1551865013856459064835.jpg Question no 3 and 4

Answers (3)

@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

@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.

 

@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.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2022 (Subscription)

Knockout JEE Main April 2022 Subscription.

₹ 5499/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions