2 = 0

p(x) = x + 2x – x – 2 = 0

p(1) =1 + 2 – 1 – 2 = 0

p(–1) = – 1 + 2 + 1 – 2 = 0

p(–2) = – 8 + 8 + 2 – 2 = 0

1, – 1, – 2 are the solutions of the equation

p(x) = 0.

Discussion

48.	Expand (x – a)(x – b). If x2 – 7x + 12 = (x – a)(x – b) then find a + b.Also find ab Calculate the values of a, b. Write the factors of (x2 – 7x +12 ). Find the solutions of (x2 – 7x + 12).
Answer» (x – a) (x – b) = x² – bx – ax + ab = x² – x (a + b) + ab = x²– 7x + 12 = (x – a)(x – b) a + b = 7 ab = 12 (a + b)² – 4ab = 7² – 4 x 12 = 49 – 48 = 1 = a – b a + b = 7 a – b = 1, 2a = 8 a = 8/2 = 4 b = 7 – 4 = 3 a = 4 and b = 3. Solution of x² – 7x + 12 is x² – 7x + 12 = (x – 4) (x – 3) Solutions = 4, 3

48.

Expand (x – a)(x – b). If x2 – 7x + 12 = (x – a)(x – b) then find a + b.Also find ab Calculate the values of a, b. Write the factors of (x2 – 7x +12 ). Find the solutions of (x2 – 7x + 12).

Answer»

(x – a) (x – b) = x² – bx – ax + ab

= x² – x (a + b) + ab

= x²– 7x + 12 = (x – a)(x – b)

a + b = 7

ab = 12

(a + b)² – 4ab = 7² – 4 x 12 = 49 – 48 = 1 = a – b

a + b = 7

a – b = 1, 2a = 8 a = 8/2 = 4

b = 7 – 4 = 3 a = 4 and b = 3.

Solution of x² – 7x + 12 is x² – 7x + 12 = (x – 4) (x – 3)

Solutions = 4, 3

Discussion

49.	Factorize:a-b-a2+b2
Answer» Given polynomial is a−b−a² +b² =(a−b)−(a² −b²) =(a−b)−(a−b)(a+b) =(a−b)[1−(a+b)]

49.

Factorize:a-b-a2+b2

Answer»

Given polynomial is

a−b−a² +b² =(a−b)−(a² −b²)

=(a−b)−(a−b)(a+b)

=(a−b)[1−(a+b)]

Discussion

50.	If (x – 1) is to be a factor of p(x) = a2x2 – 4ax + 4a – 1. What should be the value of ‘a’ ?
Answer» p(1) = 0 a² – 4a + 4a – 1 = 0 a² – 1 =0 a = +1 or –1

50.

If (x – 1) is to be a factor of p(x) = a2x2 – 4ax + 4a – 1. What should be the value of ‘a’ ?

Answer»

p(1) = 0

a² – 4a + 4a – 1 = 0

a² – 1 =0

a = +1 or –1

Discussion

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Explain the basic tags in HTML with examples (any five tags).

Find the missing y-coordinate that makes the two triangles congruent.Triangle ABC: A(8,4), B(2,6), C(5, 0)Triangle MNO: M(7,4), N(1,2), O(4, y)(1 point)−4−404−2

Which triangle is congruent to the triangle with vertices (2,−2), (7,10), and (4,6)?(1 point)(−4,2), (8,7), (1,1)(−5,3), (7,−2), (3,1)(9,6), (1,8), (−1,4)(−7,−5), (−3,−2), (3,3)

(√2 + √3)2 is a?1. Rational number2. Irrational number3. Complex Number4. Integer

Total number of distinct `x epsilon [0,1]` for which `int_(0)^(x) (t^(8)+1)/(t^(8)+t^(2)+1) dt=3x-2` is ______

If `I_(1)=int_(0)^(1)(1-(1-x^(3))^(sqrt(2)))^(sqrt(3)) x^(2)dx` and `I_(2)(1-(1-x^(3))^(sqrt(2)))^(sqrt(3)+1).x^(2)dx`. Then `((I_(1))/(I_(2))-(sqrt(3)-1)/(2sqrt(2))+0.2)/10` is equal to _________

Dividing (x80 + x40 + 5) by (x8 + 1) gives the remainder as ______.A. 5B. 8C. 12D. 131. A2. B3. D4. C

Find the total numbers between 100 and 200 which are divisible by 12.1. 62. 163. 84. 12

If f : R → R is defined by f(x) = x2 - 3x + 2, find f(f(x)).

The value of \(\frac{\sqrt{243}}{\sqrt{867}}\) is:1. \(\frac{9}{19}\)2. \(\frac{8}{17}\)3. \(\frac{9}{17}\)4. \(\frac{11}{17}\)

How to count a diameter

The set that contains atomic number of only transition element is - (1) 21, 32, 53, 64 (2) 21, 25, 42, 72 (3) 9, 17, 34, 38 (4) 37, 42, 50, 64

If a number is in the form of 810 × 97 × 78, find the total number of prime factors of the given number.1. 522. 5603. 33604. 25

If y = sin[cos{tan(cot x)}], then find dy/dx.

Prove that : tan-1 x + cot-1 x = π/2

If f : R → R be a function defined by f(x) = x2, show that the function f is many one into.

\( 2 x y^{2}-4 x^{2} y, 11 x^{2} y-x y^{2},-3 x^{2} y+5 x y^{2} \)

\( \int \frac{n^{2}+4}{n+3} d n \)

Add the following 3x2 - 4x + 5 and 4x2 + 6x + 7a.

Use a suitable identity to get each of the following products: (p - 11)(p + 11)

Find the product of (4a + 3b/5) (3a - 4b/5)

Data by itself is use less .explain

Write a polynomial p(x) in which p(1) = 0, P(-2) = 0, p(2) = 0.

In the polynomial p (x) = x2 + ax + b p (3 + √2 )= 0, p (3 – √2 ) = 0, write this polynomial after finding a and b.

Write a second degree polynomial p(x) in which \(p(\sqrt{2}+1)=p(\sqrt{2}-1)=0\)

Find the remainder and quotient obtained by dividing x3 – 5x2 + 7x + 3by (x + 2).

Write the product (x – 1) x (x + 1)Find the product of (x – 1),(x + 1),(x + 2) If the product is p(x)find p(1), (-1), p(-2) Write the solution of the equation p(x) = 0.

Expand (x – a)(x – b). If x2 – 7x + 12 = (x – a)(x – b) then find a + b.Also find ab Calculate the values of a, b. Write the factors of (x2 – 7x +12 ). Find the solutions of (x2 – 7x + 12).

Factorize:a-b-a2+b2​​​​​​

If (x – 1) is to be a factor of p(x) = a2x2 – 4ax + 4a – 1. What should be the value of ‘a’ ?

Factorize:a-b-a2+b2