Q.Show that the function given by $f(x) = e^{2x}$ is increasing on $\mathbf{R}$.

[!TLDR]

With $a<0$, $f'(x)=a(e^{ax}-e^{-ax})$ is negative precisely for $x<0$, so $f$ is strictly decreasing for $x<0$ — option (C).

Concept

A differentiable function is strictly decreasing where its derivative is negative (CBSE Class 12 “Application of Derivatives”). Note $e^{ax}+e^{-ax}=2\cosh(ax)$, an even function symmetric about $x=0$.

Solution

Differentiate:

We need $f'(x)<0$. Since $a<0$, the factor $a$ is negative, so we need $\left(e^{ax}-e^{-ax}\right)>0$, i.e. $e^{ax}>e^{-ax}$, i.e. $ax>-ax$, i.e. $2ax>0$, i.e. $ax>0$.

Because $a<0$, $ax>0$ requires $x<0$. Hence $f'(x)<0$ for all $x<0$, so $f$ is strictly decreasing there. (Equivalently, $f=2\cosh(|a|x)$ falls as $x$ moves from $0$ toward $-\infty$.)

[!ANSWER]

(C) The function is strictly decreasing for $x<0$.

We are asked to find the point $(\alpha, \beta)$ where $f(x)=x e^{-x}$ attains its maximum value. Since the domain is all real numbers, we use calculus: the maximum of a smooth function occurs at a critical point where the derivative is zero, provided the second derivative confirms it is a local maximum (and we check end behavior to ensure it is global).

Why this approach works:

For a differentiable function, the slope is zero at a peak. Here $f(x)$ is a product of $x$ and $e^{-x}$, so we apply the product rule. The exponential never vanishes, so the derivative’s sign depends on a simple linear factor. This makes finding the critical point straightforward.

Step-by-step solution:

1. Find the derivative. $f(x) = x e^{-x}$. Using the product rule:

2. Set the derivative to zero. Since $e^{-x} > 0$ for all real $x$, we have

So the only critical point is at $x = 1$.

3. Confirm it is a maximum using the second derivative. Differentiate $f'(x) = e^{-x}(1 - x)$ again:

At $x = 1$:

A negative second derivative means the graph is concave down, so $x=1$ is a local maximum.

4. Check end behavior to ensure it is global.

   As $x \to -\infty$, $e^{-x} \to \infty$ and $x \to -\infty$, so $x e^{-x} \to -\infty$ (the product goes to negative infinity).

   As $x \to \infty$, $e^{-x} \to 0$ faster than $x$ grows, so $x e^{-x} \to 0$ from above.

   Hence the function rises from $-\infty$, peaks at $x=1$, then decays to $0$. The peak is the global maximum.

5. Compute the maximum value $\beta$.

Thus $(\alpha, \beta) = (1, 1/e)$.

Concept and Intuition

When comparing $a^{f(x)}$ vs $a^{g(x)}$ with $a>1$, the inequality direction transfers directly to the exponents: $a^{f(x)}>a^{g(x)}\iff f(x)>g(x)$. The only subtlety here is that multiplying an inequality by $x$ flips it when $x<0$.

Step-by-Step Solution

1. $16(2^x) = 2^4\cdot 2^x = 2^{x+4}$.
2. $16^{-1/x} = (2^4)^{-1/x} = 2^{-4/x}$ (requires $x\neq0$).
3. Since base $2>1$: $2^{x+4}>2^{-4/x} \iff x+4 > -\dfrac{4}{x}$.
4. Case $x>0$: multiply by $x$ (positive, keep direction): $x^2+4x > -4 \Rightarrow x^2+4x+4>0 \Rightarrow (x+2)^2>0$. This holds for every real $x$ except $x=-2$; since we're in the range $x>0$, it holds for all $x>0$.
5. Case $x<0$: multiply by $x$ (negative, flip direction): $x^2+4x < -4 \Rightarrow x^2+4x+4<0\Rightarrow (x+2)^2<0$, which is impossible. So no negative $x$ satisfies it.
6. Hence the solution set is precisely $\{x\in\mathbb{R}: x>0\}$.

Common Mistakes

• Forgetting to flip the inequality when multiplying by a negative $x$ in the $x<0$ case.
• Missing the domain restriction $x\neq0$ (though it doesn't change the final answer here).

ANSWER: A

Concept and Intuition

This is a standard mathematical-induction inequality. Rather than trying to prove it for all natural numbers at once, first test small values directly to find the boundary case, then use induction to confirm the pattern holds for everything beyond that boundary.

Step-by-Step Solution

1. Test $n=1$: $2^1 = 2$; $n+1 = 2$. Is $2 > 2$? No — it's equal, so the strict inequality fails at $n=1$.
2. Test $n=2$: $2^2=4$; $n+1=3$. Is $4>3$? Yes.
3. Test $n=3$: $2^3=8$; $n+1=4$. Is $8>4$? Yes.
4. Inductive step (confirm it holds for all $n\ge2$): Assume $2^n > n+1$ for some $n\ge2$. Then $2^{n+1} = 2\cdot 2^n > 2(n+1) = 2n+2$. Since $2n+2 \ge (n+1)+1 = n+2$ for all $n \ge 0$, we get $2^{n+1} > n+2 = (n+1)+1$, so the inequality also holds for $n+1$. By induction, it holds for all $n \ge 2$.
5. So the inequality is valid precisely for $n \ge 2$ (fails only at $n=1$).

Common Mistakes

• Assuming the inequality holds for ALL natural numbers including $n=1$ without actually checking the boundary case (where it's an equality, not a strict inequality).
• Confusing "valid for $n\ge2$" with a more restrictive/different range like $1\le n\le3$.

ANSWER: B

Concept & Intuition

A function is strictly decreasing on $\mathbb{R}$ if its slope (derivative) is negative at every point. For a polynomial, that means its derivative never touches or crosses zero. Here $f'(x)$ is a quadratic: $3k x^2 - 6x - 12$. For this quadratic to be always negative, it must open downward (so the coefficient of $x^2$ is negative) and have no real roots (so its discriminant is negative). That gives two inequalities that together determine $k$.

Step-by-step reasoning

1. Find the derivative

Strictly decreasing for all $x$ means $f'(x) < 0$ for every real $x$.

2. Condition for a quadratic to be always negative

   A quadratic $ax^2 + bx + c$ is negative for all $x$ if and only if:

   • $a < 0$ (opens downward), and
   • the discriminant $\Delta = b^2 - 4ac < 0$ (no real roots, so it never crosses zero).

3. Apply the first condition

   Here $a = 3k$. We need

So $k$ must be negative. This already eliminates options (B) and (C), which require $k > -\frac14$ or $k > \frac14$.

4. Apply the second condition Compute the discriminant of $f'(x)$:

We require $\Delta < 0$:

5. Combine the conditions From step 3 we have $k < 0$; from step 4 we have $k < -\frac14$. The stricter condition is $k < -\frac14$, which automatically satisfies $k < 0$. So the necessary and sufficient condition is

Let $h(x)=2^x+5^x-3^x-4^x$.

Check obvious values: $h(0)=1+1-1-1=0$ and $h(1)=(2+5)-(3+4)=7-7=0$. So $x=0$ and $x=1$ are both roots.

To see there are no more, examine the sign of $h$:

• $h(0.5)=\sqrt2+\sqrt5-\sqrt3-2\approx3.650-3.732=-0.082<0$, so $h<0$ inside $(0,1)$.
• $h(-1)=0.5+0.2-\tfrac13-0.25\approx+0.117>0$.
• $h(2)=4+25-9-16=+4>0$.

So $h>0$ for $x<0$, $h>0$ for $x>1$, and $h<0$ strictly between, touching zero only at the endpoints $x=0,1$. Thus there are exactly two real solutions, $x=0$ (zero) and $x=1$ (non-zero).

ANSWER: B