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(B) is the correct answer

Let a, d be the first term and common difference respectively

Therefore,

T_p = a + (p – 1) d = q and       ... (1)

T_{p+ q} = a + (p + q – 1) d = 0    ... (2)

Subtracting (1), from (2) we get 

qd = – q

Substituting in (1) we get 

a = q – (p – 1) (–1) = q + p – 1

Now

T_q = a + (q – 1) d = q + p – 1 + (q – 1) (–1)

= q + p – 1 – q + 1 = p

 sin ^-1 (-1/2)

We know that

sin ^-1 (-1/2) = – sin ^-1 (1/2)

= – sin ^-1 (sin π/6)

= – π/6

(i) sin ^-1 (-1/ √2)

We know that

sin ^-1 (-1/ √2) = – sin ^-1 (sin π/4)

= – π/4

(ii) cos ^-1 (-√3/ 2)

We know that

cos ^-1 (-√3/ 2) = π – cos ^-1 (cos π/6)

= π – π/6

= 5π/6

(iii) tan ^-1 (-√3)

We know that

tan ^-1 (-√3) = – tan ^-1 (tan π/3)

= – π/3

(iv) sec ^-1 (-2)

We know that

sec ^-1 (-2) = π – sec ^-1 (sec π/3)

= π – π/3

= 2 π/3

(v) cosec ^-1 (- √2)

We know that

cosec ^-1 (- √2) = – cosec ^-1 (cosec π/4)

= – π/4

(vi) cot ^-1 (-1/ √3)

We know that

cot ^-1 (-1/ √3) = π – cot ^-1 (cot π/3)

= π – π/3

= 2 π/3

10!

= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 3628800

= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4)

= (5⁴ ) (4 × 3 × 2 × 1)

= (5⁴ ) (4!)

(i) 5!  = 5 x 4 x 3 x 2 x 1 = 120

(ii) 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040

(iii) 8!  = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320

(iv) 4! – 3! = (4 x 3 x 2 x 1) – (3 x 2 x 1) = 24-6= 18

(v) 7! – 5! = (1 x 2 x 3 x 4 x 5 x 6 x 7) -(1 x 2 x 3 x 4 x 5) = 5040 – 120 = 4920

Correct Answer is (A) \(\frac{\pi}{2}\)

Let , x = \(tan^{-1}(-1)+cos^{-1}(\frac{-1}{\sqrt{2}})\) 

⇒ x= \(tan^{-1}(-1)+(\pi-cos^{-1}(\frac{-1}{\sqrt{2}}))\) 

\(\because tan^{-1}(-\theta)=-tan^{-1}(\theta)\) and \(cos^{-1}(-\theta)=\pi-cos^{-1}(\theta))\) 

⇒ x= \(-\frac{\pi}{4}\)\(+(\pi-\frac{\pi}{4})\)

⇒ x= \(-\frac{\pi}{4}\)+\(\frac{3\pi}{4}\)

⇒ x= \(\frac{\pi}{2}\)

3! × 2!

= 3 × 2 × 1 × 2 × 1

= 12

\(\frac{6!-4!}{4!}=\frac{6\times5\times4!-4!}{4!}=\frac{4!(6\times5-1)}{4!}=29\)

(3 × 2)!

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Correct Answer is (C) 0

Let , x = \(cot (tan^{-1}x+cot^{-1}x)\)

⇒ x =\(cot(\frac{\pi}{2})\)

\((\because tan^{-1}x+cot^{-1}x=\frac{\pi}{2})\)

⇒ x= 0

Correct Answer is (C) tan^-12

Let , x = \(tan^{-1}1+tan^{-1}\frac{1}{3}\)

Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\)

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

Correct Answer is (B) \(\frac{\pi}{4}\) 

Let , x = \(tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}\)

 Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\) 

⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)=  tan^-1\((\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{3}\times\frac{2}{3})})\)= \(tan^{-1}1\) = \(\frac{\pi}{4}\)

7, बहुपद का घातांक शून्य है।

Correct Answer is (B)  tan^-1\(\frac{3}{4}\)

Let , x =  tan^-1\(\frac{1}{3}\) +  2 tan^-1\(\frac{1}{3}\) 

 Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\) 

 ⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)=  tan^-1\((\frac{\frac{1}{3}+\frac{1}{3}}{1-(\frac{1}{3}\times\frac{1}{3})})\)= \(tan^{-1}1\)\(\frac{3}{4}\)

3x⁴ + 4x² – 5x + 2, बहुपद का घातांक चार है।

Correct Answer is (A) \(\frac{3}{5}\)

Let , x = \(cos(2 tan^{-1}\frac{1}{2})\)

⇒ x = cos (tan^-1\(\frac{1}{3}\) +  2 tan^-1\(\frac{1}{3}\))

 Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\) 

  ⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)=  tan^-1\((\frac{\frac{1}{2}+\frac{1}{2}}{1-(\frac{1}{2}\times\frac{1}{2})})\)= \(tan^{-1}1\)\(\frac{4}{3}\) 

Now, let y =   \(tan^{-1}1\)\(\frac{4}{3}\) 

⇒ tan y = \(\frac{4}{3}\) = \(\frac{\text{opposite side}}{\text{adjacent side}}\)

By pythagorus theroem , 

(Hypotenuse )² = (opposite side )² + (adjacent side )² 

Therefore, Hypotenuse = 5

⇒ \(cos(tan^{-1}\frac{4}{3})\) = cos y = \(\frac{3}{5}\)

Correct Answer is (B) \(\frac{80}{89}\)

Let, x = \(sin(2tan^{-1}\frac{5}{8})\)

We know that 2 tan^-1 x= sin^-1\((\frac{2x}{1+x^2})\)

⇒ x = sin\((sin^{-1}(\frac{2(\frac{5}{8})}{1+(\frac{5}{8})2})\)= sin(sin^-1(\(\frac{80}{89}\)))= \(\frac{80}{89}\)

Correct Answer is (C) \(\frac{24}{25}\) 

Let, x = \(sin^{-1}\frac{4}{5}\)

⇒ sin x = \(\frac{4}{5}\)

We know that ,cos x =\(\sqrt{1-sin^2x}\)

\(=\sqrt{1-(\frac{4}{5})^2}\)

= \(\frac{3}{5}\)

Now since, x = \(sin^{-1}\frac{4}{5}\), hence \(sin(2sin^{-1}\frac{4}{5})\) become sin (2x)

Here, sin (2x) = 2 sin x cos x

= 2 x \(\frac{4}{5}\) x \(\frac{3}{5}\)

= \(\frac{24}{25}\)

Correct Answer is (A) \(\frac{1}{2}\)

Now, tan^-1 x = tan^-1 1 - tan-1 \(\frac{1}{3}\) \((\because tan\frac{\pi}{4}=1)\)

Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}(\frac{x+y}{1-xy})\)  

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

⇒   tan^-1 x =  \(tan^{-1}1\)\(\frac{1}{2}\)       

⇒ x = \(\frac{1}{2}\)

Correct Answer is (B) \(\frac{\pi}{3}\)

Given:  sin^-1 x + sin^-1 y = \(\frac{2\pi}{3}\) 

Since we know that sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)

⇒ cos^-1 x = \(\frac{\pi}{2}\)- sin^-1 x 

Similarly, cos^-1 y = \(\frac{\pi}{2}\)- sin^-1 y

Now consider cos^-1 x + cos^-1 y = \(\frac{\pi}{2}\)- sin^-1 x + \(\frac{\pi}{2}\) - sin^-1 y

= \(\frac{2\pi}{2}\) - [sin^-1 x + sin^-1 y]

= π - \(\frac{2\pi}{2}\)

= \(\frac{\pi}{3}\)

Correct Answer is (C) 0

Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}(\frac{x+y}{1-xy})\) 

⇒ \(tan^{-1}(1+x)+tan^{-1}(1-x)\)\(=tan^{-1}(\frac{(1+x)+(1-x)}{1-(1+x)(1-x)})\)

= tan^-1 \((\frac{2}{1-(1-x^2)})\)

= tan^-1\((\frac{2}{x^2})\)

Here since \(tan^{-1}(1+x)+tan^{-1}(1-x)=\frac{\pi}{2}\) 

⇒  tan^-1\((\frac{2}{x^2})\) = \(\frac{\pi}{2}\)

⇒ tan^-1\((\frac{2}{x^2})\)= tan^-1\((\infty)\) \((\because tan\frac{\pi}{2}=\infty)\)

⇒ \(\frac{2}{x^2}=\infty\)

⇒ \(x^2=\frac{2}{\infty}\)

⇒ x = 0

It is not possible to predict when and where the next earthquake might occur. But we can predict that there is a possibility of the earthquake when a volcano erupts, or a meteor hits the earth, or an underground nuclear explosion takes place.

Electric charge gets accumulated on the screen. On touching the screen the charge discharges through our body. Thus, we get a slight shock.

A is the lightning conductor and B is the copper plate.

No. There is no need to install lightning conductor in the building.

No, it will not occur. The charge separation cannot take place in conductors. Therefore charges will not accumulate on clouds and so lightning cannot take place.

Lightning will not strike the building because charge separation cannot take place in conductors and so, all the lightning falling on the building will reside on the surface of the building and will pass to the ground through the conductor.

So, there is no need of installing any lightning conductor because all the work of a lightning conductor is done by the conducting material itself.

B) Unlike fractions

Correct option is  B) same

The correct answer is 9/22 and 5/22

(i) (D)  (ii) (A)  (iii) (E)  (iv) (B)

C) Mixed fraction

The correct answer is True

The correct answer is 5/6, 6/6

The correct answer is 3/7, 4/7 7/7

Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Virtual, upright and diminished.

Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Sickle-cell anaemia 

i. It is an autosome-linked recessive trait. 

ii. The disease is controlled by a single pair of allele Hb^A and HbS. 

iii. Only the homozygous individuals for Hb^S, i.e., Hb^SHb^S show the diseased phenotype. 

iv. The heterozygous individuals are carriers (Hb^AHb^S). 

v. Due to point mutation, glutamic acid (Glu) is replaced by valine (Val) at the sixth position of β-globin chain of haemoglobin molecule. 

vi. HbS behaves as normal haemoglobin except under oxygen stress where erythrocytes lose their circular shape and become sickle-shaped. As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affect blood supply to different organs

Thalassemia 

i. It is an autosome-linked recessive disease. 

ii. It occurs due to either mutation or deletion resulting in reduced rate of synthesis of one of globin chains of haemoglobin. 

iii. Anaemia is the characteristic of this disease. 

iv. Thalassemia is classified into two types: 

▪ α-thalassemia—Production of α-globin chain is affected. It is controlled by the closely linked genes HbA1 and HbA2 on chromosome 16. It occurs due to mutation or deletion of one or more of the four genes. 

▪ β-thalassemia—Production of β-globin chain is affected. It occurs due to mutation of one or both HbB genes on chromosome 11.

No, the image formed by a concave lens. 

(i) Is always formed between the optical centre and the focus of the lens.

(ii) Is always virtual, erect and diminished in size. This is true for all positions of the object on the principal axis. 

 Lens is convex; object to be placed at 2F.

 No, the image formed by a concave lens is: 

(i) Always formed between the optical centre and the focus of the lens. 

(ii) Is always virtual, erect and diminished in size. This holds good for all positions of the object.

The lens for which magnification can be 1 is convex lens. For this, the object must be at 2F (i.e.r at distance twice the focal length, in front of the lens). 

A magnifying glass is used by the watch makers to see small screws, etc.

Concave lens.

A 4-chambered heart can pump blood more powerfully and efficiently. This helps in better oxygenation of the blood, better circulation and better purification of the blood.

Sodium chloride.

Ammonium acetate, when dissolved in water is hydrolysed to form ammonium hydroxide and acetic acid. Ammonium hydroxide and acetic acid both are weak alkali and acid respectively, hence the solution is neutral in nature.

When it is mixed with water, crystals of gypsum are produced and set into hard mass

CaSO₄.1/2H₂O+3/2H₂O → CaSO₄.2H₂O

CaSO₄.1/2H₂O-Plaster of Paris

CaSO₄.2H₂O-Gypsum