Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Application of Derivatives question

2017 · Shift 2 · Q26
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Application of Derivatives
  5. /2017 · Shift 2 · Q26

Application of Derivatives question

2017 · Shift 2 · Q26

JEE AdvancedMathematicsApplication of DerivativesMultiple correct+4 / −2
f : R →\to→ R is a differentiable function such that f'(x) > 2f(x) for all x ∈\in∈ R, and f(0) = 1 then
  1. A
    f(x) > e2x in (0, ∞\infty∞)
  2. B
    f'(x) < e2x in (0, ∞\infty∞)
  3. C
    f(x) is increasing in (0, ∞\infty∞)
  4. D
    f(x) is decreasing in (0, ∞\infty∞)
View written solutionFree

Correct answer: A, C

  1. We are given
\quad \forall x \in \mathbb{R}, \qquad f(0)=1.$$ We must determine which options are true for $x>0$. --- 2. Consider the function $$g(x)=f(x)e^{-2x}.$$ Differentiate: $$g'(x)=e^{-2x}\big(f'(x)-2f(x)\big).$$ Since $e^{-2x}>0$ for all $x$, and given $$f'(x)-2f(x)>0,$$ we get $$g'(x)>0 \quad \forall x\in\mathbb R.$$ Hence $g(x)$ is strictly increasing on $\mathbb R$. --- 3. Use the initial condition: $$g(0)=f(0)e^0=1.$$ Because $g$ is strictly increasing, for every $x>0$, $$g(x)>g(0)=1.$$ So, $$f(x)e^{-2x}>1$$ which gives $$f(x)>e^{2x} \quad \text{for } x>0.$$ Therefore, **Option A is correct**. --- 4. Now check whether $f(x)$ is increasing. From Option A, for $x>0$ we have $$f(x)>e^{2x}>0.$$ Using the given inequality, $$f'(x)>2f(x).$$ Since $f(x)>0$ for $x>0$, we get $$f'(x)>2f(x)>0.$$ Thus, $$f'(x)>0 \quad \text{for } x>0,$$ so $f(x)$ is increasing on $(0,\infty)$. Therefore, **Option C is correct**. --- 5. Check Option B: $f'(x)<e^{2x}$ in $(0,\infty)$. But we have $$f'(x)>2f(x)>2e^{2x}>e^{2x} \quad (x>0).$$ So Option B is false. --- 6. Check Option D: $f(x)$ is decreasing in $(0,\infty)$. Since $f'(x)>0$ for $x>0$, $f$ is increasing, not decreasing. So Option D is false. --- 7. Final conclusion: Correct options are $$\boxed{A,\ C}.$$
PreviousNext

More from Application of Derivatives

  • If f(x)=​cos2x−cosxsinx​cos2xcosxsinx​sin2x−sinxcosx​​, then2017 · Multiple correct
  • The least value of a ∈R for which 4ax2+x1​≥1,, for all x>0. is2016 · MCQ
  • Let f: R →(0,∞) and g : R → R be twice differentiable functions such that f'' and g'' are continuous functions on R. Suppose f'(2)= g (2)=0, f''(2)e0 and g'(2)e0. If x→2lim​f′(x)g′(x)f(x)g(x)​=1,…2016 · Multiple correct
  • A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of Vmm3, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid…2015 · Numerical
  • Let f,g: [−1,2]→R be continuous functions which are twice differentiable on the interval (−1,2). Let the values of f and g at the points −1,0 and 2 be as given in the following table: In each of the… Includes table2015 · Multiple correct
  • The slope of the tangent to the curve (y−x5)2=x(1+x2)2 at the point (1,3) is2014 · Numerical
  • A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed…2013 · Multiple correct
  • Let f:[0,1]→R(the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥.ex,x∈[0,1]…2013 · MCQ