Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Kunal Sharma 7 years, 6 months ago
- 1 answers
Posted by Ishfaq Gani Bhat 7 years, 6 months ago
- 0 answers
Posted by Alok Verma 7 years, 6 months ago
- 0 answers
Posted by Shreya Sharma 7 years, 6 months ago
- 4 answers
Kannu Kranti Yadav 7 years, 6 months ago
Posted by Ashish Agarwal 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
{tex}2 \mathrm { x } ^ { 3 } + \mathrm { x } ^ { 2 } - 5 \mathrm { x } + 2 ; \frac { 1 } { 2 } , 1 , - 2{/tex}
Comparing the given polynomial with
{tex}a x ^ { 3 } + b x ^ { 2 } + c x + d{/tex}, we get
a = 2, b = 1, c = 5, d = 2
Let {tex}p ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}
Then,
{tex}P \left( \frac { 1 } { 2 } \right) = 2 \left( \frac { 1 } { 2 } \right) ^ { 3 } + \left( \frac { 1 } { 2 } \right) ^ { 2 } - 5 \left( \frac { 1 } { 2 } \right) + 2{/tex}
{tex}= \frac { 1 } { 4 } + \frac { 1 } { 4 } - \frac { 5 } { 2 } + 2 = 0{/tex}
{tex}p ( 1 ) = 2 ( 1 ) ^ { 3 } + ( 1 ) ^ { 2 } - 5 ( 1 ) + 2{/tex}
= 2 + 1 - 5 + 2 = 0
{tex}p ( - 2 ) = 2 ( - 2 ) ^ { 3 } + ( - 2 ) ^ { 2 } - 5 ( - 2 ) + 2{/tex}
= -16 + 4 + 10 + 2 = 0
Therefore, {tex}\frac{1}{2}{/tex}, 1 and -2 are the zeroes of
{tex}2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}
So, {tex}\alpha = \frac { 1 } { 2 } , \beta = 1 \text { and } \gamma = - 2{/tex}
Therefore,
{tex}\alpha + \beta + \gamma = \frac { 1 } { 2 } + 1 + ( - 2 ) = - \frac { 1 } { 2 } = - \frac { b } { a }{/tex}
{tex}\alpha \beta + \beta \gamma + \gamma \alpha = \left( \frac { 1 } { 2 } \right) \times ( 1 ) + ( 1 ) \times ( - 2 ) + ( - 2 ) \times \left( \frac { 1 } { 2 } \right){/tex}
{tex}= \frac { 1 } { 2 } - 2 - 1 = - \frac { 5 } { 2 } = \frac { c } { a }{/tex}
{tex}\alpha \beta \gamma = \left( \frac { 1 } { 2 } \right) \times ( 1 ) \times ( - 2 ) = - 1 = \frac { - 2 } { 2 } = \frac { - d } { a }{/tex}
Posted by Pratham Ratan 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
All multiples of 7 lying between 500 and 900 are
504,511,518,...,896
This is an AP in which a = 504, d=7 and l = 896.
Let the given AP contain n terms. Then,
Tn = 896 {tex}\Rightarrow{/tex} a + (n - 1)d = 896 {tex}\Rightarrow{/tex}504 + (n -1) {tex}\times{/tex} 7 = 896 {tex}\Rightarrow{/tex}497 + 7n = 896
{tex}\Rightarrow{/tex}7n = 399 {tex}\Rightarrow{/tex}n = 57.
{tex}\therefore{/tex}required sum = {tex}\frac{n}{2}{/tex}(a + l)
={tex}\frac{{57}}{2}{/tex}{tex}\cdot{/tex}(504 + 896) = ({tex}\frac{{57}}{2}{/tex}{tex}\times{/tex}1400) = 39900.
Hence, the required sum is 39900.
Posted by Keshav Khandelwal 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Given that the pth, qth and rth terms of an AP be a, b, c respectively.
Suppose, x be the first term and d be the common difference of the given arithmetic progression. Therefore,
Tp = x+(p-1)d, Tq =x+(q-1)d and Tr =x+(r-1)d
Now, Tp = a {tex}\Rightarrow{/tex} x+(p-1)d = a ....(i)
Tq = b {tex}\Rightarrow{/tex} x+(q-1)d = b....(ii)
Tr = c {tex}\Rightarrow{/tex} x+(r-1)d = c....(iii)
On multiplying (i) by (q-r), (ii) by (r-p) and (iii) by (p-q), we get,
(q-r)[x+(p-1)d]=a(q-r)....(iv)
(r-p)[x+(q-1)d]=b(r-p)....(v)
(p-q)[x+(r-1)d]=c(p-q)....(vi)
Now adding we get,
a(q - r) + b(r - p) + c(p - q) = x{(q - r) + (r - p) + (p - q)} + d{(p - 1)(q - r)+ (q - 1)(r - p)+ (r - 1)(p - q)}
= (x {tex}\times{/tex} 0) + (d {tex}\times{/tex} 0) = 0
Therefore, a(q - r) + b(r - p) + c(p - q) = 0.
Hence proved.
Posted by Keshav Khandelwal 7 years, 6 months ago
- 1 answers
Srushti Kunkolienkar 7 years, 6 months ago
Posted by Mohit Sharma 7 years, 6 months ago
- 1 answers
Posted by Amit Kumar 7 years, 6 months ago
- 3 answers
Amit Kumar 7 years, 6 months ago
Aditya Singh 7 years, 6 months ago
Posted by Anushka Garg 7 years, 6 months ago
- 2 answers
Posted by Sanjay Kumar 7 years, 7 months ago
- 4 answers
Posted by Sukhpreet Singh Saini 7 years, 7 months ago
- 2 answers
Posted by Sandeep Yadav 7 years, 7 months ago
- 2 answers
Posted by Raunak Kumar 7 years, 7 months ago
- 2 answers
Posted by Preet Singh 7 years, 7 months ago
- 0 answers
Posted by Raunak Kumar 7 years, 7 months ago
- 0 answers
Posted by Ujjwal Sharma 7 years, 7 months ago
- 2 answers
Posted by Rishabh Kaushik 7 years, 7 months ago
- 3 answers
The Devil Prince 7 years, 7 months ago
Priyanshu Kumar 7 years, 7 months ago
Ujjwal Sharma 7 years, 7 months ago
Posted by Rajpal Sahu 7 years, 7 months ago
- 4 answers
Posted by Ankit Kumar 7 years, 7 months ago
- 3 answers
Pratyush Ranjan Mohapatra 7 years, 7 months ago
Posted by Ayush Sharma 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let, the two zeroes of the polynomial f(x) = 4x2 - 8kx - 9 be α and -α.
Comparing f(x) = 4x2 - 8kx - 9 with ax2+bx+c we get
a = 4; b = -8k and c = -9.
Sum of the zeroes = α + (-α) ={tex}-\frac ba=\;-\frac{-8k}4 {/tex}
0 = 2k
k = 0
Posted by Harsh Shukla 7 years, 7 months ago
- 0 answers
Posted by Muskan Singh 7 years, 7 months ago
- 0 answers
Posted by Akki Sharma 7 years, 7 months ago
- 1 answers
Posted by Srushti Kunkolienkar 7 years, 7 months ago
- 1 answers
Posted by Yukti Mulani 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
We have,
LHS = (tanA + cosec B)2 - (cotB - sec A)2
{tex}\Rightarrow{/tex} LHS = (tan2A + cosec2B + 2tanA cosecB ) - (cot2B + sec2A - 2cotB secA)
{tex}\Rightarrow{/tex} LHS = (tan2A - sec2A)+(cosec2B - cot2B)+2tanA cosecB +2cotB secA
But, Sec2A - tan2A =1 & cosec2A - cot2 A = 1
{tex}\therefore{/tex} LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA
{tex}\Rightarrow{/tex} LHS = 2 (tanA cosecB + cotB secA)
{tex}\Rightarrow{/tex} LHS = 2 tanA cotB{tex}\left( \frac { cosec\: B } { \cot B } + \frac { \sec A } { \tan A } \right){/tex} [Dividing and multiplying by tanA cotB]
{tex}\Rightarrow{/tex} LHS = 2tan A cotB{tex}\left\{ \frac { \frac { 1 } { \sin B } } { \frac { \cos B } { \sin B } } + \frac { \frac { 1 } { \cos A } } { \frac { \sin A } { \cos A } } \right\}{/tex} [Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]
{tex}\Rightarrow{/tex} LHS = 2 tanA cotB{tex}\left( \frac { 1 } { \cos B } + \frac { 1 } { \sin A } \right){/tex} = 2tanA cotB ( secB + cosecA ) = RHS. Hence, proved.
Posted by Prapti Prajapati 7 years, 7 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Ayush Gupta Ayush Gupta 7 years, 6 months ago
0Thank You