Question

Short essay on medaram festival in hindi

Answer 1

Central government is likely to declare Medaram’s Sammakka-Sarakka/Saralamma Jatara a national festival this year. Union Ministry of Tribal Affairs released ₹2 crore as Centre’s contribution towards conducting the massive event which is held bi-annually in Jayashankar Bhupalpally district to honour the twin goddesses Sammakka and her daughter Sarakka.

Union government had in 2015 declared Vanaj, a tribal dance and music festival, as national festival. Union Minister of Tribal Affairs Jual Oram is expected to announce the Centre’s decision on Jatara’s national status on January 31 when he visits Medaram, senior officials of Telangana’s Tribal Welfare Development department told The Hindu.

Biggest festival

Sammakka-Sarakka Jatara held by forest dwelling Koya tribe of Telangana and surrounding States, is the biggest Tribal festival in Asia which is attended by one crore people on an average. This year, the four-day Jatara, scheduled to begin on January 31, is expected to have a footfall of 1.20 crore persons.

The State government has already released ₹80.5 crore for the Jatara this year. Once declared a national festival, Jatara can be considered for ‘intangible cultural heritage of humanity’ tag of United Nations Educational, Scientific and Cultural Organisation (UNESCO). UNESCO had given the tag to Kumbh Mela, which is world’s biggest festival which sees participation by up to 10 crore persons.

Out of the huge congregation which comes to Medaram, 50 % are non-adivasi making the Jatara the most popular Tribal festival in India. People from Madhya Pradesh, Odisha, Jharkhand, Chhattisgarh and Andhra Pradesh come for the festival. “Tribal Welfare Department has been requesting the Centre to recognise Jatara since the past eight years. But this year State government also requested the Centre because of which we expect a positive outcome,” Director Tribal Cultural Research and Training Institute Sarveswar Reddy told The Hindu.


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Answer 2

Explanation:

f(x) = kx³ – 8x² + 5

Roots are α – β , α & α +β

Sum of roots = – (-8)/k

Sum of roots = α – β + α + α +β = 3α

= 3α = 8/k

= k = 8/3α

or we can solve as below

f(x) = (x – (α – β)(x – α)(x – (α +β))

= (x – α)(x² – x(α+β + α – β) + (α² – β²))

= (x – α)(x² – 2xα + (α² – β²))

= x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

= x³ – 3αx² + x(3α² – β²) + αβ² – α³

= kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

kx³ – 8x² + 5

k(3α² – β²) = 0 => 3α² = β²

k(αβ² – α³) = 5

=k(3α³ – α³) = 5

= k2α³ = 5

3αk = 8 => k = 8/3α

(8/3α)2α³ = 5

=> α² = 15/16

=> α = √15 / 4