CBSE Mathematics Set I Outside Delhi 2012
Q7. Given e tan x
Write 1 sec x dx
e . satisfying above equation. Given e tan x
Answer: L.H.S. = e tan x
e sec x
1 sec x dx
e … … … … 1 1 sec x dx sec x tan x dx e sec x dx
e sec x tan x dx ,
Integrating first by parts, e sec x
e sec x tan x dx
e sec x
c e
secx, v
e c 1 On comparing, we get Q8. e sec x tan x dx
Write the value of ̂
sec . ̂ .
̂. ̂. Answer: We know that And .
Dot product of unlike unit vector is zero as θ=900. Therefore ̂
̂ .
̂. ̂
.
̂. ̂ = 1 + 0 = 1. ̂ ̂
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Q9. Find the scalar components of the vector with initial point A (2, 1) and terminal point B (‐5, 7). Answer: From the figure drawn we have Vector 2̂
7̂
̂
5̂
7 ̂ 6 ̂ Hence the scalar components are ‐7 and 6. Q10. Find the distance of the plane 3x – 4y +12z = 3 from the origin. Answer: We know that perpendicular distance of Therefore Perpendicular distance from the origin (0, 0, 0) and the plane 3x – 4y +12z – 3 = 0. ,
,
√
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