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We have, 

f(x) = x³(x – 1)² 

Differentiate w.r.t x, we get, 

f ‘(x) = 3x²(x – 1)² + 2x³(x – 1) 

= (x – 1)(3x²(x – 1) + 2x³) 

= (x – 1)(3x³ – 3x² + 2x³) 

= (x – 1)(5x³ – 3x²) 

= x² (x – 1)(5x – 3)

 For all maxima and minima, 

f ’(x) = 0 = x²(x – 1)(5x – 3) = 0

= x = 0, 1,\(\frac{3}{5}\)

At x = \(\frac{3}{5}\) 

f ’(x) changes from –ve to + ve

Since, 

x = \(\frac{3}{5}\) is a point of Minima

At x = 1 

f ‘(x) changes from –ve to + ve 

Since, 

x = 1 is point of maxima.

Given as f(x) = 1/(x² + 2)

Differentiate the above with respect to x

f'(x) = -2x/(x² + 2)²

For the local minima and local maxima, f'(x) = 0

-2x/(x² + 2)² = 0 

So, x = 0, now for the values close to x = 0 and to the left of 0, f'(x) > 0

Also for the values x = 0 and to the right of 0, f’(x) < 0

So, by first derivative test, x = 0 is a point of local maxima and local minima value of f (x) is 1/2.

Correct option is (D) A = GH

We have, 

g (x) = x³ – 3x 

Differentiate w.r.t x then we get, 

g’ (x) = 3x² – 3 

Now, 

g‘(x) =0 = 3x² = 3 

⇒ x = ±1 

Again, differentiate g’(x) = 3x² – 3 

g’’(x)= 6x 

g’’(1) = 6 > 0 

g’’( – 1) = – 6 > 0 

By second derivative test, x = 1 is a point of local minima and local minimum value of g at 

x =1 is g(1) = 1³ – 3 

= 1 – 3 = – 2 

However, 

x = – 1 is a point of local maxima and local maxima value of g at 

x = – 1 is g( – 1) = ( – 1)³ – 3( – 1) 

= – 1 + 3 = 2 

Hence, 

The value of Minima is – 2 and Maxima is 2.

f(x) = (x – 5)⁴ 

Differentiate w.r.t x 

f ’(x) = 4(x – 5)³ for local maxima and minima 

f ‘(x) = 0 

= 4(x – 5)³ = 0 

= x – 5 = 0 

x = 5 

f ‘(x) changes from –ve to + ve as passes through 5. 

So, 

x = 5 is the point of local minima 

Thus, local minima value is f(5) = 0

Given as f(x) = xe^x

f'(x) = e^x + x e^x = e^x(x + 1)

f”(x) = e^x(x + 1) + e^x

= e^x(x + 2)

For the maxima and minima,

f'(x) = 0

e^x(x + 1) = 0

x = – 1

f’’ (– 1) = e ^{– 1} = 1/e > 0

x = – 1 is point of local minima

Thus, local min = f (– 1) = – 1/e

Given,

S_n = 2n² + 7n

We know that,

a_n = S_n − S_n-1

∴ a_n = 2n² + 7n − {2(n − 1)² + 7(n − 1)}

= 2n² + 7n − {2(n² − 2n + 1) + 7n − 7}

= 2n² + 7n − {2n² − 4n + 2 + 7n − 7}

= 4n + 5

Given series,

x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞ is in G.P with common ratio \(\frac{1}{r}\).

and series y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ is in G.P with common ratio − \(\frac{1}{r}\).

and series z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ is in G.P with common ratio − \(\frac{1}{r^2}\)

Sum of infinite terms of series

x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞ is x = \(\frac{a}{1-\frac{1}{r}}\)

Sum of infinite term of series

y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ is y = \(\frac{b}{1-\big(\frac{1}{r}\big)}\) = \(\frac{b}{1+\frac{1}{r}}\)

Sum of infinite terms of series

z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ is z = \(\frac{c}{1-\frac{1}{r^2}}\)

Now, LHS = \(\frac{xy}{z}\)

= \(\frac{\bigg\{\frac{a}{1 − \frac{1}{ r}} \bigg\} \bigg\{\frac{b}{1 + \frac{1}{ r}}\bigg\}}{\frac{c}{1-\frac{1}{r^2}}}\)

= \(\frac{\frac{ab}{\big(1-\frac{1}{r^2}\big)}}{\frac{c}{\big(1-\frac{1}{r^2}\big)}}\)

= \(\frac{ab}{c}\) = R.H.S

Let three terms of G.P. \(\frac{a}{r}\), a, ar

Given \(\frac{a}{r}\). a. ar = 27

⇒ a³ = 27 = (3)³

⇒ a = 3

∴ middle term = 3

We have T_n = a + (n – 1)d 

20^th term of series 

= [20^th term of 2,4, 6….] x [20^th term of 4, 6, 8….. ] 

= [2 + (20 -1)²] x [4+ (20-1)²] 

= (2+ 38)(4 + 38) 

= (40)(42) = 1680

It is given that p,q,r are in G.P.

∴ q² + pr

Now, px² + 2qx + r = 0

⇒ px² + 2\(\sqrt{prx}\) + r = 0

⇒ \((\sqrt{px} +\sqrt{r})^2\) = 0

⇒ \((\sqrt{px} +\sqrt{r})\) = 0

⇒ x = \(-\sqrt{\frac{r}{p}}\)

It is given that the equations px², + 2px + r = 0 and dx² + 2ex + f − 0 have a common root and the equation px², + 2qx + r = 0 has equal roots equal to −\(\sqrt{\frac{r}{p}}\)

∴ −\(\sqrt{\frac{r}{p}}\) is a root of the equation dx² + 2ex + f = 0

⇒ \(d\frac{r}{p} − 2e\sqrt{\frac{r}{p}} + f\) = 0

⇒ \(\frac{d}{p} − 2e\sqrt{\frac{1}{pe}} + \frac{f}{r}\)             [Dividing though out by r]

⇒ \(\frac{d}{p} − \frac{2e}{q} +\frac{f}{r}\) = 0              [∴ q² = pr]

⇒ 2\(\frac{e}{q}\) = \(\frac{d}{p}\) + \(\frac{f}{r}\)

⇒ \(\frac{d}{p}\), \(\frac{e}{q}\), \(\frac{f}{r}\) are in A. P

Hence proved

\(\frac{1}{\sqrt{b}+\sqrt{c}} \), \(\frac{1}{\sqrt{c}+\sqrt{a}} \), \(\frac{1}{\sqrt{a}+\sqrt{b}} \) will be in A. P.

If \(\frac{1}{\sqrt{c}+\sqrt{a}} \) − \(\frac{1}{\sqrt{b}+\sqrt{c}} \) = \(\frac{1}{\sqrt{a}+\sqrt{b}} \) − \(\frac{1}{\sqrt{c}+\sqrt{a}} \)

i.e if \(\frac{\sqrt{b}-\sqrt{a}}{(\sqrt{c}+\sqrt{a})(\sqrt{b}+\sqrt{c})} \) = \(\frac{\sqrt{c}-\sqrt{b}}{(\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{a})} \)

i.e if \(\frac{\sqrt{b}-\sqrt{a}}{(\sqrt{b}+\sqrt{c})} \) = \(\frac{\sqrt{c}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} \)

i.e if b − a = c + b

i.e if 2b = a + c

i.e if a, b, c are in A.P.

Thus, a, b, c are in A.P. ⇒ \(\frac{1}{\sqrt{b}+\sqrt{c}} \), \(\frac{1}{\sqrt{c}+\sqrt{a}} \), \(\frac{1}{\sqrt{a}+\sqrt{b}} \) are in A. P.

The two digit natural numbers which leave a remainder 5, when divided by 7 are 12, 19, 26 ...., 89, 96. 

∴ 12, 19, 26, ...., 89, 96 is an A.P. whose first term a = 12 and common difference d = 7. 

Let the last or nth term be T_n. 

Then, T_n = a + (n – 1) d, where n is the number of terms in A.P. 

⇒ 96 = 12 + (n – 1) 7

⇒ 84 = (n – 1) 7 ⇒ n – 1 = 12 ⇒ n = 13

∴ Required Sum = \(\frac{n}{2}\) (a + l) = \(\frac{13}{2}\)(12 + 96) 

= \(\frac{13}{2}\) × 108 

=13 × 54 = 702.

Let the first term of C.P. be ′a′ and the common ratio be ′r′ where −1 < r < 1

The G.P. is a, ar, ar² …

Therefore the sum of the infinite terms of the G.P. is \(\frac{a}{1−r}\) = 57 … (i)

If taking the cube of each terms the new G.P. is a³, a³r³, a³r⁶ , …

Therefore the sum of their cube is \(\frac{a^3}{1−r^3}\) = 9747 … (ii)

Taking the cube of the (i)

\(\frac{a^3}{1−r^3}\) = (57)³ ⇒ a³ = (57)³ (1 − r)³ … (iii)

Substituting the value of ′a′ in terms of r in (ii)

\(\frac{(57)^3 (1 − r)^3}{1 − r^3}\) = 9747

\(\frac{(1 − r)^3}{(1 − r)(1 − r^2 + r)}\) = \(\frac{9747}{ (57)^3}\)

\(\frac{(1 − r)^2}{(1 + r^2 + r)}\) = \(\frac{1}{19}\)

\(\frac{1 + r^2 − 2r}{1 + r^2 + r}\) = \(\frac{1}{19}\)

19(1 + r² − 2r) = 1 + r² + r

18 r² − 39r + 18 = 0

6 r² − 13r + 6 = 0

6 r² − 9r − 4r + 6 = 0

(2r − 3)(3r − 2) = 0

Therefore, r =\(\frac{3}{2}\) or r = \(\frac{2}{3}\) since r < 1.

∴ r = ^{\(\frac{2}{3}\)}

Substitute in eq. (i),

\(\frac{a}{1 − \frac{2}{3}}\) = 57

⇒ 3a = 57

⇒ a = \(\frac{57}{3}\) = 19

Thus, the first term of the GP. is 19 and the common ratio is \(\frac{2}{3}\).

The G.P. is 19, \(\frac {38}{3}\), \(\frac {76}{6}\) and so on.

Answer : (d) 3

log₃ 2, log₃ (2^x – 5) and log₃ (2^x - \(\frac{7}{2}\)) are in A.P

⇒ 2 log₃ (2^x – 5) = log₃ 2 + log₃ (2^x – 7/2) (∵ a, b, c in A.P ⇒ 2b = a + c)

⇒ log₃ (2^x – 5)² = log₃[2(2^x – 7/2)] (∵ a log b = log b^a and log a + log b = log ab)

⇒ (2^x – 5)² = 2^x+1 – 7

Let 2^x = y. Then, 

(y – 5)² = 2y – 7 

⇒ y² – 10y + 25 = 2y – 7 

⇒ y² – 12y + 32 = 0 

⇒ (y – 8) (y – 4) = 0 

⇒ y = 8 or 4 

⇒ 2^x = 8 or 2^x = 4 

2^x = 8 

⇒ x = 3 (∵ 2^x = 4 shall make the term log₃(2^x – 5) negative which is not possible)

(d) 16

Let 

S = 4 + \(\frac{4+d}{5}\) + \(\frac{4+2d}{5^2}\) + \(\frac{4+3d}{5^3}\)+ ........∞

∴ \(\frac{1}{5}\)S = \(\frac{4}{5}\) + \(\frac{4+d}{5^2}\) + \(\frac{4+2d}{5^3}\)+ ........∞

⇒ S - \(\frac{1}{5}\)S = 4 + \(\frac{d}{5}\)\(\bigg[1+\frac{1}{5}+\frac{1}{5^2+}.......\infty\bigg]\)

⇒ \(\frac{4}{5}\)S = 4 + \(\frac{d}{5}\)x \(\frac{1}{1-\frac{1}{5}}\) ⇒ \(\frac{4}{5}\)S = 4 + \(\frac{d}{5}\) x \(\frac{5}{4}\)

⇒ \(\frac{4}{5}\)S = 4 + \(\frac{d}{4}\) ⇒ S = 5 + \(\frac{5d}{16}\)

Given S = 10

∴ 5 + \(\frac{5d}{16}\) = 10 ⇒ \(\frac{5d}{16}\) = 5 ⇒ d = 16.

Given 

a, b, c are in A.P. ⇒ 2b = a + c

Now ,

  \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P., if 

\(\frac{1}{\sqrt{c} +\sqrt{a}}\)- \(\frac{1}{\sqrt{b} +\sqrt{c}}\) = \(\frac{1}{\sqrt{a} +\sqrt{b}}\)- \(\frac{1}{\sqrt{c} +\sqrt{a}}\) 

⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{1}{\sqrt{b} +\sqrt{c}}\)+ \(\frac{1}{\sqrt{a} +\sqrt{b}}\)

⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{c}}{(\sqrt{b} + \sqrt{c}) (\sqrt{a} + \sqrt{b})}\)

⇒ 2(√b + √c)(√a +√b) = (√c +√a)(√a +2√b +√c)

⇒ 2(√ab + b + √ac + √bc) = (√ac + 2√bc + c + a + 2√ba + √ac)

⇒ 2√ab +2b +2√ac +2√bc = 2√ac + 2√bc + c + a + 2√ba

⇒ 2b = a + c, which is true as a, b, c are in A.P.

⇒  \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P

Let a₁ = a₃ – 2d, a₂ = a₃ – d, a₃ = a₃, a₄ = a₃ + d, a₅ = a₃ + 2d 

Then a₁ + a₃ + a₅ = –12, (given) 

⇒ a₃ – 2d + a₃ + a₃ + 2d = –12 

⇒ 3a₃ = –12

⇒ a₃ = – 4 

Also, a₁ . a₂ . a₃ = 8 (given) 

⇒ a₁ . a₂ = –2 (∵ a₃ = – 4) 

⇒ (a₃ – 2d) (a₃ – d) = –2 

⇒ (– 4 – 2d) (– 4 – d) = – 2 

⇒ (2 + d) (4 + d) = 1 

⇒ d² + 6d + 9 = 0  ⇒ (d + 3)² = 0 ⇒ d = – 3 

∴ a₁ = a₃ – 2d = – 4 –2 (–3) = 2.

 \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\)  are in A.P.

⇒ \(\frac{b+c-a}{a}\)+ 2, \(\frac{c+a-b}{b}\)+ 2, \(\frac{a+b-c}{c}\)+ 2 are in A.P. (Adding 2 to each term of A.P.)

⇒   \(\frac{b+c+a}{a}\), \(\frac{c+a+b}{b}\), \(\frac{a+b+c}{c}\)  are in A.P.

⇒  \(\frac{1}{a}\), \(\frac{1}{b}\),\(\frac{1}{c}\) are in A.P. (Dividing each term by a + b + c)

It is given that a is the A.M. ot b and c.

∴ a = \(\frac{b + c}{2}\) ⇒ b + c = 2a … (i)

Since G₁ and G₂ are two geometric means between b and c. Therefore, b, G₁, G₂ c is a G.P.

with common ratio r = \(\big(\frac{c}{b}\big)^\frac{1}{3}\)

∴ G₁ = br = b\(\big(\frac{c}{b}\big)^\frac{1}{3}\) = \(C^\frac{1}{3}b^\frac{1}{3}\) and

G₂ = br² = b\(\big(\frac{c}{b}\big)^\frac{2}{3}\) = \(b^\frac{1}{3}b^\frac{2}{3}\)

⇒ \(G_1^3\) = b²c and ⇒ \(G_2^3\) = bc²

⇒ \(G_1^3\) + \(G_2^3\) = b²c  bc² ⇒ \(G_1^3\) + \(G_2^3\) = bc(b + c)

⇒ \(G_1^3\) + \(G_2^3\)  = 2abc

[using (i)]

Hence proved

N term of 3, 5, 7,… is                 2n + 1

n^th term of 7, 8, 9,… is               6 + n

n^th term of 11² , 12² , 13², … is          (10 + n)²

∴ n^th term of the given series = \(\frac{2n\,+\,1}{(6+n)(10+n)^2}\)

We know that, A.M.> G.M.

∴ \(\frac{a\,+\,b}{2}\) > \(\sqrt{{a^2b^2}}\)

⇒ \(\frac{a^2\,+\,b^2}{2}\) > ab … (i)

Similarly \(\frac{b^2\,+\,c^2}{2}\) > \(\sqrt{{b^2c^2}}\)

⇒ \(\frac{b^2\,+\,c^2}{2}\) > bc ... (ii)

and \(\frac{c^2\,+\,a^2}{2}\) > \(\sqrt{{c^2a^2}}\)

⇒ \(\frac{c^2\,+\,a^2}{2}\) > ca  …. (iii)

On adding eqs. (i), and (iii) we get

\(\frac{a^2\,+\,b^2}{2}\) + \(\frac{b^2\,+\,c^2}{2}\) + \(\frac{c^2\,+\,a^2}{2}\) > ab + bc + ca

⇒ a² + b² + c² > ab + bc + ca

Hence proved

Given that, a, b, c are in G.P

⇒ b² = ac

⇒ b² − ac = 0

⇒ (b² − ac)² = 0

⇒ b⁴ + a²c² − 2b²ac = 0

⇒ a²b² + b²c² + 2b²ac = a²b² + b²c² + a²c² + b⁴

(Adding a²b² + b²c² both sides)

⇒ (ab + bc)² = (a² + b²)(b² + c²)

⇒ a² + b², ab + bc, b² + c² are in G.P.

Hence Proved

Given, log2, log(2ⁿ − 1) and log (2ⁿ + 3) are in A.P.

∴ log(2ⁿ + 3) − log(2ⁿ − 1) = log(2ⁿ − 1) - log2

⇒ log \(\big(\frac{2^n+3}{2^n−1}\big)\) = log \(\big(\frac{2^n-1}{2}\big)\)

Taking antilog, we get

⇒ \(\frac{2^n+3}{2^n−1}\) = \(\frac{2^n-1}{2}\)

⇒ 2ⁿ − 4.2ⁿ − 5 = 0

⇒ y² − 4y − 5 = 0 where y = 2ⁿ

⇒ y − 5 and y = −1 (reject)

Hence, 2ⁿ = 5

or n log2 = log5 or n = \(\frac{log5}{log2}\)

(i) Given a, b, c are in A.P

⇒ \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P.

[On dividing each term by a, b, c]

⇒ \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P.

(ii) b + c, c + a, a + b will be in A.P.

If (c + a) − (b + c) = (a + b) − (c + a)

i.e if a − b = b − c

i.e if 2b = a + c

i.e if a, b, c are in A.P

Thus, a, b, c are in A.P ⇒ b + c, c + a, a + b are in A.P.

Given equation is 13x – 12y = 25

Given equation is x + y = 0

Given equation is y = 6x ⇒ 6x – y = 0

Given equation is 10x + 11y = 21

Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions. 

As per the given condition 

0.90x + 0.97y = 21 × 0.95 

⇒ 0.90x + 0.97y = 21 × 0.95 ……….(i) 

And 

x + y = 21 

From (ii), substitute y = 21 – x in (i) to get 

0.90x + 0.97(21 – x) = 21 × 0.95 

⇒ 0.90x + 0.97× 21 – 0.97x = 21 × 0.95

⇒ 0.07x = 0.97× 21 – 21 × 0.95 

⇒ x = 21 ×0.02/0.07 = 6 

Now, putting x = 6 in (ii), we have 

6 + y = 21 ⇒ y = 15 

Hence, the request quantities are 6 litres and 15 litres.

Trees are cut down for many reasons.

Trees are cut to make space to build new houses and to clear land to grow crops for human consumption, grass for animals like domestic and milch animals so they can produce dairy foods. The trees are also cut down for production of paper and making of furniture. Sometimes trees are cut to make roads and buildings.

Carbon dioxide Carbon dioxide is absorbed during photosynthesis by plants.

b) restocking of the destroyed forests by planting new trees

17 fully grown trees are required to make tonnes of paper.

Let f (x) = x³ - 3x²a + 2a²x + b 

Since, x - a is a factor of f(x) 

So, 

f (a) = 0 

a³- 3 a² (a) + 2a² (a) + b = 0 

a³ - 3a³ + 2a³ + b= 0 

b = 0

We have, 

(3x – 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵+….a₁x + a₀ 

Putting x = 1, we get 

(3 x 1 – 1)⁷ = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀ 

(2)⁷ = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀ 

a₇+a₆+a₅+….+a₁+a₀ = 128

A lens is portion of transparent refracting medium having at least one of its two surfaces spherical. One of the surfaces may also be plane.

Let, p (x) = 3x³ + 8 (x)² + 8x + 3 + 5k 

g (x) = x² + x + 1 

Given g (x) is a factor of p (x) so remainder will be 0 

Remainder= -2 + 5k 

Therefore, 

- 2 + 5k = 0 

k = 2/5

The correct answer is (c) 

 (c) 2αIΔt 

The power of a lens is the reciprocal of its focal length in metre. It may also be defined as the power of convergence or divergence of a lens. If f is the focal length of a lens in metre and P is the power of the lens. Then:

P = 1/f(in metre).

(i) Since the incident parallel rays are diverging after refraction, the lens represented by the dotted line must be a concave lens. 

(ii) Since the rays, starting from the object point ‘O’, are being made to converge to another point ‘I’, the lens used must be a convex lens.

c = (F – 32) × \(\frac{5}{9}\)

= (185 -32) × \(\frac{5}{9}\)

= 153 × \(\frac{5}{9}\)

185F° = 85°C

C = k – 273 

= 0 – 273 

0 K  = – 273°C

F =\(\frac {9c}{5}\) + 32

\(\frac{9 \times 45}{8}\)

= 81 +32

45°c = 113°F

(b) bend with copper on convex side

In both cases the two surfaces are convex towards air. In case of double convex lens the two surfaces may have any values for the two radii of curvature of the two surfaces while in bi-convex lens the radii of curvature of the two surfaces, has the same value. Thus, every bi-convex lens is a double convex lens about every double convex may not be bi-convex. 

K = C + 273 

= – 30 + 273

The correct answer is: (c), (d)

(c) Celsius scale and ideal gas scale 
(d) ideal gas scale and absolute scale.

The correct answer is (a), (b), (c)

(a) kinetic 

(b) total 

(c) mechanical

The correct answer is all 

(a) kinetic 

(b) total 

(c) mechanical 

(d) internal.