NMTC BHASKARA Junior Level 2024 Solved Question Paper PDF

$$x^2 = 1 - x$$

We can find

 $$x^3 = x - x^2 = x - (1-x) = 2x-1$$

$$x^4 = 2x^2 - x = 2(1-x) -x =2-3x $$, 

$$x^5 = 2x-3x^2 = 2x-3(1-x) = 5x - 3$$, 

$$x^6 = 5x^2-3x = 5(1-x)-3x = 5 - 8x$$ 

$$x^7 = 5x-8x^2 = 5x-8(1-x) = 13x - 8$$. 

Hence $$\dfrac{x^7 + 34}{x + 2} = \dfrac{13x - 8 + 34}{x + 2} = \dfrac{13(x + 2)}{x + 2} = 13$$.

Minute hand comples $$360\degree$$ in $$60$$ minutes. 

This means in a minute it covers $$\dfrac{360}{60} = 6\degree$$

The minute hand at $$38$$ minutes is at $$38 \times 6\degree = 228\degree$$ from the $$12$$ mark. 

The hour hand covers $$360\degree$$ in $$12$$ hours. In one hour it covers $$ \dfrac{360}{12} = 30\degree$$. And in one minute it covers $$\dfrac{30\degree}{60} = 0.5\degree$$

Thus, the hour hand is at $$9 \times 30\degree + 38 \times 0.5\degree = 270\degree + 19\degree = 289\degree$$. 

The angle between them is $$289\degree - 228\degree = 61\degree$$.

TO BE FILLED - figure required

Transposingto get the same bases on either side of the equation we get,

$$2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$$, that is $$\dfrac{3}{2} \cdot 2^{2x} = \dfrac{4}{\sqrt{3}} \cdot 3^{x}$$. 

This simplifies to 

$$2^{2x - 3} = 3^{\frac{2x-3}{2}}$$

$$\left(\dfrac{2}{\sqrt{3}}\right)^{2x - 3} = 1 \implies 2x - 3 = 0$$

$$x = \dfrac{3}{2}$$. 

This gives $$a = 3$$ and $$b = 2$$

$$\dfrac{a+b}{a-b} = \dfrac{5}{1} = 5$$.

If $$x^2 + 1$$ divides the cubic, the other factor must be $$x + d$$,

Let

$$x^3 + ax^2 + bx + c = (x^2 + 1)(x + d) = x^3 + dx^2 + x + d$$. 

Comparing coefficients gives $$a =d, b = 1$$ and $$c = d$$. Since $$a$$ can be any of the 12 allowed values and then $$c$$ gets fixed, there are 12 such polynomials.

The denominator cannot be zero, so $$x\neq2,3,4,5$$.

We are given $$\dfrac{(x+2)(x+3)(x+4)(x+5)}{(x-2)(x-3)(x-4)(x-5)}=1$$.

Multiplying both sides by the denominator, we get $$ (x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5) $$.

We can pair the factors on both sides to get the same coefficient for $$x$$ as follows:

$$ (x+2)(x+5)=x^2+7x+10 $$

$$ (x+3)(x+4)=x^2+7x+12 $$.

Similarly,

$$ (x-2)(x-5)=x^2-7x+10 $$

$$ (x-3)(x-4)=x^2-7x+12 $$.

Therefore, the equation becomes $$ (x^2+7x+10)(x^2+7x+12)=(x^2-7x+10)(x^2-7x+12) $$.

Let $$ A=x^2+10 $$.

Then we get $$ (A+7x)(A+7x+2)=(A-7x)(A-7x+2) $$.

Expanding both sides gives $$ A^2+14Ax+49x^2+2A+14x=A^2-14Ax+49x^2+2A-14x $$.

Cancelling the common terms, we get $$ 28Ax+28x=0 $$.

Factoring, $$ 28x(A+1)=0 $$.

Since $$ A=x^2+10 $$, we get $$ 28x(x^2+11)=0 $$.

Therefore, $$ x(x^2+11)=0 $$.

So either $$ x=0 $$ or $$ x^2+11=0 $$.

But $$ x^2+11=0 $$ gives $$ x^2=-11 $$, which has no real solution because the square of a real number can never be negative.

Hence, the only real solution is $$ x=0 $$.

Therefore, the number of real solutions is $$ 1$$.

Squaring the equations we get

$$a^2 = 23a + b$$ and $$b^2 = 23b + a$$. 

Subtracting the two gives us

$$a^2 - b^2 = 22(a-b) \implies (a-b)(a+b) = 22(a-b)  $$

As $$a \neq b$$ this gives

$$a + b = 22$$.

Adding the two equations we get, 

$$a^2 + b^2 = 24(a + b) = 528$$

$$\sqrt{a^2 + b^2 + 48} = \sqrt{576} = 24$$.

 $$PA = PB$$ and the angle at $$P$$ is $$72\degree$$

The base angles are 

$$\angle PAB = \angle PBA = \dfrac{1}{2} (180-72) =54\degree$$ (Angle sum property of an isosceles triangle. 

As $$AC$$ is parallel to $$PB$$

$$\angle CAB = \angle ABP = 54\degree$$(Interior alternate angles)

Now, using the alternate segment theorem,

$$\angle ACB = \angle ABP = 54\degree$$

$$\implies \angle CBA = 180\degree - 54\degree - 54\degree = 72\degree$$

Now, ABCD forms a cyclic quadrilateral, which implies that opposite angles are supplementary.

$$ \angle CDA = 180\degree- \angle CBA = 180\degree- 72\degree = 108\degree$$

Squaring the equation on both sides we get

 $$19^8 + 19^x = 19^4(19^x + 1)$$

$$19^8 - 19^4 = 19^x(19^4 - 1)$$,

$$19^4(19^4 - 1) = 19^x(19^4 - 1)$$ 

Hence $$ x = 4$$. 

Substituting $$x = 4$$ in the options, only  option 4 satisfies the condition

$$3(16) - 11(4) - 4 = 48 - 48 = 0$$ 

The general term can be expressed as

$$n(n+3)^2 = n^3 + 6n^2 + 9n$$ for $$n$$ from 1 to 25. 

$$\sum n(n+3)^2 = \sum n^3 + 6\sum n^2 + 9\sum n$$

$$\sum n^3 = \dfrac{n^2(n+1)^2}{4} =   \dfrac{25^2(25+1)^2}{4}= 105625$$

$$\sum n^2 =\dfrac{n(n+1)(2n+1)}{6} =\dfrac{25(25+1)(50+1)}{6} = 5525$$ 

$$\sum n =\dfrac{n(n+1)}{2}=\dfrac{25(26+1)}{2}=   325$$, 

We get $$S = 105625 + 6(5525) + 9(325) = 141700$$. 

Hence $$\dfrac{S}{325} = 436$$.

We are given

$$a_1=\dfrac{1}{2}$$

and

$$a_n=\dfrac{a_{n-1}}{2na_{n-1}+1}.$$

We want to find

$$a_1+a_2+a_3+\cdots+a_{2024}.$$

First, take the reciprocal of both sides of the given equation:

$$\dfrac{1}{a_n}=\dfrac{2na_{n-1}+1}{a_{n-1}}.$$

Splitting the fraction,

$$\dfrac{1}{a_n}=2n+\dfrac{1}{a_{n-1}}.$$

Therefore,

$$\dfrac{1}{a_n}=\dfrac{1}{a_{n-1}}+2n.$$

Now we know that

$$a_1=\dfrac{1}{2},$$

$$\dfrac{1}{a_1}=2\times 1$$

$$\dfrac{1}{a_2}=\dfrac{1}{a_1}+2 \times 2$$

$$\dfrac{1}{a_3}=\dfrac{1}{a_2}+2 \times 3$$

$$\dfrac{1}{a_4}=\dfrac{1}{a_3}+2 \times 4$$

.

.

.

$$\dfrac{1}{a_n}=\dfrac{1}{a_{n-1}}+2 \times n$$

Adding all the equations we get

$$\dfrac{1}{a_n}=2\sum n = n(n+1)$$

Hence,

$$a_n=\dfrac{1}{n(n+1)}.$$

Now split this fraction:

$$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}.$$

Therefore,

$$a_n=\dfrac{1}{n}-\dfrac{1}{n+1}.$$

Now substitute this into the required sum:

$$a_1+a_2+a_3+\cdots+a_{2024}$$

$$=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\cdots+\left(\dfrac{1}{2024}-\dfrac{1}{2025}\right).$$

Notice that most terms cancel:

$$\cancel{1-\dfrac{1}{2}}+\cancel{\dfrac{1}{2}-\dfrac{1}{3}}+\cancel{\dfrac{1}{3}-\dfrac{1}{4}}+\cdots+\left(\dfrac{1}{2024}-\dfrac{1}{2025}\right).$$

Everything in the middle cancels, leaving only

$$1-\dfrac{1}{2025}.$$

Therefore,

$$1-\dfrac1{2025}=\dfrac{2025-1}{2025}$$

$$=\dfrac{2024}{2025}$$

Applying logarithms to base 10 on both sides we get

$$\log_{10}{x^{1 + \log_{10} x} }= \log_{10}{10x}$$

$$({1 + \log_{10} x})\log_{10}{x }= \log_{10}{10} + \log_{10}{x} $$

$$({1 + \log_{10} x})\log_{10}{x }= 1 + \log_{10}{x} $$

writing $$t = \log_{10} x$$, the equation becomes

$$(1 + t)t = 1 + t$$, 

$$t^2 = 1$$ 

$$t = \pm 1$$. 

$$log_{10}{x} = \pm 1$$

$$x = 10, x = \dfrac{1}{10}$$

Since we know that $$\alpha>\beta$$, the roots are $$\alpha = 10$$ and $$\beta = \dfrac{1}{10}$$, 

This gives us

$$\alpha + \dfrac{1}{\beta} = 10 + 10 = 20$$.

Consider 

$$\triangle OCE, \triangle ODF$$

$$ \angle OCE = \angle ODF = 90\degree$$ (Angle made by radius with the tangent)

$$\angle COE = \angle DOF$$

Thus, by AA similarity criterion, the triangles are similar to each other. We can extend this to all the circles to find that the triangles made in such a way by the line passing through their centres, the radius and the tangent are all similar.

Let $$OE = x$$

Thus, using similarity we can write

$$ \dfrac{x}{r_1} = \dfrac{x + r_1+r_2}{r_2} $$

We can further extend this to all four circles to write

$$ \dfrac{x}{r_1} = \dfrac{x + r_1+r_2}{r_2} = \dfrac{x+r_1+r_2+r_3}{r_3} = \dfrac{x+r_1+r_2+r_3+r_4}{r_4}$$

Let the common value be $$k$$.

Then,

$$\dfrac{x}{r_1}=k$$

$$x=kr_1.$$

Also,

$$\dfrac{x+r_1}{r_2}=k$$

$$x+r_1=kr_2.$$

Since $$x=kr_1$$,

$$kr_1+r_1=kr_2$$

$$r_1(k+1)=kr_2$$

Therefore,

$$\dfrac{r_2}{r_1}=\dfrac{k+1}{k}.$$

Now consider the next pair:

$$\dfrac{x+r_1+r_2}{r_3}=k.$$

Since $$x+r_1=kr_2$$,

$$x+r_1+r_2=kr_2+r_2$$

$$=r_2(k+1).$$

Therefore,

$$kr_3=r_2(k+1)$$

and hence

$$\dfrac{r_3}{r_2}=\dfrac{k+1}{k}.$$

Similarly,

$$\dfrac{r_4}{r_3}=\dfrac{k+1}{k}.$$

Thus,

$$\boxed{\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}=\dfrac{r_4}{r_3}}$$

Therefore, the radii form a geometric progression.

$$\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}=\dfrac{r_4}{r_3}$$

Let this common ratio be $$k$$.

Thus, the radii form a geometric progression:

$$r_1,\ r_2,\ r_3,\ r_4$$

We are given

$$r_1=7$$ and $$r_3=28$$

Therefore,

$$r_3=r_1k^2$$

$$28=7k^2$$

$$\implies k^2=4$$

Since the radii are increasing, $$k=2$$.

Therefore,

$$r_4=r_3k$$

$$r_4=28(2)$$

$${r_4=56}$$

The constant $$q$$ was taken correctly which can be found from the prouct of the wrong roots obtained. 

$$(-2)(-15) = 30$$, so $$q = 30$$.

The original equation is 

$$x^2 + 13x + 30 = 0$$ 

Factorising we get

$$ (x+3)(x+10) = 0$$

with roots $$\alpha = -3$$ and $$\beta = -10$$. 

Then

$$\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{9 + 100}{30} = \dfrac{109}{30}$$ and the product is 1.

Hence, the required equation is 

$$x^2 - (\text{sum of roots})x + \text{(product of roots)}= 0$$.

$$x^2 - \left(\dfrac{109}{30}\right)x + 1= 0$$.

$$30x^2 - 109x + 30 = 0$$.

$$(1 + xy + x + y)^2 - (1 - xy + x - y)^2 = (1+xy+x+y+1-xy+x-y)(1+xy+x+y -1+xy-x+y)$$

Simplifying the RHS by cancelling out terms we get 

$$(2+2x)(2xy+2y) = 4y(1+x)(1+x) = 4y(1+x)^2$$

Thus, $$ k = 4$$

$$x^{10} + 1 = (x^2)^5 + 1$$ 

Using the property that $$ a^n + b^n$$ is exactly divisible by $$a+b$$ when $$n$$ is odd, we can say that

 $$x^2 + 1$$ divides $$x^{10}+1$$ exactly 

We can simply multiply $$ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k$$ with $$x^2 +1$$ and compare with $$x^{10}+1$$

$$(ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k)(x^2+1) = ax^{10}+bx^9 + cx^8 + dx^7 + ex^6 + fx^5 + gx^4 + hx^3 + kx^2 + ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k$$

Grouping the coefficients of the terms we get

$$ax^{10} + bx^9 + (a+c)x^8 + (b+d)x^7 + (c+e)x^6 + (d+f)x^5 + (e+g)x^4 + (f+h)x^3 + (g+k)x^2 + hx + k = x^{10}+1$$

Equating the coefficients on both sides we get

$$ a= 1, b = 0, c=-1, d=0, e=1, f=0, g=-1, h=0,k=1$$

 $$a^{2024} + b^{2024} + c^{2024} + d^{2024} + e^{2024} + f^{2024} + g^{2024} + h^{2024} + k^{2024} = 1+0+1+0+1+0+1+0+1 =5 $$ 

The four positive roots have sum 4 and product 1, so their arithmetic mean is 1 and their geometric mean is also 1. 

Equality in the AM-GM inequality forces all four roots to equal 1, so the equation is $$(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$$. 

Hence $$a = 6$$, $$b = -4$$ and $$a + b = 2$$.

Since $$AB$$ is perpendicular to the diameter at $$B$$ hence it is also a tangent at $$B$$

Now the two tangents from the same point to the circle are equal which implies

$$EB = ED$$ and $$\angle EDO = 90\degree$$

Consider $$\triangle EDO, \triangle EBO$$

$$ED = EB, BO =DO = r, EO = EO$$

Thus, the triangles are congruent to each other

$$\angle EOB = \angle EOD = x$$

Consider $$\triangle DOC$$

$$\angle DOC = 180 -2x$$

$$ CO = DO = r$$. 

Thus, it's an isosceles triangle

$$\angle DCO=\dfrac{1}{2}(180-(180-2x))=x$$

Now, consider $$\triangle EOB, \triangle ACB$$

$$\angle ACB = \angle EOB = x$$

$$\angle ABC = \angle EBO = 90\degree$$

Thus, by AA similarity criterion, the triangles are similar to each other

$$ \dfrac{AE}{AB} = \dfrac{BO}{BC} = \dfrac{r}{2r} = \dfrac{1}{2}$$

$$\implies AB = 2AE$$

$$ k = 2$$

In any triangle,

$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$

We are given that the ratios of the tangents of the three angles are

$$\tan A:\tan B:\tan C=1:2:3.$$

Using the ratio, we can write them as

$$\tan A=t,\qquad \tan B=2t,\qquad \tan C=3t.$$

Substituting these values in the triangle identity,

$$t+2t+3t=t(2t)(3t)$$

$$6t=6t^3$$

Dividing both sides by $$6t$$,

$$1=t^2$$

Since the angles of a triangle are positive, their tangents are positive in this case. Hence,

$$t=1.$$

Therefore,

$$\tan A=1,\qquad \tan B=2,\qquad \tan C=3.$$

$$\tan B=\dfrac{\text{opposite}}{\text{adjacent}}=2=\dfrac{2}{1},$$

we can consider a right triangle with sides $$2,1,\sqrt{5}$$. Therefore,

$$\sin B=\dfrac{2}{\sqrt{5}}.$$

Similarly,

$$\tan C=3=\dfrac{3}{1},$$

so the hypotenuse is

$$\sqrt{3^2+1^2}=\sqrt{10}.$$

Therefore,

$$\sin C=\dfrac{3}{\sqrt{10}}.$$

By the sine rule,

$$\dfrac{AC}{AB}=\dfrac{\sin B}{\sin C}.$$

Substituting the values of $$\sin B$$ and $$\sin C$$,

$$\dfrac{AC}{AB}=\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{3}{\sqrt{10}}}.$$

Dividing by a fraction is the same as multiplying by its reciprocal:

$$\dfrac{AC}{AB}=\dfrac{2}{\sqrt{5}}\times\dfrac{\sqrt{10}}{3}.$$

Therefore,

$$\dfrac{AC}{AB} =\dfrac{2\sqrt{10}}{3\sqrt{5}}.$$

Since

$$\dfrac{\sqrt{10}}{\sqrt{5}}=\sqrt{\dfrac{10}{5}}=\sqrt{2},$$

we get

$$\dfrac{AC}{AB}=\dfrac{2\sqrt{2}}{3}.$$

Thus,

$${\dfrac{AC}{AB}=\dfrac{2\sqrt{2}}{3}}.$$

$$p=2,\qquad q=2,\qquad r=3,$$

then

$$p+q+r=2+2+3=7$$

Let the number be $$N$$.

When the number is successively divided by $$5,6,$$ and $$7$$, the remainders are $$3,2,$$ and $$2$$ respectively.

Therefore,

$$N=5q_1+3$$

The quotient $$q_1$$ is then divided by $$6$$:

$$q_1=6q_2+2$$

The quotient $$q_2$$ is then divided by $$7$$:

$$q_2=7q_3+2$$

Now substitute backwards.

From

$$q_2=7q_3+2,$$

we get

$$q_1=6(7q_3+2)+2$$

$$=42q_3+12+2$$

$$=42q_3+14.$$

Therefore,

$$N=5(42q_3+14)+3$$

$$=210q_3+70+3$$

$$=210q_3+73.$$

So the number can be written as

$$N=210q_3+73.$$

Let the required number be $$d$$.

Since $$d$$ leaves the same remainder when dividing $$30,53,$$ and $$99$$, we can write

$$30=dq_1+r$$

$$53=dq_2+r$$

$$99=dq_3+r$$

where $$r$$ is the same remainder in all three cases.

Now, if two numbers leave the same remainder when divided by the same number, their difference will be exactly divisible by that number. 

$$53-30=(dq_2+r)-(dq_1+r)$$

The remainders cancel:

$$53-30=d(q_2-q_1)$$

Therefore, $$53-30$$ is exactly divisible by $$d$$.

Similarly,

$$99-53$$ is also exactly divisible by $$d$$.

$$53-30=23$$

$$99-53=46.$$

Therefore, $$d$$ must be a common divisor of $$23$$ and $$46$$.

The greatest common divisor is

$$\gcd(23,46)=23.$$

Hence, the greatest possible value of $$d$$ is $$ 23$$

Putting $$y = x + 1$$, so that $$x = y - 1$$, gives $$f(y) = (y-1)^2 - 3(y-1) + 2 = y^2 - 5y + 6$$. The roots of $$y^2 - 5y + 6 = 0$$ are 2 and 3, so $$\alpha^2 + \beta^2 = 4 + 9 = 13$$.

The largest cylinder has radius $$\dfrac{a}{2}$$ and height $$a$$, so its volume is 

$$\pi \dfrac{a^3}{4} = \dfrac{22}{7} \times \dfrac{a^3}{4} = \dfrac{11a^3}{14}$$. 

The remaining solid has volume

$$a^3 - \dfrac{11a^3}{14} = \dfrac{3a^3}{14}$$, so

$$k = \dfrac{3}{14}$$ and $$p + q = 3 + 14 = 17$$.

Through trial and error we can see that $$x = 1$$ and $$x = -2$$ are roots of the polynomial

We can factorise it as 

$$(x - 1)(x + 2)(5x^2 + 4x + 4)$$.

The quadratic $$5x^2 + 4x + 4$$ has discriminant $$16 - 80 < 0$$, so it is the irreducible factor. 

Hence $$a = 5$$, $$b = 4$$, $$c = 4$$ and 

$$a^2 + b^2 - c^2 = 25 + 16 - 16 = 25$$.

From the area of the square, we can deduce that length of the side of the square is $$6$$ cm.

$$OB$$ is the radius of the circle and can be found as it's also the diagonal of the square which is $$6\sqrt{2}$$ cm.

$$LQ = \sqrt{OQ^2 + OL^2} = \sqrt{6^2 + (6\sqrt{2})^2} =\sqrt{108} cm$$

Consider $$\triangle PQD, \triangle LQO$$

$$\angle QOL = \angle QDP = 90\degree$$ (Angle made by the diameter on any point of the circle is 90 \degree)

$$\angle DQP = \angle OQL$$ (Common)

Thus, by AA similarity criterion, $$\triangle PQD \sim \triangle LQO$$

$$\dfrac{DQ}{OQ} = \dfrac{PQ}{QL} = \dfrac{2\times 6\sqrt{2}}{\sqrt{108}} $$

$$ DQ =  6\sqrt{2}\dfrac{2\times 6\sqrt{2}}{\sqrt{108}} = 8\sqrt{3}$$ cm

$$ x = 24$$

$$1 + \dfrac{1}{4}\left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)^2$$ can be expanded using $$(a+b)^2$$ formula to get

 $$1 + \dfrac{1}{4}\left(\dfrac{a}{b} + \dfrac{b}{a} - 2\right) = \dfrac{4ab + a^2 + b^2 - 2ab}{4ab} = \dfrac{(a+b)^2}{4ab}$$. 

As $$a,b >0$$, its square root is 

$$\dfrac{a+b}{2\sqrt{ab}}$$

Now, the expression reduces to 

$$\dfrac{2\sqrt{ab}}{a+b} \times \dfrac{a+b}{2\sqrt{ab}} = 1$$

With first term $$A$$ and ratio $$r$$, the sum of infinite terms of a geometric progression is

$$\dfrac{A}{1-r}$$

From the statement in the question we can write

$$\dfrac{A}{1-r} = \dfrac{1}{8} \cdot \dfrac{A^2}{1-r^2}$$ 

$$A = 8(1 + r)$$

We are also given that second term is $$6$$

$$\implies Ar = 6$$ 

Substituting in the above equation we get 

$$4r^2 + 4r - 3 = 0 \implies r = \dfrac{1}{2}  \text{ or}  r = \dfrac{-3}{2} $$

For the sum of infinite terms to be a finite value, we need $$|r|<1$$

$$r = \dfrac{1}{2}$$ and $$A = 12$$. 

The first and fourth terms add to 

$$A+ Ar^3 = 12\left(1+\dfrac{1}{8}\right) = \dfrac{27}{2}$$, so $$\left[\dfrac{p}{q}\right] = \left[13.5\right] = 13$$.

We are given the expression

$$10^{-\frac{1}{2}(\log_{10}9-2)}$$

First, distribute $$-\frac{1}2$$ inside the bracket:

$$10^{-\frac12\log_{10}9+1}$$

Using the rule

$$a^{m+n}=a^m\times a^n,$$

we can write this as

$$10^{-\frac12\log_{10}9}\times10$$

Now use the logarithm rule

$$10^{\log_{10}9}=9.$$

Therefore,

$$10^{-\frac12\log_{10}9}=\left(10^{\log_{10}9}\right)^{-\frac12}.$$

So,

$$\left(10^{\log_{10}9}\right)^{-\frac12}=9^{-\frac12}.$$

$$9^{-\frac12}=\dfrac{1}{3}.$$

Therefore, the original expression becomes

$$10\times\dfrac{1}{3}=\dfrac{10}{3}.$$

Hence,

$$a=10,\qquad b=3.$$

Therefore,

$$a-b=10-3=7$$

Let the side of the square be $$2r$$ and $$DE = t$$. 

Tangents drawn to a circle from the same point are equal. Let $$T$$ be the point of contact of $$BE$$ with the semicircle

$$BA = BT, DE = DT$$

$$\implies BE = BA + DE = 2r + t$$

From the right angled triangle $$BCE$$ we get 

$$(2r + t)^2 = (2r)^2 + (2r - t)^2$$, 

which simplifies to $$t = \dfrac{r}{2}$$.

Then the area is $$\dfrac{1}{2} \times 2r \times \left(2r - \dfrac{r}{2}\right) = \dfrac{3r^2}{2} = 216$$

$$r^2 = 144 \implies r = 12$$.

Every power appearing in $$f$$ is odd, so $$f$$ is an odd function and $$f(4) = -f(-4) = -18$$, giving $$|f(4)| = 18$$. Since $$|2\cos x|$$ can be at most 2, the maximum value of the sum is $$18 + 2 = 20$$.