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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2101. |
√5 prove it is irrational |
| Answer» Let take √5 as a rational numberIf a and b are two co-prime number and b is not equal to 0.We can write √5 = a/bMultiply by b both side we getb√5 = aTo remove root, Squaring on both sides, we get5b2 = a2 ……………(1)Therefore, 5 divides a2 and according to a theorem of rational number, for any prime number p which is divided \'a2\' then it will divide \'a\' also.That means 5 will divide \'a\'. So we can writea = 5cand putting the value of a in equation (1) we get5b2 = (5c)25b2 = 25c2Divide by 25 we getb2/5 = c2again using the same theorem we get that b will divide by 5and we have already get that a is divided by 5but a and b are co-prime number. so it is contradicting.Hence √5 is an irrational number | |
| 2102. |
Find the coordinates of the point on y-axis which is nearest to the point (–2, 5) |
| Answer» The point on y-axis that is nearest to the point(-2,5) is (0,5). | |
| 2103. |
What is a lemma |
| Answer» Lemma:It is a small proof and a stepping stone in proving a theorem. It is a small hypothesis which is proven and used to demonstrate a principal proposition. | |
| 2104. |
Solve the question(-4)+(-1)+2+5+.....+x=437Find x |
| Answer» Ans:50 | |
| 2105. |
Which term of the progression 19,181/5, |
| Answer» \xa0According to condition the given arithmetic progression is 19,18{tex}\\frac{1}{5}{/tex},17{tex}\\frac{2}{5}{/tex}.........(i)Here, T2\xa0- T1\xa0=\xa0{tex}\\frac { 91 } { 5 } - 19 = \\frac { 91 - 95 } { 5 } = - \\frac { 4 } { 5 }{/tex}T3\xa0- T2\xa0=\xa0{tex}\\frac { 87 } { 5 } - \\frac { 91 } { 5 } = - \\frac { 4 } { 5 }{/tex}Therefore, (i) is an arithmetic progression with a = 19, d =\xa0{tex}-\\frac{4}{5}{/tex}Suppose, the nth term of the given arithmetic progression be the first negative term. Then, nth term < 0.{tex}\\Rightarrow{/tex}\xa0Tn\xa0< 0\xa0{tex}\\Rightarrow{/tex}\xa0[ 19 + (n - 1){tex}\\left( - \\frac { 4 } { 5 } \\right){/tex}] < 0{tex}\\Rightarrow{/tex}\xa0(99 - 4n) < 0\xa0{tex}\\Rightarrow{/tex}\xa04n > 99\xa0{tex}\\Rightarrow{/tex}\xa0n >\xa0{tex}24 \\frac { 3 } { 4 }{/tex}.{tex}\\therefore{/tex}\xa0n = 25,i.e., 25th is the first negative term in the given AP. | |
| 2106. |
Find the value of k in quadratic equation is=3x+k√3x_4=0 |
| Answer» D=b2-4ac SE solve karo where D=0 | |
| 2107. |
X^4+2x^3-13x^2-38x-24. Zeroes= -2,-1,?,? Find other 0s |
| Answer» | |
| 2108. |
Evaluate sin 30 ÷ cos 30+1÷cos30 |
| Answer» May be sin30 +1 | |
| 2109. |
Cos*4A/Cos*2B+Sin*4A/Sin*2B=1To prove Cos*4B/Cos*2A+Sin*4B/Sin*2A=1 |
| Answer» | |
| 2110. |
2×7×9×8÷3 |
| Answer» 336 | |
| 2111. |
(a-b) whole cube identity |
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Answer» a cube -b cube -3ab (a-b) (a-b)³=a³-b³-3ab(a-b) |
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| 2112. |
How can we find h in statistics when there is uneven class interval |
| Answer» | |
| 2113. |
Prove volume of frustum of cone???????? |
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| 2114. |
In a leap year what is the probability of 53 Sundays? |
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Answer» 1/7 Always 2/7 wheather it is Sunday Monday or something else Answer is 2/7. 0 |
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| 2115. |
What is the value of a in statistics |
| Answer» Assumed no. | |
| 2116. |
Pythagoras theoram |
| Answer» According to Pythagoras property, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the base and perpendicular. | |
| 2117. |
Find the coordinates of the point equidistant from the point A(5,3),B(5,-5)and C(1,-5) |
| Answer» Let the required points be P(x, y), thenPA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively{tex} \\Rightarrow{/tex}\xa0PA2\xa0= PB2\xa0= PC2{tex}\\Rightarrow{/tex}\xa0PA2 = PB2 and PB2 = PC2PA2 = PB2{tex}\\Rightarrow{/tex}\xa0(5 - x)2\xa0+ (3 - y)2\xa0= (5 - x)2\xa0+ (-5 -y)225 + x2\xa0- 10x + 9 + y2 - 6y = 25 + x2\xa0\xa0- 10x + 25 + y2\xa0+ 10y-6y - 10y = 25 - 9\xa0{tex}\\Rightarrow{/tex}\xa0-16y = 16y = -1and PB2 = PC2{tex}\\Rightarrow{/tex}\xa0(5 - x)2\xa0+ (-5 - y)2\xa0= (1 - x) + (-5 - y)225 + x2 - 10x + 25 + y2 + 10y = 1 + x2 - 2x + 25 + y2 + 10y-10x + 2x = -24\xa0{tex}\\Rightarrow{/tex}\xa0-8x = -24{tex}x = \\frac { - 24 } { - 8 } = 3{/tex}Hence, the point P is (3, -1) | |
| 2118. |
Find the hcf and LCM 90 , 144 by prime factorization |
| Answer» 90= 3×3×2×5144= 2×2×2×2×3×3HCF=18LCM=720 | |
| 2119. |
PP\'and qq\' are the two common tangents of the two circles. Then show that pp\' =qq\' |
| Answer» | |
| 2120. |
tanA=ntanB and sinA=msinB prove that m2-1/n2-1 |
| Answer» | |
| 2121. |
The two vertices of equilateral triangle are (-2,0) and (2,0) find the third vertex |
| Answer» Let ABC be the ∆ then AB=AC=BC => AB^2= BC^2 = AC^2 .let x and y be the third vertex. AC^2 =AB^2.by distance formula (x+2)^2 +y^2 =(x-2)^2+y^2.x^2+4x+4=x^2-4x+4.8x=0x=0Now AB^2 =(2+2)^2+0 =16AB^2=AC^2 (0+2)^2+y^2 =16y^2=12. y=2√3. The third vertex =(0,2√3) | |
| 2122. |
52 card vale question samaj me ni ar re he |
| Answer» | |
| 2123. |
Compelet squar method |
| Answer» | |
| 2124. |
Find the AP a=3,d=6,l=68 |
| Answer» | |
| 2125. |
If Sn, denote the sum of first n, terms of an A. P, prove that S30=3(S20-s10) |
| Answer» Let a be first term and d be common difference, thenS30\xa0={tex}\\frac{30}{2}{/tex}( 2a + 29d){tex}= 15(2a + 29d){/tex}.....(i)3(S20- S10) = 3 [{tex}\\;\\frac{20}2{/tex} (2a + 19d) - {tex}\\;\\frac{10}2{/tex}(2a + 9d)]{tex}= 3[10 (2a + 19d) - 5(2a + 9d)]{/tex}{tex}= 3[20a + 190d - 10a - 45d]{/tex}{tex}= 3[10a + 145d]{/tex}{tex} = 15[2a + 29d]{/tex} ...(ii)From (i) and (ii){tex}\\therefore{/tex}\xa0Hence S30 = 3(S20 - S10) | |
| 2126. |
(a2+b2) |
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Answer» a^2+b^2+2ab ( a+ b)2 -2ab |
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| 2127. |
The zeroes of the quardratic polynomial x² +99x+98 are? |
| Answer» (X-98)(x+99) | |
| 2128. |
If alpha -bita,alpha and alpha +vita are zeroes of xcube -6xsquare+ 8x,then find value of bitta |
| Answer» | |
| 2129. |
How many questions comes from ncert book |
| Answer» | |
| 2130. |
In a right triangle abc,right -angled at B, if tan A= 1 ,then verify that2 sin A cos A =1. |
| Answer» In {tex} \\triangle A B C{/tex},{tex}\\tan A = 1{/tex}{tex}\\Rightarrow \\quad \\frac { B C } { A C } = 1{/tex}{tex}\\Rightarrow {/tex}\xa0BC = x and\xa0AC = xUsing Pythagoras theorem,\xa0{tex}\\Rightarrow A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 2 } x{/tex}{tex}\\therefore \\quad \\sin A = \\frac { B C } { A B } = \\frac { x } { \\sqrt { 2 } x } = \\frac { 1 } { \\sqrt { 2 } } \\text { and } \\cos A = \\frac { A C } { \\sqrt { 2 } x } = \\frac { x } { \\sqrt { 2 } x } = \\frac { 1 } { \\sqrt { 2 } } {/tex}2 sin A cos A\xa0{tex}= 2 \\times \\frac { 1 } { \\sqrt { 2 } } \\times \\frac { 1 } { \\sqrt { 2 } } = 1{/tex} | |
| 2131. |
Find the discrimiant of quadratic equations 6ײ-7×+2=0 |
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Answer» discriminant is (7)^2 -4(6*2) = 0 49-48=5>0then x= -(7)+- under root of 5 /2*6= 1.. Its simple yrr |
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| 2132. |
Find the largest number which divides 378 and 510 leaving remainder 6 in each case |
| Answer» 12 | |
| 2133. |
SinA=cosA to A ka man nikaliye |
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Answer» Thank you 45 |
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| 2134. |
12×46 |
| Answer» 552 | |
| 2135. |
The area of the shaded segment made by the cord PQ |
| Answer» 210 sq.m | |
| 2136. |
show that n,n+2,n+4 js divisible by 3. |
| Answer» Let n =3kthen n + 2 = 3k + 2and n + 4 = 3k + 4Case 1: When n=3k ,n is divisible by 3 ............(1)n + 2 = 3k + 2or, n + 2 is not divisible by 3n + 4 = 3k + 4= 3(k + 1) + 1or, n + 4 is not divisible by 3Case 2:When n=3k+1, n is not divisible by 3\xa0n + 2 = (3k + 1) + 2=3k + 3 = 3(k + 1){tex} \\Rightarrow{/tex}\xa0n+ 2 is clearly divisible by 3..........................(2)n + 4 = (3k + 1) + 4= 3k + 5= 3(k + 1) + 2{tex}\\Rightarrow{/tex}\xa0n + 4 is not divisible by 3Case 3:When n=3k+2,n is not divisible by 3\xa0n + 2 = (3k + 2) + 2= 3k + 4(n + 2) is not divisible by 3x + 4 = 3k + 6 = 3(k + 2){tex}\\Rightarrow{/tex}\xa0n + 4 is divisible by 3........................(3)Hence, from (1),(2) and (3) it is clear that\xa0exactly one of the numbers n, n + 2, n + 4, is divisible by 3. | |
| 2137. |
Parabola |
| Answer» | |
| 2138. |
Write the greatest integer smaller than 2root5 and smallest integer greater than 2root5. |
| Answer» Cg | |
| 2139. |
4x²+6x² |
| Answer» 10x² | |
| 2140. |
Prove that line drawn from centre to the point of contact of a tangent is perpendicular to it |
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Answer» 1.Join center to that point from the tangent are drawn and prove both triangle congruent.2.By SAS congruence Tangents drawn from an external points to a circle is perpendicular to the radius Perpendicular is equal to the tangents |
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| 2141. |
Show that point (4,1),(6,6),(7,2),(5,1) are the vertices of square or not. |
| Answer» | |
| 2142. |
Fghggb |
| Answer» fuc.... | |
| 2143. |
2+2=5 Can anyone rove it |
| Answer» | |
| 2144. |
Divide 264 ÷25 with prove |
| Answer» 265/25=step 1 we will say the table of 25 upto 10 their 25 ×10 equals to 250 then subtratct 250 from.264 so remainder will be 14 and quotient will be 10Step2 then taking pointvin quotient add zero to 14 it will become 140 so multply 25 by 5 then remainder will be 15 and quotiet will be 10.5 Step 3 then multiply 25 by 6 and subtract 150 from 150 remainder will be 0 and quotient will be 10.56 # proovedFinal answer will be 10.56 | |
| 2145. |
Tan+sin/tan-sin=sec+1/sec-1 |
| Answer» {tex}\\text { L.H.S. } = \\frac { \\tan \\theta + \\sin \\theta } { \\tan \\theta - \\sin \\theta } = \\frac { \\frac { \\sin \\theta } { \\cos \\theta } + \\sin \\theta } { \\frac { \\sin \\theta } { \\cos \\theta } - \\sin \\theta }{/tex}{tex}= \\frac { \\sin \\theta \\left( \\frac { 1 } { \\cos \\theta } + 1 \\right) } { \\sin \\theta \\left( \\frac { 1 } { \\cos \\theta } - 1 \\right) } = \\frac { \\sec \\theta + 1 } { \\sec \\theta - 1 } = R . H S{/tex} | |
| 2146. |
I am week in math |
| Answer» | |
| 2147. |
Hello where are u??? |
| Answer» Who?? | |
| 2148. |
The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm find the area of the sector |
| Answer» Let OAB be the given sector.It is given that Perimeter of sector OAB = 16.4 cm{tex}\\Rightarrow{/tex}OA + OB + arc AB = 16.4 cm{tex}\\Rightarrow{/tex}5.2 + 5.2 + arc AB = 16.4{tex}\\Rightarrow{/tex}arc AB = 6 cm\xa0{tex}\\Rightarrow{/tex}l = 6 cm{tex}\\therefore{/tex}Area of sector OAB ={tex}\\frac { 1 } { 2 } l r = \\frac { 1 } { 2 } \\times 6 \\times 5.2 \\mathrm { cm } ^ { 2 }{/tex}= 15.6 cm2 | |
| 2149. |
Prove that pie (22/7) is an irrational number. |
| Answer» | |
| 2150. |
Which term of AP 134, 129,124,....... is the first negative term |
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Answer» a=134, d=-5Let the nth term of the given AP be the first negative term. ThenTn<0{a+(n-1)d}<0{134+(n-1)(-5)}<0134-5n+5<0139-5n<05n>139n>139/5n>27.8n=2828th term is the first negative term Let its xth term is first negative term so A=a+(n-1)d. A=129+(n-1)(-9) A=129-9n+9. A=138-9n. Aage kudh try karo 28 |
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