Class intervals 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 P 12 q 8 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-20...
Class intervals 0-20 20-40 40-60 60-80 80-100 Frequency 17 P 32 q 19 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-20 17 10...
Daily pocket allowance in Rs. 11-13 13-15 15-17 17-19 19-21 21-23 23-25 No. of children 3 6 9 13 f 5 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Daily pocket allowance in...
Class intervals 0-8 8-16 16-24 24-32 32-40 40-48 Frequency 5 3 10 P 4 2 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-8 5 4...
No. of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 No. of students 11 10 7 4 4 3 1 Solution: Class mark, xi = (upper class limit + lower class limit)/2 No. of days Frequency fi Class mark xi fixi...
Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80 No. of students 2 6 10 12 9 7 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45.5 Marks No. of students (fi)...
Wages in Rs. 45-50 50-55 55-60 60-65 65-70 70-75 75-80 No. of workers 5 8 30 25 14 12 6 Calculate their mean by short cut method. Solution: Class mark, xi = (upper class limit + lower class limit)/2...
Class intervals 0-50 50-100 100-150 150-200 200-250 250-300 frequency 4 8 16 13 6 3 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi...
Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 10 9 25 30 16 10 Solution: Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 25 Class size (h) = 10 Class Interval No....
Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 8...
Class interval 0-10 10-20 20-30 30-40 40-50 Frequency 10 6 8 12 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 10 5 50...
Marks 5 6 7 8 9 No. of students 6 a 16 13 b If the mean of the distribution is 7.2, find a and b. Solution: Marks (x) No. of students (f) fx 5 6 30 6 a 6a 7 16 112 8 13 104 9 b 9b Total Ʃf = 35+a+b...
Variate (xi) 5 7 9 11 _ 15 20 Frequency (fi) 4 4 4 7 3 2 1 Solution: Let the missing variate be x. Variate (xi) Frequency (fi) fixi 5 4 20 7 4 28 9 4 36 11 7 77 x 3 3x 15 2 30 20 1 20 Total Ʃfi =25...
Variate 5 6 7 8 9 10 11 12 Frequency 20 17 f 10 8 6 7 6 Solution: Variate (x) Frequency (f) fx 5 20 100 6 17 102 7 f 7f 8 10 80 9 8 72 10 6 60 11 7 77 12 6 72 Total Ʃf = 74+f Ʃfx = 563+7f Given mean...
Category A B C D E F G Wages (in Rs) per day 50 60 70 80 90 100 110 No. of workers 2 4 8 12 10 6 8 Category Wages in Rs. (x) No. of workers f fx A 50 2 100 B 60 4 240 C 70 8 560 D 80 12 960 E 90 10...
Numbers 60 61 62 63 64 65 66 Cumulative frequency 8 18 33 40 49 55 60 Solution: Numbers (x) Cumulative frequency Frequency (f) fx 60 8 8 60×8 = 480 61 18 18-8 = 10 61×10 = 610 62 33 33-18 = 15 62×15...
No. of heads 6 5 4 3 2 1 0 No. of tosses 20 25 160 283 338 140 34 Calculate the mean for this distribution. Solution: No. of heads (x) No. of tosses (f) fx 6 20 6×20 = 120 5 25 5×25 = 125 4 160...
Pocket money (in Rs) 60 70 80 90 100 110 120 No. of students 2 6 13 22 24 10 3 Solution: Pocket money in Rs (x) Number of students (f) fx 60 2 60×2 = 120 70 6 70×6 = 420 80 13 80×13 = 1040 90 22...
No. of matches 35 36 37 38 39 40 41 No. of boxes 6 10 18 25 21 12 8 (i) Calculate, correct to one decimal place, the mean number of matches per box. (ii) Determine how many extra matches would have...
Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1 Solution: Number (x) Frequency (f) fx 5 1 5×1 = 5 10 2 10×2 = 20 15 5 15×5 = 75 20 6 20×6 = 120 25 3 25×3 = 75 30 2 30×2 = 60 35 1 35×1 = 35 Total...
Solution: Mean of 10 numbers = 13 Sum of numbers = 13×10 = 130 Mean of remaining 15 numbers = 18 Sum of numbers = 15×18 = 270 Sum of all numbers = 130+270 = 400 Mean = sum of numbers/25 = 400/25 =...
Solution: (a)Given mean age = 13 Number of students = 33 Sum of ages = mean ×number of students = 13×33 = 429 After a girl leaves, the mean of 32 students becomes = 207/16 Now sum of ages =...
Solution: (a)Given observations are 6, y, 7, x, 14. Mean = 8 Number of observations = 5 Mean = Sum of observations/number of observations 8 = (6+y+7+x+14)/5 40 = 27+x+y 40-27 = x+y 13 = x+y y = 13-x...
Solution: (i) Marks obtained by students are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20. Number of students = 15 Mean = sum of observations / number of observations...
Solution: (a) Given observations are 5.7, 6.6, 7.2, 9.3, 6.2. Number of observations = 5 Mean = sum of observations / number of observations Mean = (5.7+6.6+7.2+9.3+6.2)/5 = 35/5 = 7 Hence the mean...
Solution: Steps to construct: Step 1: Draw a regular hexagon of sides 4cm. Step 2: Draw the angle bisector of A and B. which intersects each other at point O. Step 3: Draw OL perpendicular to AB....
Solution: Steps to construct: Step 1: Draw a line segment AB = 4cm. Step 2: At points A and B draw rays making on angle of 120o each and cut off AF = BC = 4cm. Step 3: At point C and F draw rays...
Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: At point B, draw a perpendicular and cut off BE = 2.5cm. Step 3: From, E, draw a line EF parallel to BC. Step 4: From...
Solution: (i) The point O where the angle bisectors meet is called the incenter of the triangle. (ii) The perpendicular drawn from point O to AB and CA are equal. i.e., OR and OQ. (iii) ∠ACO = ∠BCO....
Solution: Steps to construct: Step 1: Draw a line segment BC = 6.4cm. Step 2: Construct an angle of 60o at point B. Step 3: With C as center and radius CA = 5.8cm, draw an arc cutting BD at A. Step...
Solution: Steps to construct: Step 1: Draw a line segment BC = 5cm. Step 2: With Center as B and radius 5cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
Solution: Steps to construct: Step 1: Draw a line segment AB = 6cm. Step 2: At point A, draw a ray making an angle of 60o. Step 3: With B as the center and radius 5.2cm, draw an arc which intersects...
Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: With Center as B and radius 3cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
Solution: Steps to construct: Step 1: Draw a line segment AB = 4cm. Step 2: At point B, draw a ray BX making an angle of 90o and cut off BC = 6cm. Step 3: Join AC. Step 4: Draw the perpendicular...
Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: With centers B and C, draw two arcs of radius 4cm which intersects each other at point A. Step 3: Join AB and AC. Step 4:...
Solution: Steps to construct: Step 1: Draw a line segment AB = 8cm. Step 2: With center as A and radius 4cm, with center as B and radius 3cm, draw circles. Step 3: Draw the third circle AB as...
Solution: Steps to construct: Step 1: Consider a point O on a line, with center O, and radius 3cm, draw a circle. Step 2: Extend its diameters on both sides and cut off OP = OQ = 7cm. Step 3: Mark...
Solution: Steps to construct: Step 1: Mark a point O. Step 2: With center O and radius 4cm and 6cm, draw two concentric circles. Step 3: Join OA and mark its mid-point as M. Step 4: With center M...
Solution: Steps to construct: Step 1: Draw a line segment AB = 6cm. Step 2: Draw its perpendicular bisector bisecting it at point O. Step 3: With center O and radius OB, draw a circle. Step 4:...
Solution: Steps to construct: Step 1: Draw a circle with center O and radius 4cm and mark that point as A. Step 2: Take a point P such that OP = 7cm. Step 3: Bisect OB at M. Step 4: With center M...
Solution : (a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle) ∠ABC = 180o – (∠ACB + ∠BAC)….(i) In the...
(B) inthe figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD. Solution: (a) Join DB,...
Solution: (a) triangle ABC is an equilateral triangle Each angle = 60o ∠A = 60o But ∠A = ∠D (Angles in the same segment) ∠D = 600 Now ABEC is a cyclic quadrilateral, ∠A = ∠E = 180o 60o + ∠E = 180o...
Solution: Three circles with centers A, B and C touch each other externally at P, Q and R respectively and the radii of these circles are 2 cm, 3 cm and 4 cm. By joining the centers of triangle ABC...
Solution: (a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively, where BC = a, CA = b and AB = c (as showing in the given figure). As the...
Solution: (i) Join OB ∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent) OB2 = OA2 – AB2 r2 = (r + 7.5)2 – 152 r2 = r2 + 56.25 + 15r – 225 15r = 168.75 r = 11.25 Hence,...
Solution: (a) From A, AP and AS are the tangents to the circle ∴AS = AP = 6 From B, BP and BQ are the tangents ∴BQ = BP = 5 From C, CQ and CR are the tangents CR = CQ From D, DS and DR are the...
(a) In figure (i) given below, triangle ABC is circumscribed, find x. (b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x. Solution: (a) From A, AP and AQ are the tangents...
Solution: Radii of the circles are 5 cm and 2.8 cm. i.e. OP = 5 cm and CP = 2.8 cm. (i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8 = 7.8 cm. (ii) When...
Solution: Two concentric circles with center O OP and OB are the radii of the circles respectively, then OP = 5 cm, OB = 13 cm. Ab is the chord of outer circle which touches the inner circle at P....
Solution: Radius of the circle = 6 cm and length of tangent = 8 cm Let OP be the distance i.e. OA = 6 cm, AP = 8 cm . OA is the radius OA ⊥ AP Now In right ∆OAP, OP2 = OA2 + AP2 (By Pythagoras...
Solution: CT is the radius CP = 13 cm and tangent PT = 12 cm CT is the radius and TP is the tangent CT is perpendicular TP Now in right angled triangle CPT, CP2 = CT2 + PT2 [using Pythagoras axiom]...
Solution: In a circle with center O and radius 3cm and p is at a distance of 5cm. That is OT = 3 cm, OP = 5 cm OT is the radius of the circle OT ⊥ PT Now in right ∆ OTP, by Pythagoras axiom, OP2 =...
Solution: (a) Construction: Join BC, and AC then ABCD is a cyclic quadrilateral. Now in ∆DCF Ext. ∠2 = x + z and in ∆CBE Ext. ∠1 = x + y Adding (i) and (ii) x + y + x + z = ∠1 + ∠2 2 x + y + z =...
(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007) Solution: (a) In ∆PQR, ∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58° ∠RPQ = 90° – ∠PQR = 90° –...
(a) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate (i) ∠QBC (ii) ∠BCP Solution: (i) ABCD is a cyclic quadrilateral ∠A + ∠C = 180o 30o + p =...
Solution: (a) ADFE is a cyclic quadrilateral Ext. ∠FEB = ∠ADF ⇒ ∠ADF = 80° ABCD is a parallelogram ∠B = ∠D = ∠ADF = 80° or ∠ABC = 80° (b)In trapezium ABCD, AD || BC (i) ∠B + ∠A = 180° ⇒ 70° + ∠A =...
(b) In the figure given below, O is the center of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC. Solution: (a) In the given figure, ABCD is a cyclic quadrilateral ∠ADC =...
(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find: (i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008) Solution: (a) ∠DBC = 58° BD is diameter ∠DCB = 90° (Angle in...
Solution: (a) Given, ∠AOC = 150° and AD = CD We know that an angle subtends by an arc of a circle at the center is twice the angle subtended by the same arc at any point on the remaining part of the...
Solution: From the figure (i) ABCD is a cyclic quadrilateral Ext. ∠DCE = ∠BAD ∠BAD = xo Now arc BD subtends ∠BOD at the center And ∠BAD at the remaining part of the circle. ∠BOD = 2 ∠BAD = 2 x 2 x =...
(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate (i)∠CAB (ii)∠OAC Solution: Given that (a) Arc AB subtends ∠APB at the center...
Solution (a) (i) ∠PRB = ∠BAP (Angles in the same segment of the circle) ∴ ∠PRB = 35° (∵ ∠BAP = 35° given)
Solution: In ∆APB, ∠APB = 90° (Angle in a semi-circle) But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle) ∠A + 90° + 42°= 180° ∠A + 132° = 180° ⇒ ∠A = 180° – 132° = 48° But ∠A = ∠PQB (Angles in the...
Solution (a) ∠NYB = 50°, ∠YNB = 20°. In ∆YNB, ∠NYB + ∠YNB + ∠YBN = 180o 50o + 20o + ∠YBN = 180o ∠YBN + 70o = 180o ∠YBN = 180o – 70o = 110o But ∠MAN = ∠YBN (Angles in the same segment) ∠MAN = 110o...
(a) ABCD is cyclic Quadrilateral ∠B + ∠D = 1800 Y + 400 + 45o = 180o (y + 85o = 180o) Y = 180o – 85o = 95o ∠ACB = ∠ADB xo = 40 (a) Arc ADC Subtends ∠AOC at the centre and ∠ ABC at the remaining part...
Solution: (a) Construction: Join AB ∠A = ∠C = 350 (Alt Angles) ∠ABC = 35o (b) ∠AOC + reflex ∠AOC = 360o 130o + Reflex ∠AOC = 360o Reflex ∠AOC = 360o – 130o = 230o Now arc BC Subtends reflex ∠AOC at...
Solution: (i) ∠ACB = ∠ADB (Angles in the same segment of a circle) But ∠ADB = x° ∠ABC = xo Now in ∆ABC ∠CAB + ∠ABC + ∠ACB = 180o 40o + 900 + xo = 180o (AC is the diameter) 130o + xo = 180o xo =...
Solution: (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° = 42o + x + 50o = 180o = 92o + x = 180o x = 180o – 92o x = 88o (ii) In the given figure we...
Correct answer is $\mathrm{v}$ Explanation: We know the formula of force experienced by a moving charge in magnetic field : $ \begin{array}{l} \mathrm{F}=\mathrm{q}(\mathrm{v} \times \mathrm{B}) \\...
Solution: Assume $f(x)=\sec x$ So, $f(x)=\frac{1}{\cos x}$ $f(x)$ is not defined when $\cos x=0$ And $\cos x=0$ when, $x=\frac{\pi}{2}$ and odd multiples of $\frac{\pi}{2}$ like $-\frac{\pi}{2}$...
Solution: It is given that: $f(x)=\left\{\begin{array}{c} \frac{x^{n}-1}{x-1}, \text { when } x \neq 1 \\ n, \text { when } x=1 \end{array}\right.$ L.H.L. and $\mathrm{x}=1$ $\begin{array}{l} \lim...
On comparing the corresponding terms, we have \[\begin{array}{*{35}{l}} 2a\text{ }\text{ }4\text{ }=\text{ }0 \\ 2a\text{ }=\text{ }4 \\ a\text{ }=\text{ }4/2 \\ a\text{ }=\text{ }2 \\...
Given, On comparing the corresponding elements, we have \[\begin{array}{*{35}{l}} -a\text{ }=\text{ }1\Rightarrow a\text{ }=\text{ }-1 \\ -b\text{ }=\text{ }0\Rightarrow b\text{ }=\text{ }0 \\...
On comparing the corresponding elements, we have \[\begin{array}{*{35}{l}} 9\text{ }+\text{ }3p\text{ }=\text{ }0 \\ 3p\text{ }=\text{ }-9 \\ p\text{ }=\text{ }-9/3 \\ p\text{ }=\text{...
Given,
Given, Hence, its clearly seen that \[\left( A\text{ }+\text{ }B \right)\text{ }\left( A\text{ }\text{ }B \right)\text{ }\ne \text{ }{{A}^{2}}~\text{ }{{B}^{2}}\].
Comparing the corresponding elements, we have \[\begin{array}{*{35}{l}} 4\text{ }+\text{ }2a\text{ }=\text{ }18 \\ 2a\text{ }=\text{ }18\text{ }\text{ }4\text{ }=\text{ }14 \\ a\text{...
Given,
Given \[3\left[ \begin{align} & a\,\,\,\,\,\,b \\ & c\,\,\,\,\,\,\,d \\ \end{align} \right]=\left[ \begin{align} & 4\,\,\,\,\,\,\,\,\,\,a+b \\ & c+d\,\,\,\,\,\,\,3 \\ \end{align}...
Given \[\left[ \begin{align} & a+3\,\,\,\,\,\,\,{{b}^{2}}+2 \\ & \,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6 \\ \end{align} \right]=\left[ \begin{align} & 2a+1\,\,\,\,\,\,\,\,\,\,\,\,3b \\...
Given, On comparing the corresponding elements, we have \[2x\text{ }+\text{ }y\text{ }=\text{ }3\]… (i) \[3x\text{ }+\text{ }y\text{ }=\text{ }2\]… (ii) Subtracting, we get \[-x\text{ }=\text{...
Comparing the corresponding elements, \[\begin{array}{*{35}{l}} \text{ }3x\text{ }+\text{ }4\text{ }=\text{ }-5 \\ -3x\text{ }=\text{ }-5\text{ }\text{ }4\text{ }=\text{ }-9 \\ x\text{ }=\text{...
Given, On comparing, \[\begin{array}{*{35}{l}} 1\text{ }+\text{ }x\text{ }=\text{ }0 \\ \therefore x\text{ }=\text{ }-1 \\ \end{array}\]
Given
Given,
Given, – Hence proved
Given, From above, its clearly seen that \[{{A}^{3}}~=\text{ }A\].
Given,
Given
Now consider,
Yes, the product is possible because of number of column in A = number of row in B That is order of matrix is \[2\text{ }\times \text{ }1\]
On comparing the corresponding terms, we get \[\begin{array}{*{35}{l}} 5a\text{ }+\text{ }6\text{ }=\text{ }9 \\ 5a\text{ }=\text{ }9\text{ }\text{ }6 \\ 5a\text{ }=\text{ }3 \\ a\text{ }=\text{...
Next, on comparing the corresponding terms, we have \[\begin{array}{*{35}{l}} a\text{ }+\text{ }1\text{ }=\text{ }5\Rightarrow a\text{ }=\text{ }4 \\ b\text{ }+\text{ }2\text{ }=\text{...
Now, comparing the corresponding terms, we get \[\begin{array}{*{35}{l}} 4\text{ }\text{ }4x\text{ }=\text{ }-8 \\ 4\text{ }+\text{ }8\text{ }=\text{ }4x \\ 12\text{ }=\text{ }4x \\ x\text{...
On comparing the corresponding terms, we have \[\begin{array}{*{35}{l}} 2x\text{ }+\text{ }1\text{ }=\text{ }5 \\ 2x\text{ }=\text{ }5\text{ }-1\text{ }=\text{ }4 \\ x\text{ }=\text{ }4/2\text{...
On comparing the corresponding elements, we have \[\begin{array}{*{35}{l}} 8\text{ }+\text{ }y\text{ }=\text{ }0 \\ Then,\text{ }y\text{ }=\text{ }-8 \\ And,\text{ }2x\text{ }+\text{ }1\text{...
Given \[\left[ \begin{align} & 2\,\,\,\,\,\,\,1 \\ & 5\,\,\,\,\,\,0 \\ \end{align} \right]-3X=\left[ \begin{align} & -7\,\,\,\,\,\,\,4 \\ & 2\,\,\,\,\,\,\,\,\,\,6 \\ \end{align}...
Given,
Given \[\left[ \begin{align} & a+b\,\,\,\,\,3 \\ & 5+c\,\,\,\,\,ab \\ \end{align} \right]=\left[ \begin{align} & 6\,\,\,\,d \\ & -1\,\,\,8 \\ \end{align} \right]\] Comparing the...
Given \[\left[ \begin{align} & x-2\,\,\,\,\,\,y \\ & a+2b\,\,\,\,\,3a-b \\ \end{align} \right]=\left[ \begin{align} & 3\,\,\,\,\,1 \\ & 5\,\,\,\,\,1 \\ \end{align} \right]\]...
Given \[\left[ \begin{align} & x+2\,\,\,\,\,\,\,\,6 \\ & 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,5z \\ \end{align} \right]=\left[ \begin{align} & -5\,\,\,\,\,\,{{y}^{2}}+y \\ &...
Given \[\left[ \begin{align} & x+3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \\ & y-4\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+y \\ \end{align} \right]=\left[ \begin{align} & 5\,\,\,\,4 \\ & 3\,\,\,\,9 \\...
Given \[\left[ \begin{align} & 3x+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-y \\ & 2y-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 \\ \end{align} \right]=\left[ \begin{align} & 1\,\,\,\,\,2 \\ &...
Given \[\left[ \begin{align} & 2x+y \\ & 3x-2y \\ \end{align} \right]=\left[ \begin{align} & 5 \\ & 4 \\ \end{align} \right]\] Now by comparing the corresponding elements, \[2x\text{...
(i) Given \[{{\mathbf{a}}_{\mathbf{ij}}}~=\text{ }\mathbf{2i}\text{ }\text{ }\mathbf{j}\] Therefore matrix of order \[\mathbf{2}\text{ }\times \text{ }\mathbf{2}\]is \[\left[ \begin{align} &...
It can have \[1\text{ }\times \text{ }4,\text{ }4\text{ }\times \text{ }1\text{ }or\text{ }2\text{ }\times \text{ }2\] order. (ii) If a matrix has \[4\] elements, what are the possible orders it can...
It is a matrix of order \[2\text{ }\times \text{ }3\] (vi) \[\left[ \begin{align} & 0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,0\, \\ & 0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,0 \\ \end{align} \right]\] Solution:...
It is column matrix of order \[3\text{ }\times \text{ }1\] (iv) \[\left( \begin{align} & 2\,\,\,-4 \\ & 0\,\,\,\,\,\,0 \\ & 1\,\,\,\,\,\,\,7 \\ \end{align} \right)\] Solution: It is a...
It is square matrix of order \[2\] (ii)\[[2,3,-7]\] Solution: It is row matrix of order \[1\text{ }\times \text{ }3\]
$\begin{array}{l} x^{2}+6 x-\left(a^{2}+2 a-8\right)=0 \\ \Rightarrow x^{2}+6 x-(a+4)(a-2)=0 \\ \Rightarrow x^{2}+[(a+4)-(a-2)] x-(a+4)(a-2)=0 \\ \Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0 \\...
$\begin{array}{l} 4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0 \\ \Rightarrow 4 x^{2}+4 b x-(a-b)(a+b)=0 \\ \Rightarrow 4 x^{2}+2[(a+b)-(a-b)] x-(a-b)(a+b)=0 \\ \Rightarrow 4 x^{2}+2(a+b) x-2(a-b)...
$\begin{array}{l} \sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x^{2}-3 \sqrt{2} x+\sqrt{2} x-2 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})=0 \\...
$\begin{array}{l} 3 x^{2}+5 \sqrt{5} x-10=0 \\ \Rightarrow 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-10=0 \\ \Rightarrow 3 x(x+2 \sqrt{5})-\sqrt{5}(x+2 \sqrt{5})=0 \\ \Rightarrow(x+2 \sqrt{5})(3 x-\sqrt{5})=0...
$\begin{array}{l} x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0 \\ \Rightarrow x^{2}-\sqrt{3} x-x+\sqrt{3}=0 \\ \Rightarrow x(x-\sqrt{3})-1(x-\sqrt{3})=0 \\ \Rightarrow(x-\sqrt{3})(x-1)=0 \\ \Rightarrow...
It is given that the quadratic equation $9 x^{2}-3 k x+k=0$ has equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-3 k)^{2}-4 \times 9 \times k=0 \\ \Rightarrow 9 k^{2}-36 k=0 \\...
It is given that the roots of the quadratic equation $p x^{2}-2 p x+6=0$ are equal. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-2 p)^{2}-4 \times p \times 6=0 \\ \Rightarrow 4 p^{2}-24 p=0 \\...
Let the other zero of the given polynomial be $\alpha$. Now, Sum of the zeroes of the given polynomial $=\frac{-(-4)}{1}=4$ $\begin{array}{l} \therefore \alpha+(2+\sqrt{3})=4 \\ \Rightarrow...
It is given that the quadratic equation $p x^{2}-2 \sqrt{5} p x+15=0$ has two equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-2 \sqrt{5} p)^{2}-4 \times p \times 15=0 \\ \Rightarrow 20...
The given quadratic equation is $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$ $\begin{array}{l} 3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0 \\ \Rightarrow 3 \sqrt{3} x^{2}+9 x+x+\sqrt{3}=0 \\ \Rightarrow 3 \sqrt{3}...
From the question it is given that, Polynomial f(x) is divided by \[(x-1)\], Remainder = \[5\] Let us assume \[x-1=0\] x = \[1\] \[f\left( 1 \right)\text{ }=\text{ }5\] and the divided be \[(x-2)\],...
From the question it is given that, f(x) = \[{{\mathbf{x}}^{\mathbf{4}}}-\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\text{ }\mathbf{ax}\text{ }+\text{...
Let us assume \[2x\text{ }+\text{ }1\text{ }=\text{ }0\] Then, \[2x\text{ }=\text{ }-1\] \[x\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2}\] Given, p(x) =...
Let us assume \[x+3=0\] Then, x = \[-3\] Given, f(x) = \[{{x}^{3}}~+\text{ }a{{x}^{2}}~\text{ }bx\text{ }+\text{ }24\] Now, substitute the value of x in f(x), \[\begin{array}{*{35}{l}} f\left( -3...
From the question it is given that, \[(x+2)\] is a factor of the expression \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }+\text{...
Let us assume x = \[-1\], Given, f(x) = \[\mathbf{4}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{9x}\text{ }\text{ }\mathbf{9}~\] Now, substitute the...
Let us assume \[\left( 5x\text{ }+\text{ }4 \right)\text{ }=\text{ }0\] Then, \[5x=-4\] x = \[-4/5\] Given, f(x) = \[\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{...
Let us assume \[x\text{ }\text{ }2\text{ }=\text{ }0\] Then, x = \[2\] Given, f(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }+\text{ }\mathbf{p}\] Now, substitute the...
Let us assume 2x – 3 = 0 Then, \[2x=3\] \[x=3/2\] Given, f(x) = \[6{{x}^{2}}~+\text{ }x\text{ }+\text{ }a\] Now, substitute the value of x in f(x), \[\begin{array}{*{35}{l}} f\left( 3/2...
Let us assume \[x+1=0\] Then, \[x=-1\] Given, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{10x}\text{ }\text{ }\mathbf{p}\] Now,...
From the question it is given that, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{7}\] (iii) consider...
From the question it is given that, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{7}\] (i) Consider...
From the question it is given that, \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{bx}\text{ }+\text{ }\mathbf{2}\text{...
Let us assume, \[\begin{array}{*{35}{l}} ~2x\text{ }+\text{ }3\text{ }=\text{ }0 \\ Then,\text{ }2x\text{ }=\text{ }-3 \\ x\text{ }=\text{ }-3/2 \\ \end{array}\] Given, f(x) =...
From the question it is given that, \[(x-2)\] is a factor of the expression \[2{{x}^{3}}~+\text{ }a{{x}^{2}}~+\text{ }bx\text{ }\text{ }14\] Then, f(x) = \[2{{x}^{3}}~+\text{ }a{{x}^{2}}~+\text{...
From the question it is given that, \[(x-2)\] is a factor of the expression \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{bx}\text{ }+\text{...
Let us assume \[x\text{ }+\text{ }2\text{ }=\text{ }0\] Then, x = \[-2\] Given, f(x) = \[{{x}^{3}}~+\text{ }ax\text{ }+\text{ }b\] Now, substitute the value of x in f(x), \[\begin{array}{*{35}{l}}...
Let us assume \[x\text{ }\text{ }2\text{ }=\text{ }0\] Then, x = \[2\] Given, f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{bx}\text{ }\text{...
Let us assume the number to be subtracted from \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{5x}\] be p. Then, f(x) =...
Let us assume \[x\text{ }-2\text{ }=\text{ }0\] Then, \[x=2\] Given, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{...
Let us assume \[3x\text{ }\text{ }2\text{ }=\text{ }0\] Then, \[\begin{array}{*{35}{l}} 3x\text{ }=\text{ }2 \\ X\text{ }=\text{ }2/3 \\ \end{array}\] Given, f(x) =...
Let us assume \[2x+1=0\] Then, \[2x=-1\] \[X\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2}\] Given, f(x) = \[\mathbf{6}{{\mathbf{x}}^{\mathbf{3}}}~+\text{...
Given, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{9x}\text{ }\text{ }\mathbf{10}\] Let us assume, x = \[-1\]...
Let us assume x = \[2\], Then, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\] Now, substitute the value of x in...
Let us assume \[x=-1\], Given, f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{6}\] Now, substitute the value of x in...
Let us assume \[2x+7=0\] Then, \[\begin{array}{*{35}{l}} 2x\text{ }=\text{ }-7 \\ X\text{ }=\text{ }-7/2 \\ \end{array}\] Given, f(x) = \[2{{x}^{3}}~+\text{ }5{{x}^{2}}~\text{ }11x\text{ }\text{...
Let us assume \[x\text{ }\text{ }2\text{ }=\text{ }0\] Then, x = \[2\] Given, f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{4x}\text{ }\text{...
Let us assume \[x\text{ }\text{ }2\text{ }=\text{ }0\] Then, x = \[2\] Given, f(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{x}\text{ }\text{ }\mathbf{10}\] Now, substitute the value...
Consider \[{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{2x}\text{ }\text{ }\mathbf{3}\] By factor method, \[{{x}^{2}}~+\text{ }3x\text{ }\text{ }x\text{ }\text{ }3\] \[\begin{array}{*{35}{l}}...
Let us assume, \[x\text{ }+\text{ }3\text{ }=\text{ }0\] Then, \[x\text{ }=\text{ }\text{ }3\] Given, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{3}\]...
Let us assume p(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{kx}\text{ }+\text{ }\mathbf{8} \right)\mathbf{x}\text{ }+\text{ }\mathbf{k}\] From the question it is given that, the sum...
Let us assume p(x) = \[\mathbf{a}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\] and q(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{...
From the question it is given that, by dividing \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{p}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{6}\]and...
let us assume ‘p’ be subtracted from \[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\] So, dividing \[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\] by \[\mathbf{2x}\text{...
Let us assume \[x\text{ }-1\text{ }=\text{ }0\] Then, x = \[1\] Given, f(x) = \[{{x}^{3}}~+\text{ }5{{x}^{2}}~\text{ }ax\text{ }+\text{ }6\] Now, substitute the value of x in f(x),...
Let us assume, \[x\text{ }\text{ }2\text{ }=\text{ }0\] Then, x = \[2\] Given, \[2{{x}^{3}}~+\text{ }3{{x}^{2}}~\text{ }kx\text{ }+\text{ }5\] Now, substitute the value of x in f(x),...
Let us assume \[~2x\text{ }+\text{ }1\text{ }=\text{ }0\] Then, \[2x\text{ }=\text{ }-1\] \[X\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2}\] Given, f(x) = \[4{{x}^{2}}~+\text{ }5x\text{...
Let us assume \[x\text{ }+\text{ }3\text{ }=\text{ }0\] Then, x = \[-3\] Given, f(x) =\[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }+\text{ }\mathbf{1}\] Now, substitute the...
Let us assume \[x\text{ }\text{ }2\text{ }=\text{ }0\] Then, x = \[2\] Given, f(x) = \[\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{7x}\text{ }+\text{ }\mathbf{4}\] Now, substitute the...
The given quadratic equation is $2 x^{2}-x-6=0$ $\begin{array}{l} 2 x^{2}-x-6=0 \\ \Rightarrow 2 x^{2}-4 x+3 x-6=0 \\ \Rightarrow 2 x(x-2)+3(x-2)=0 \\ \Rightarrow(x-2)(2 x+3)=0 \\ \Rightarrow x-2=0...
The given equation is $3 x^{2}+13 x+14=0$. Putting $x=-2$ in the given equation, we get $L H S-3 \times(-2)^{2}+13 \times(-2)+14=12-26+14=0=R H S$ $\therefore x=-2$ is a solution of the given...
Answer is (b) $2, \frac{-3}{2}$ The given quadratic equation is $2 x^{2}-x-6=0$. $\begin{array}{l} 2 x^{2}-x-6=0 \\ \Rightarrow 2 x^{2}-4 x+3 x-6=0 \\ \Rightarrow 2 x(x-2)+3(x-2)=0 \\...
Answer is (c) $\frac{-8}{5}<k<\frac{8}{5}$ It is given that the equation $\left(x^{2}+5 k x+16=0\right)$ has no real roots. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)<0 \\...
Answer is (b) real and unequal We know that when discriminant, $D>0$, the roots of the given quadratic cquation are real and uncqual.
Answer is (a) 1 or 4 It is given that the roots of the equation $\left(x^{2}+2(k+2) x+9 k=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow\{2(k+2)\}^{2}-4...
Answer is (d) $\frac{b^{2}}{4 a}$ It is given that the roots of the equation $\left(a x^{2}+b x+c=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow b^{2}=4 a...
Answer is (c) $-4$ It is given that $\alpha$ and $\beta$ are the roots of the equation $3 x^{2}+8 x+2=0$ $\therefore \alpha+\beta=-\frac{8}{3}$ and $\alpha \beta=\frac{2}{3}$...
Answer is (d) $\frac{-2}{3}$ Given: $k x^{2}+2 x+3 k=0$ Sum of the roots $=$ Product of the roots $\begin{array}{l} \Rightarrow \frac{-2}{k}=\frac{3 k}{k} \\ \Rightarrow 3 k=-2 \\ \Rightarrow...
Answer is (d) $2: 3$ Given: $7 x^{2}-12 x+18=0$ $\therefore \alpha+\beta=\frac{12}{7}$ and $\beta=\frac{18}{7}$, where $\alpha$ and $\beta$ are the roots of the equation $\therefore$ Ratio of the...
Answer is (b) -2 Sum of the roots of the equation $x^{2}-6 x+2=0$ is $\alpha+\beta=\frac{-b}{a}=\frac{-(-6)}{1}=6$, where $\alpha$ and $\beta$ are the roots of the equation.
Answer is (b) $-11$ It is given that $x=3$ is a solution of $3 x^{2}+(k-1) x+9=0$; therefore, we have: $\begin{array}{l} 3(3)^{2}+(k-1) \times 3+9=0 \\ \Rightarrow 27+3(k-1)+9=0 \\ \Rightarrow...
Answer is (c) $(\sqrt{2} x+3)^{2}=2 x^{2}+6$ $\begin{array}{l} \because(\sqrt{2} x+3)^{2}=2 x^{2}+6 \\ \Rightarrow 2 x^{2}+9+6 \sqrt{2} x=2 x^{2}+6 \end{array}$ $\Rightarrow 6 \sqrt{2} x+3=0$, which...
Let the shortest side be $x \mathrm{~m}$. Therefore, according to the question: Hypotenuse $=(2 x-1) m$ Third side $=(x+1) m$ On applying Pythagoras theorem, we get: $\begin{array}{l} (2...
Let one side of the right-angled triangle be $x \mathrm{~m}$ and the other side be $(x+4) \mathrm{m}$. On applying Pythagoras theorem, we have: $\begin{array}{l} 20^{2}=(x+4)^{2}+x^{2} \\...
Let the altitude of the triangle be $x \mathrm{~m}$. Therefore, the base will be $3 x \mathrm{~m}$. $\begin{array}{l} \text { Area of a triangle }=\frac{1}{2} \times \text { Base } \times \text {...
Let the length and breadth of the rectangular garden be $x$ and $y$ meter, respectively. Given: $x y=180 \mathrm{sq} \mathrm{m}$$\ldots(i)$ and $\begin{array}{l} 2 y+x=39 \\ \Rightarrow x=39-2 y...
Let the length of the side of the first and the second square be $x$ and $y \cdot$ respectively. According to the question: $x^{2}+y^{2}=640$ Also, $\begin{array}{l} 4 x-4 y=64 \\ \Rightarrow x-y=16...
Let the length and breadth of the rectangular plot be $x$ and $y$ meter, respectively. Therefore, we have: $\begin{array}{l} \text { Perimeter }=2(x+y)=62 \quad \ldots . .(i) \text { and } \\ \text...
Let the breath of the rectangular hall be $x$ meter. Therefore, the length of the rectangular hall will be $(x+3)$ meter. According to the question: $\begin{array}{l} x(x+3)=238 \\ \Rightarrow...
Let the tap of smaller diameter fill the tank in $x$ hours. $\therefore$ Time taken by the tap of larger diameter to fill the tank $=(x-9) h$ Suppose the volume of the tank be $V$. Volume of the...
Let one pipe fills the cistern in $x$ mins. Therefore, the other pipe will fill the cistern in $(x+3)$ mins. Time taken by both, running together, to fill the cistern $=3 \frac{1}{13} \min...
Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. $\therefore$ Downstream speed $=(9+x) \mathrm{km} / \mathrm{hr}$ Upstream speed $=(9-x) \mathrm{km} / \mathrm{hr}$ Distance covered...