PREVIOUS YEAR
NDA 2020 PREVIOUS YEAR QUESTION PAPERS
UPSC conducts NDA & NA Entrance Examination every year twice in the month of April and September but this year due to COVID-19 both the examinations were held on the same day i.e. 6 september 2020. We have provided you previous year papers (from 2009-20) to prepare for the NDA 2021 Examination . Previous year papers help you analyse the type of questions asked in…
Read more
NCERT
NCERT Solutions for Class 12th Maths Chapter 4 Exercise 4.3
SOLUTION VIEW PDF DOWNLOAD PDF
Read more
NCERT
NCERT Solutions for Class 12th Maths Chapter 4 Exercise 4.1
SOLUTION VIEW PDF DOWNLOAD PDF
Read more
NCERT
NCERT Solutions for Class 12th Maths Chapter 3 Exercise 3.1
SOLUTION VIEW PDF DOWNLOAD PDF
Read more
Differentiation
The function f(x)=2x³-15x²+36x+4 is maximum at
The function f(x)=2x³-15x²+36x+4 is maximum at (a)x=2 (b)x=4 (c)x=0 (d)x=3 solution : (a)
Read more
Trigonometry
If m=sin⁶x+cos⁶x,then m belongs to the interval
If m=sin⁶x+cos⁶x,then m belongs to the interval (a)[7/8,5/4] (b)[1/2,5/8] (c)[1/4,1] (d)[3/4,1] solution (c)
Read more
NCERT
NCERT Solutions for Class 12th Maths Chapter 2 Miscellaneous Exercise
SOLUTIONS VIEW PDF DOWNLOAD PDF
Read more
NCERT
NCERT Solutions for Class 12th Maths Chapter 2 Exercise 2.1
SOLUTIONS VIEW PDF DOWNLOAD PDF
Read more![The minimum value of f(x)=sin⁴x +cos⁴x in [0,Π/2 ] is](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYq6Bs988DkDMsafYweeGpJNjMrlU7U40VE13jWnCDFvnW8y10VgLEgnG1t-rXjN09h94XewPYWPXNb0YLnt9vNn-7tG7dAiiBIYwrTa6X8XPVUTLicLblAA7Yd9LTAAz3j9nx51AjxTc-/w680/MIN+VALUE+1.jpeg)
Differentiation
The minimum value of f(x)=sin⁴x +cos⁴x in [0,Π/2 ] is
The minimum value of f(x)=sin⁴x +cos⁴x in [0,Π/2 ] is (a)1/2√2 (b)1/4 (c)-1/2 (d)1/2 solution : (d)
Read more
INVERSE TRIGONOMETRY
The value of cosec⁻¹(1+x²/2x) is
The value of cosec⁻¹(1+x²/2x) is (a)x (b)π/2 -tan⁻¹x (c)2tan⁻¹x (d)None of these SOLUTION : (C)
Read more
12th
The roots of the equation |x²-x-6|=x+2 are
The roots of the equation |x²-x-6|=x+2 are (a)-2,1,4 (b)0,1,4 (c)0,2,4 (d)-2,2,4 solution : (d)
Read more
The smallest positive integer for which (1+i)²ⁿ=(1-i)²ⁿ is
The smallest positive integer for which (1+i)²ⁿ=(1-i)²ⁿ is: (a) 4 (b) 2 (c) 12 (d) 8 solution: (b)
Read more
