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Question: If a \> b \> 0 and f(q) = \(\frac { \left( a ^ { 2 } - b ^ { 2 } \right) \cos \theta } { a - b \sin...

If a > b > 0 and f(q) = (a2−b2)cos⁡θa−bsin⁡θ\frac { \left( a ^ { 2 } - b ^ { 2 } \right) \cos \theta } { a - b \sin \theta } , then the maximum value of f(q) is –

A
B

a2−b2\sqrt { a ^ { 2 } - b ^ { 2 } }

C

a – b

D

) a + b

Answer

a2−b2\sqrt { a ^ { 2 } - b ^ { 2 } }

Explanation

Solution

f(q) = (a2−b2)cos⁡θa−bsin⁡θ\frac { \left( a ^ { 2 } - b ^ { 2 } \right) \cos \theta } { a - b \sin \theta } = a2−b2asec⁡θ−btan⁡θ\frac { a ^ { 2 } - b ^ { 2 } } { a \sec \theta - b \tan \theta }

f(q) = a2−b2 h(θ)\frac { \mathrm { a } ^ { 2 } - \mathrm { b } ^ { 2 } } { \mathrm {~h} ( \theta ) } , where h(q) = a sec q – b tan q

f(q) = depends on hq

h(q) = a sec q – b tan q

h'(q) = a sec q tan q – b sec2q

for max. and min. of h(q), h'(q) = 0

sec q [a tan q – b sec q] = 0

sin q = b/a as sec q ¹ 0

h''(q) = sec q tan q (a tan q – b sec q)

+ (a sec2 q – b sec q tan q) sec q

= a sec3 q + a sec q tan2 q – 2b sec2 q tan q

h''q = a+asin⁡2θ−2bsin⁡θcos⁡3θ\frac { a + a \sin ^ { 2 } \theta - 2 b \sin \theta } { \cos ^ { 3 } \theta } {sin⁡θ=b/aa>b>0\left\{ \begin{array} { c } \sin \theta = b / a \\ a > b > 0 \end{array} \right.

= a+ab2a2−2bbacos⁡3θ\frac { a + a \frac { b ^ { 2 } } { a ^ { 2 } } - 2 b \frac { b } { a } } { \cos ^ { 3 } \theta } Ž a2−b2acos⁡2θ\frac { a ^ { 2 } - b ^ { 2 } } { a \cos ^ { 2 } \theta } > 0

h(q) is min. when sinq = b/a

\ f(q) is max. when sin q = b/a

max. f(q) = (a2−b2)a2−b2aa−b(ba)\frac { \left( a ^ { 2 } - b ^ { 2 } \right) \frac { \sqrt { a ^ { 2 } - b ^ { 2 } } } { a } } { a - b \left( \frac { b } { a } \right) }

= (a2−b2)a2−b2(a2−b2)\frac { \left( a ^ { 2 } - b ^ { 2 } \right) \sqrt { a ^ { 2 } - b ^ { 2 } } } { \left( a ^ { 2 } - b ^ { 2 } \right) }

f(q) = a2−b2\sqrt { a ^ { 2 } - b ^ { 2 } }