Let f(x)=sin23x−cos22x and g(x)=1+12tan−1|x|. Then, the number of values of x in interval [−10π,20π] satisfying the equation f(x)=sgn(g(x)), is 5a Then, a is
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Solution
g(x)=12tan−1|x|+1 ⇒sgn(g(x))=1 sin23x−cos22x=1 sin23x=1+cos22x which is possible ,if sinx=1 and cosx=0⇒sinx=1,x=2nπ+π2 Hence, −10π≤2nπ+π2≤20π ⇒−212≤2n≤392 ⇒−214≤n≤394⇒−5≤n≤9 Hence, number of values of x=15⇒a=3 Ans: 3