Solveeit Logo
Login

Question

Question: Let \(f:\left[ {\dfrac{1}{2},1} \right] \to R\) (the set of all real numbers) be a positive, non-con...

Let f:[12,1]→Rf:\left[ {\dfrac{1}{2},1} \right] \to R (the set of all real numbers) be a positive, non-constant and differentiable function such that f’(x) < 2 f (x) and f(12)f\left( {\dfrac{1}{2}} \right) = 1. Then the value of ∫121f(x)dx\int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} lies in the interval
(a)(2e−1,2e)\left( a \right)\left( {2e - 1,2e} \right)
(b)(e−1,2e−1)\left( b \right)\left( {e - 1,2e - 1} \right)
(c)(e−12,e−1)\left( c \right)\left( {\dfrac{{e - 1}}{2},e - 1} \right)
(d)(0,e−12)\left( d \right)\left( {0,\dfrac{{e - 1}}{2}} \right)

Explanation

Solution

Given data:
f:[12,1]→Rf:\left[ {\dfrac{1}{2},1} \right] \to R (The set of all real numbers) be a positive, non-constant and differentiable function.
Therefore, f (x) > 0, (because f (x) is a positive function).
Now it is given that, f′(x)<2f(x)f'\left( x \right) < 2f\left( x \right)
Therefore, f′(x)−2f(x)<0f'\left( x \right) - 2f\left( x \right) < 0.................. (1)
Now as we know that ex>0{e^x} > 0 and e−x>0{e^{ - x}} > 0, [x∈R]\left[ {x \in R} \right]
So multiply e−2x{e^{ - 2x}} in equation (1) we have,
⇒e−2xf′(x)−2f(x)e−2x<0\Rightarrow {e^{ - 2x}}f'\left( x \right) - 2f\left( x \right){e^{ - 2x}} < 0

Complete step-by-step solution:
Now as we know that ddxe−2xf(x)=e−2xf′(x)+f(x)ddxe−2x=e−2xf′(x)+f(x)e−2x(−2)=e−2xf′(x)−2f(x)e−2x\dfrac{d}{{dx}}{e^{ - 2x}}f\left( x \right) = {e^{ - 2x}}f'\left( x \right) + f\left( x \right)\dfrac{d}{{dx}}{e^{ - 2x}} = {e^{ - 2x}}f'\left( x \right) + f\left( x \right){e^{ - 2x}}\left( { - 2} \right) = {e^{ - 2x}}f'\left( x \right) - 2f\left( x \right){e^{ - 2x}} so we have,
⇒ddxe−2xf(x)<0\Rightarrow \dfrac{d}{{dx}}{e^{ - 2x}}f\left( x \right) < 0
So as the first derivative of e−2xf(x){e^{ - 2x}}f\left( x \right) is less than zero, so e−2xf(x){e^{ - 2x}}f\left( x \right) is a decreasing function.
So we can say that e−2xf(x)<(e−2xf(x))x=12{e^{ - 2x}}f\left( x \right) < {\left( {{e^{ - 2x}}f\left( x \right)} \right)_{x = \dfrac{1}{2}}} as f:[12,1]→Rf:\left[ {\dfrac{1}{2},1} \right] \to R
⇒e−2xf(x)<e−2(12)f(12)\Rightarrow {e^{ - 2x}}f\left( x \right) < {e^{ - 2\left( {\dfrac{1}{2}} \right)}}f\left( {\dfrac{1}{2}} \right)
⇒e−2xf(x)<e−1f(12)\Rightarrow {e^{ - 2x}}f\left( x \right) < {e^{ - 1}}f\left( {\dfrac{1}{2}} \right)
Now it is given that f(12)f\left( {\dfrac{1}{2}} \right) = 1, so we have,
⇒e−2xf(x)<e−1\Rightarrow {e^{ - 2x}}f\left( x \right) < {e^{ - 1}}
⇒f(x)<e−1e−2x\Rightarrow f\left( x \right) < \dfrac{{{e^{ - 1}}}}{{{e^{ - 2x}}}}
⇒f(x)<e−1+2x\Rightarrow f\left( x \right) < {e^{ - 1 + 2x}}
Now as we calculated that f (x) > 0
⇒0<f(x)<e−1+2x\Rightarrow 0 < f\left( x \right) < {e^{ - 1 + 2x}}
Now integrate the above equation from (12\dfrac{1}{2} to 1) we have,
⇒∫1210dx<∫121f(x)dx<∫121(e−1+2x)dx\Rightarrow \int_{\dfrac{1}{2}}^1 {0dx} < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \int_{\dfrac{1}{2}}^1 {\left( {{e^{ - 1 + 2x}}} \right)dx}
Now integrate it using the property that integration of zero is zero, and ∫eax+bdx=eax+ba+c\int {{e^{ax + b}}dx} = \dfrac{{{e^{ax + b}}}}{a} + c, where C is some integration constant so we have,
⇒0<∫121f(x)dx<[e−1+2x2]121\Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{{{e^{ - 1 + 2x}}}}{2}} \right]_{\dfrac{1}{2}}^1
Now apply integrating limits we have,
⇒0<∫121f(x)dx<[e−1+22−e−1+2(12)2]\Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{{{e^{ - 1 + 2}}}}{2} - \dfrac{{{e^{ - 1 + 2\left( {\dfrac{1}{2}} \right)}}}}{2}} \right]
⇒0<∫121f(x)dx<[e2−e02]\Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{e}{2} - \dfrac{{{e^0}}}{2}} \right]
Now as we know that something to the power zero is always 1, so we have,
⇒0<∫121f(x)dx<[e2−12]\Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{e}{2} - \dfrac{1}{2}} \right]
⇒0<∫121f(x)dx<[e−12]\Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{{e - 1}}{2}} \right]
So, ∫121f(x)dx\int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} lies in the interval, (0,e−12)\left( {0,\dfrac{{e - 1}}{2}} \right).
So this is the required answer.
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property such as, ddxmn=mddxn+nddxm,ddxenx=nenx\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m,\dfrac{d}{{dx}}{e^{nx}} = n{e^{nx}} and always recall the basic integration property such as ∫eax+bdx=eax+ba+c\int {{e^{ax + b}}dx} = \dfrac{{{e^{ax + b}}}}{a} + c, where C is some integration constant.