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1551865013856459064835.jpg Question no 3 and 4

Answers (3)

best_answer

@Nitin

use\:this\:concept\:to\:solve\\f(x)=\log_ax;\:\:x,a>0\:\:and\:a\neq1

use this concept to solve logax; x, a > 0 anda
Posted by

himanshu.meshram

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

@Nitin Sinha

3.

[x+1/2]>0 \ and \ [x+1/2] \neq 1

x+1/2 \geq 2 \\ x \geq 3/2

ALSO, x^2 - 5x +6 \neq 0

=> x \neq {2,3}

which gives option a.

 

Posted by

Kshitij

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@Nitin Sinha

4.

For even function, f(x) = f(-x)

For odd function, f(x) = - f(-x)

lets take 1/f(x) + 1/f(-x) as we can simplify in cos easier than sec.

now, Cos[log(\sqrt{1+x^2} + x)] + Cos[log(\sqrt{1+x^2} - x)]

=2 * Cos[\frac{log(\sqrt{1+x^2} + x)+log(\sqrt{1+x^2} - x) }{2}]*Cos[\frac{log(\sqrt{1+x^2} + x)-log(\sqrt{1+x^2} - x) }{2}]

=2 * Cos[\frac{log[(\sqrt{1+x^2} + x)*(\sqrt{1+x^2} - x)] }{2}]*Cos[\frac{log[\frac{(\sqrt{1+x^2} + x)}{(\sqrt{1+x^2} - x) }]}{2}]

=2*1*Cos[log(\sqrt{1+x^2} + x)]

=> \frac{1}{f(x)} + \frac{1}{f(-x)} = \frac{2}{f(x)}

=> f(x)=f(-x) which shows f(x) is an even function.

Posted by

Kshitij

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