# Thread to share Good Questions from Calculus which requires deep thinking

#452

For (21) Let \displaystyle \color{blue}I =\int\frac{x-1}{(x+x\sqrt{x}+\sqrt{x})\sqrt{\sqrt{x}(x+1)}}dx.

Now put x=t^2 and dx=2tdt

So we have \displaystyle I =2\int\frac{t^2-1}{(t^2+t+1)\sqrt{t^3+t}}dx

\displaystyle \Rightarrow I =2\int\frac{1-\frac{1}{t^2}}{\bigg(t+\frac{1}{t}+1\bigg)\sqrt{t+\frac{1}{t}}}dt.

Now put \displaystyle t+\frac{1}{t}=u^2. Then \displaystyle \bigg(1-\frac{1}{t^2}\bigg)dt=2udu.

So We have \displaystyle I =4\int\frac{1}{u^2+1}du=4\arctan(u)+\mathcal{C}.

So finally we get \displaystyle \color{brown}I =4\arctan\bigg(\sqrt{\frac{x+1}{\sqrt{x}}}\bigg)+\mathcal{C}.

10 Likes
#453

Let g(x)=max|y^2 -xy| (0 \leq y \leq 1, x,y \in \mathbb{R}), then the minimum value of g(x) is ??

3 Likes
#454

Is that ans correct.

#455

3-2\sqrt 2?

4 Likes
#456

Yes sir , can u share your method sir ??

#457

Between why not we r considering g(x) = 0 ? putting y=0 , why can't it be minimum value of g(x) ?? @Hari_Shankar sir

1 Like
Doubt from functions
#458

We have to take cases: The cases are

(a) x<0, then |y^2-xy|=y^2-xy=y(y-x) \le y-x \le 1-x and hence g(x)=1-x in this interval

(b) x>2. then x>y \Rightarrow xy>y^2 \Rightarrow |y^2-xy| = xy-y^2 = h(y), h'(y) =x-2y>0 and so the maximum is attained at y=1 and g(x)=x-1

(c) 1<x<2, we have h(y)=y(x-y) \le \dfrac{x^2}{4} by AM-GM and so g(x)= \dfrac{x^2}{4}

(d) 0<x<1, we have two maxima, one at y=\dfrac{x}{2} which is \dfrac{x^2}{4} and one at x=1 which is 1-x

Now \dfrac{x^2}{4} \ge 1-x \Rightarrow x \le 2\sqrt 2 -2

Its easy to see from drawing the graph that the least value is therefore 1-(2\sqrt 2 -2) = 3 - 2\sqrt 2

12 Likes
#459

Nice approach @Hari_Shankar sir , also sir can u tell what's the fault in g(x)=0 , when y = 0 ? Why can't it be minimum ? As for minima we r free to choose values of y right ??

1 Like
#460

because g(x) is the maximum of y^2-xy as y varies from 0 to 1

4 Likes
#461

How to do these types of questions sir ?

#462

I meant this question.

1 Like
#463

Assume a€{0,4} where 4a-a^2=t.
Breaking integrals in 0 to a, a to 4-a, 4-a to 4. By drawing graph mod can be opened easily. Then substitute t=4a-a^2 and integrate. Then it's normal method of finding minimum. Similar question is in reso if u want I will share solution

1 Like
#464

@Gajraj_Singh hint: form perfect square answer is 1/16 . (I have got it bro. What answer r u getting @Siddhant_Mudholkar bro )

3 Likes
#465
1 Like
#466

@Jagdish_Singh sir isn't answer should be 1/16 ??

1 Like
#467

Right. may be some calculation Error.

1 Like
#468
5 Likes
#469

\bf{Another\; way:}

Let we assume \displaystyle I =\int^{1}_{0}x\bigg[xf(x)-(f(x))^2\bigg]dx

Then \displaystyle I =\frac{1}{4}\int^{1}_{0}x\bigg[4xf(x)-4(f(x))^2\bigg]dx

So we have

\displaystyle I =\frac{1}{4}\int^{1}_{0}x\bigg[x^2-\bigg(2f(x)-x\bigg)^2\bigg]dx\leq \frac{1}{4}\int^{1}_{0}x^3dx=\color{blue}\frac{1}{16}.

And equality hold when \displaystyle \color{green}f(x)=\frac{x}{2}.

9 Likes
#470

@Hari_Shankar Bhatt Sir plz see that problem Sir.

2 Likes
#472

Thank you @Azimuddin_Sheikh @Jagdish_Singh Sir

1 Like