Question

Please answer the above question only if u knew the answer..

⭐Explaination needed
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Answer 1

If a graph of a function in $x$ is symmetric about the y axis, the function is said to be an even function.

Because, suppose a function $f:A\to B$ is symmetric about y axis, then we have,

$\longrightarrow f(-h)=f(h)\quad\!\forall h\in A$

This is the condition for a function being even. Hence $f$ is an even function.

Let us consider first function.

$\longrightarrow f(x)=\log_e\left(x+\sqrt{x^2+1}\right)$

Replacing $x$ by $-x,$

$\longrightarrow f(-x)=\log_e\left(-x+\sqrt{(-x)^2+1}\right)$

$\longrightarrow f(-x)=\log_e\left(-x+\sqrt{x^2+1}\right)$

But, on multiplying and dividing $x+\sqrt{x^2+1}$ with $-x+\sqrt{x^2+1},$

$\longrightarrow f(-x)=\log_e\left(\dfrac{(-x+\sqrt{x^2+1})(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}\right)$

$\longrightarrow f(-x)=\log_e\left(\dfrac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{x+\sqrt{x^2+1}}\right)$

$\longrightarrow f(-x)=\log_e\left(\dfrac{x^2+1-x^2}{x+\sqrt{x^2+1}}\right)$

$\longrightarrow f(-x)=\log_e\left(\dfrac{1}{x+\sqrt{x^2+1}}\right)$

$\longrightarrow f(-x)=\log_e\left(x+\sqrt{x^2+1}\right)^{-1}$

$\longrightarrow f(-x)=-\log_e\left(x+\sqrt{x^2+1}\right)$

That is,

$\longrightarrow f(-x)=-f(x)$

This means $f(x)$ is an odd function. So its graph is not symmetric about y axis.

Hence (A) is wrong.

Let us consider second function.

$\longrightarrow f(x+y)=f(x)+f(y)\quad\!\forall x,\,y\in\mathbb{R}$

Take $y=0.$

$\longrightarrow f(x+0)=f(x)+f(0)$

$\longrightarrow f(x)=f(x)+f(0)$

$\longrightarrow f(0)=0$

And take $y=-x.$

$\longrightarrow f(x-x)=f(x)+f(-x)$

$\longrightarrow f(0)=f(x)+f(-x)$

$\longrightarrow 0=f(x)+f(-x)$

$\longrightarrow f(-x)=-f(x)$

This implies $f(x)$ is an odd function. So its graph is not symmetric about y axis.

Hence (B) is wrong.

Let us consider third function.

$\longrightarrow f(x)=\cos x+\sin x\quad\quad\dots(1)$

Replacing $x$ by $-x,$

$\longrightarrow f(-x)=\cos (-x)+\sin (-x)$

$\longrightarrow f(-x)=\cos x-\sin x\quad\quad\dots(2)$

Subtracting (2) from (1),

$\longrightarrow f(x)-f(-x)=\cos x+\sin x-\cos x+\sin x$

$\longrightarrow f(x)-f(-x)=2\sin x$

$\longrightarrow f(-x)=f(x)-2\sin x$

If $\sin x=0$ only,

$\longrightarrow f(-x)=f(x)$

Since $\sin x\neq0$ for every $x$ (not considering values of $x$ satisfying the equality particularly), $f(x)$ is not an even function and so the graph is not symmetric about y axis.

Hence (C) is wrong.

Therefore, the correct answer is (D) None of these.