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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. |
sinx**3 - cosx**3 is equals to |
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| 2102. |
Maths ch 3 extras questions |
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| 2103. |
Unit 1 miscellanous exercise |
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| 2104. |
why sinx+sin5x=sin6x |
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Answer» Sin (x+5x)=sin6x Sin (x+5x)Sin6x Sina +sinb indentity |
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| 2105. |
Find the value of cos 765° |
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Answer» 1/√2 -1/root 2 Cos (360×2+45)Cos (45)Cos (1/root2 1/√2 |
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| 2106. |
Venn diagram |
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| 2107. |
Prove tan3x tan2x tanx= tan3x - tan2x - tanx |
| Answer» Eg. Dekh le bc | |
| 2108. |
How can solve mode question f(x)=-|x| |
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Answer» F (x)= -x Agar mod ke under koi sign rehe toh woh plus ho jata hai ,,agar -( minus) rahega toh plus agar plus toh plus hi hoga .Agar bahar koi sign rahe toh wahi sign hoga -x hoga |
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| 2109. |
Cot2π/6+cosec5π/6+3ttan2π/6 |
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| 2110. |
Chapter 8 elements solutions in |
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| 2111. |
Underoot 2x-1 upon x-1 less than zero |
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| 2112. |
f(x)=√x-3 find domain and range |
| Answer» x-3≥0x≥3Domain = [3,infinity)For range let y=√x-3ysquare =x-3x=y square+3 So Range=[0,infinity) | |
| 2113. |
kya koi pratibha school se hai |
| Answer» ?????are zubair tum..kaise ho | |
| 2114. |
Please explain the questions of 3.3 question number 23 and 24 by a video |
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| 2115. |
Prove this eq cosx(pi/2-x)=sinx |
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| 2116. |
(2n+7)>(n+3)² prove by mathematical induction |
| Answer» Let P (n) =(2n + 7) < (n + 3)2For n = 1{tex}P(1) = (2 \\times 1 + 7) < {(1 + 3)^2} \\Rightarrow 9 < 16{/tex}{tex}\\therefore {/tex}\xa0P ( 1) is trueLet P(n) be true for n = k{tex}\\therefore P(k) = (2k + 7) < {(k + 3)^2}{/tex}\xa0....(1)For n = k + 1\xa0P (k + 1) = 2 (k + 1) + 7 < (k + 1 + 3)2{tex} \\Rightarrow {/tex}\xa02(k + 1) + 7 < (k + 4)2From (1)2k + 7 < (k + 3)2Adding 2 on both sides2k + 7 + 2 < (k + 3)2 + 2\xa0{tex} \\Rightarrow {/tex}\xa02(k + 1) + 7 < k2 + 9 + 6k + 2{tex} \\Rightarrow {/tex}\xa02(k + 1) + 7 < k2 +6k + 11 < k2 + 8k + 162(k + 1) + 7 <(k + 4)2{tex}\\therefore {/tex}\xa0P (k + 1) is trueThus P(k) is true {tex} \\Rightarrow {/tex}\xa0P(k + 1) is trueHence by principle of mathematical induction, P (n) is true for all {tex}n \\in N{/tex}. | |
| 2117. |
Exercise 9.2 question number 9 |
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| 2118. |
The angle between the minute and hour hands of a clock at 8:30is |
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Answer» Sorry answer is wrong In 30 minutes =180° So 255-180=75° 12hours = 360° 1 hour = 360÷128:30 hours = 8+1÷2×360÷12 = 255°-----------(i) Now 60minute = 360°1minute =360÷60=1÷630 minute = 30×1÷6=5°----------------------(ii) Substract equation ii by i255°-5°=250° May 60 degree or 300 degree 270 75° 75 degree |
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| 2119. |
Find the range of f(x)= 1/√9-x^2 |
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| 2120. |
Solve the equation-|z| = z+1+2i |
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| 2121. |
What is the definition of set ?? |
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Answer» Well defined collections of distinct objects is called set Well defined collection of objects A set is a collection of well - defined objects or elements |
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| 2122. |
Tan3x tan2x tanx=tan3x-tan2x-tanx |
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Answer» Abhijeet Sharma...can u pls explain ur steps in the ans u submitted... >>Tan3x=tan(2x+x)>>Tan3x=tan2x+tanx/1-tan2x•tanx>>Tan3x(1-tan2x.tanx)=tan2x +tanx>>Tan3x-Tan3xTan2xtanx=tan2x+tanx>>Tan3xtan2xtanx=tan3x-tan2x-tanx |
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| 2123. |
Prove that :a+b/c=cos(a-b/2)/sinc/2 |
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| 2124. |
Find square root of ( -i) of complex Number |
| Answer» Let {tex}x + yi = \\sqrt { - i} {/tex}Squaring both sides, we get(x + yi)2 = -ix2 - y2 + 2xyi = -iEquating the real and imaginary partsx2 - y2 = 0 . . . (i)and 2xy = -1{tex}\\therefore xy = - \\frac{1}{2}{/tex}Now using the identity(x2 + y2)2 = (x2 - y2)2 + 4x2y2{tex} = {(0)^2} + 4{\\left( { - \\frac{1}{2}} \\right)^2}{/tex}= 1{tex}\\therefore {/tex}\xa0x2 + y2 = 1 . . . (ii) [Neglecting (-) sign as x2 + y2 > 0]Solving (i) and (ii), we get{tex}{x^2} = \\frac{1}{2}{/tex}\xa0and {tex}{y^2} = \\frac{1}{2}{/tex}{tex}\\therefore x = \\pm \\frac{1}{{\\sqrt 2 }}{/tex}\xa0and {tex}y = \\pm \\frac{1}{{\\sqrt 2 }}{/tex}Since the sign of xy is (-) then if{tex}x = \\frac{1}{{\\sqrt 2 }}{/tex},\xa0{tex}y = - \\frac{1}{{\\sqrt 2 }}{/tex}If {tex}x = - \\frac{1}{{\\sqrt 2 }}\\;y = \\frac{1}{{\\sqrt 2 }}{/tex}{tex}\\style{font-family:Tahoma}{\\style{font-size:8px}{\\therefore\\sqrt{-i}=\\pm\\left(\\frac1{\\sqrt2}-\\frac1{\\sqrt2}i\\right)}}{/tex} | |
| 2125. |
COS20 COS40 COS80-1/8 |
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Answer» So 1/8 -1/8 is 0 The value of cos20 cos40 cos80 is 1/8 0 |
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| 2126. |
If (x/3+1,y-2/3)=(5/3,1/3),find the values of x and y. |
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Answer» X=2 and y=1 X=2 and y=-1/2. |
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| 2127. |
Cos x =3/5,x lies in second quadrant |
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Answer» Cos x= -3/5 hoga because cos thetha is negative in 2nd quadrant. Instead of cosx it should be sin x |
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| 2128. |
3f(x)+2f(1/x)=x*x |
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| 2129. |
(A+b)( a-b) |
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Answer» a2- b2 a^2-b^2 (a + b)( a - b) = a2 - b2 A square-B square A square ×B square |
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| 2130. |
2sinx +2 cosx =1 |
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| 2131. |
Prove that 0!=1Where (!)=factorial |
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| 2132. |
Make modulus function graph for -4,4 |
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| 2133. |
since (B+C) ÷2=(b-c) ÷a×cos(A) ÷2 |
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| 2134. |
Let Sn denotes the sum of n terms of an A.P. if S2n =3Sn , the ratio S3n / sn= |
| Answer» Let 1st term be a and common difference be d.Given, S2n\xa0= 3sn{tex}\\therefore{/tex}\xa0{tex}\\frac { 2 n } { 2 }{/tex}\xa0[2a + (2n - 1) d] = 3 {{tex}\\frac { n } { 2 }{/tex}\xa0[2a + (n - 1) d]}{tex}\\Rightarrow{/tex}\xa04a + (4n - 2) d = 6a + (3n - 3)d{tex}\\Rightarrow{/tex}\xa02a = (n + 1)d\xa0...(i)Now,\xa0{tex}\\frac { S _ { 3 n } } { S _ { n } } = \\frac { \\frac { 3 n } { 2 } [ 2 a + ( 3 n - 1 ) d ] } { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] }{/tex}=\xa0{tex}\\frac { 3 \\{ ( n + 1 ) d + ( 3 n - 1 ) d \\} } { ( n + 1 ) d + ( n - 1 ) d }{/tex}\xa0[from Eq. (i)]=\xa0{tex}\\frac { 3 ( 4 n d ) } { 2 n d }{/tex} = 6:1 | |
| 2135. |
The sum of numbers from 250 to 1000 which is divisible by 3 is |
| Answer» 45° | |
| 2136. |
Write the value of tan 3π/12 |
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Answer» Tan 3π/12=3π/12×180/π=45°tan45°=. 1 Answer is 1 Answer is 1 It is 1, if you will write it as tan(π/4) then ot will look easy.... Hope it may help you... |
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| 2137. |
Find tan5x in term of tan |
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| 2138. |
Find the rank of the word "VARTIKA" and "ANNU" |
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| 2139. |
Cos(60+a) sin(60-a)=1/4(root3-2sin a) |
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| 2140. |
Trigononetric formulas used in chapter 3 of ncert |
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Answer» Chapter is Trigonometric Functions Dekh iss app me maths me ja aur lesson trigo choose kar all formula diya hua hai . Ok Can you please tell me the full name of the chapter so that I can help you because I am not having the book with me It may be formula of compound angle |
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| 2141. |
In how many ways can an examinee answer a set of 10 true false type questions |
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| 2142. |
6π÷0 = |
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Answer» Friend it actually it is not defined but we consider it as infinity. Hope it may help you... Zero |
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| 2143. |
Prove that 1+cossquare 2theta = 2 ( cos power 4theta +sinpower 4 theta) |
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| 2144. |
The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find the progression. |
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| 2145. |
1+(1+3)+(1+3+5)+...+ n terms=n(n+1)(2n+1)/6 |
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| 2146. |
a+(a+d)+(a+2d)+...+ a+(n-1)d =n/2 2a+(n-1)d |
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| 2147. |
Exercise - 4.1 , Q - 1 |
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| 2148. |
(1+w)^3-(1-w^2)^3 |
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| 2149. |
1w((1+w) |
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| 2150. |
Class 11 maths chapter 1 mescellious exercise question 8 , 10 ,11,12 please answer |
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