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39. Let f (x) = cos (x³) sin² (x⁵).

f' (x) = cos (x³) $\frac{d}{d x}$ sin² (x⁵) + sin² (x⁵) $\frac{d}{d x}$ cos (x³)

= cos (x³) 2sin (x⁵) $\frac{d}{d x}$ sin (x⁵) + sin² (x⁵) [sin (x³)] $\frac{d}{d x}$ x³.

= 2 cos (x³) sin (x⁵). cos (x⁵) $\frac{d}{d x}$ (x⁵) - sin² (x⁵) sin (x³). 3x²

= 2. cos (x³) sin (x⁵) cos (x⁵). 5 - 3x²sin² (x⁵) sin (x³)

= x² sin (x⁵). [2x² cos (x³) cos (x⁵) - 3 sin (x⁵) sin x³].

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26. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:

$\begin{array}{l} \end{array} (a) 2 x + 3 y + 4 z - 12 = 0$

$(b) 3 y + 4 z - 6 = 0$

$(c) x + y + z = 1$

$(d) 5 y + 8 = 0$

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(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1}, y_{1}, z_{1}) .$

$\begin{array}{l} 2 x + 3 y + 4 z - 12 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 x + 3 y + 4 z = 12 \dots (1) \end{array}$

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25. Find the Cartesian equation of the following planes:

$\begin{array}{l} (a) \vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2 \\ (b) \vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1 \\ (c) \vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \end{array}$

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Trigonometric Functions Maths

Contributor-Level 10

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

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21. $2 s i n^{2} \frac{3 π}{4} + 2 c o s^{2} \frac{π}{4} + 2 s e c^{2} \frac{π}{3} = 10$

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Payal Gupta

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Continuity and Differentiability Continuity and Differentiability

38. Kindly consider the following

$\frac{s i n (a x + b)}{c o s (c x + d)}$

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Fortune Institute of International Business Colleges in Delhi

Contributor-Level 10

38. Let f(x) = $\frac{s i n (a x + b)}{c o s (c x + d)} .$

f'(x) = $\frac{c o s (c x + d) \frac{d}{d x} s i n (a x + b) - s i n (a x + b) \frac{d}{d x} c o s (c x + d)}{c o s^{2} (c x + d) .}$

= $\frac{c o s (c x + d) c o s (a x + b) \frac{d}{d x} (a x + b) + s i n (a x + b) s i n (c x + d) \frac{d (x + d)}{d x}}{c o s^{2} (c x + d)}$

= $= \frac{c o s (c x + d) c o s (a x + b) \cdot a + s i n (a x + b) s i n (c x + d) \cdot c}{c o s^{2} (c x + d)}$

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What are the entrance exams accepted for PGDM admission at FIIB Delhi?

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Himanshi Pandey

Contributor-Level 10

FIIB Delhi accepts scores of various national-level Management entrance exams to shortlist candidates for PGDM admission. These exams are CAT, GMAT, XAT, CMAT, MAT and ATMA. Candidates willing to take admission to the two-year programme must apply and appear for at least one of these exams. They must score at least 60 percentile in the selected entrance exam. However, this does not guarantee their admission at the institute. FIIB Delhi conducts a WAT and PI for final selection. It also gives consideration to the applicants' academic background, work experience (if any) and extracurricular achievements.

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Nursing Nursing in Hyderabad

Are there any government Nursing colleges in Hyderabad?

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Tasbiya Khan

Contributor-Level 10

Yes, there are many government Nursing colleges in Hyderabad. Some of them are mentioned below along with their tuition fees:

College Name	Tuition Fee	Eligibility / Exams
NIMS Admission	INR 45,000 – INR 1.38 lakh	Merit-Based
UoH Admission	-	Merit-Based
AIIMS Hyderabad Admission	-	Merit-Based
Bharathi School of Nursing Hyderabad Admission	-	Merit-Based
Swapna School of Nursing Admission	-	Merit-Based

Disclaimer: This information is sourced from official website and may vary.

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NIILM University Pharmacy

Is admission open for the BPharm programme at NIILM University?

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Himanshu Singh

Contributor-Level 10

Yes, admissions for the BPharm programme at NIILM University for the 2025 academic session are currently open. Students who meet the eligibility criteria, including passing Class 12 from a recognised board, can apply. For the most accurate and updated information on the application process, students are encouraged to visit the official university website regularly.

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St. Anne's Degree College for Women Degree College

Which boards are recognised by St Anne's Degree College for Women for BBA admissions?

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Aayush Sharma

Contributor-Level 10

St Anne's Degree College for Women recognises different state and central boards like CBSE, ICSE, BIEAP etc. for its BBA course. Candidates looking to get admission into the BBA programme at St Anne's Degree College for Women should enrol for either a state or central board of education available in their state and apply for admissions.

Candidates who have a state or centrally recognised board can easily get admission into St Anne's Degree College for Women provided they meet the eligibility criteria for BBA at the university. The eligibility criteria for BBA is a 35% minimum aggregate in Class 12 exam.

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Continuity and Differentiability Continuity and Differentiability

37. Kindly consider the following

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37. Kindly go through the solution

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24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$

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Continuity and Differentiability Maths

Contributor-Level 10

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

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10 months ago

36. Kindly consider the following

$s i n (a x + b)$

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36. Let f (x) = sin (ax + b)

f' (x) = $\frac{d}{d x}$ sin (ax + b)

= cos (ax + b) $\frac{d}{d x}$ (ax + b)

= a cos (ax + b).

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23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$(a$ $)$ $z = 2$

$(b) x + y + z = 1$

$(c) 2 x + 3 y - z = 5$

$(d) 5 y + 8 = 0$

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Can I get direct admission for BTech at DIT University through management quota? I got 70% in class 12th.

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(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

Dividing both sides of equation (1) by 1, we obtain

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n

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(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

Dividing both sides of equation (1) by 1, we obtain

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are $\frac{1}{}$ , $\frac{1}{}$ and $\frac{1}{}$ the distance of normal from the origin is $\frac{1}{}$ units.

(c) $2 x + 3 y - z = 5 . . . . . . . . . . (1)$

The direction ratios of normal are 2, 3, and −1.

$∴ + (3) ² + (- 1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{2}{x} + \frac{3}{y} - \frac{1}{z} = \frac{5}{}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane $\frac{2}{x} + \frac{3}{y} - \frac{1}{z}$ and the distance of normal from the origin is $\frac{5}{}$ units.

(d)

$\begin{array}{l} 5 y + 8 = 0 & \Rightarrow 0 x - 5 y + 0 z = 8 . . . . . . . . . . (1) \end{array}$

The direction ratios of normal are 0, −5, and 0.

$∴ + (- 5) ² + 0 = 5$

Dividing both sides of equation (1) by 5, we obtain

$- y = \frac{8}{5}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.

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DIT University Engineering

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Continuity and Differentiability Maths

35. Kindly consider the following

$c o s (s i n x)$

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Contributor-Level 10

35. Let f (x) = cos (sin x).

f' (x) $\frac{d}{d x}$ cos (sin x)

= - sin (sin x) $\frac{d}{d x}$ sin x

= - sin (sin x) cos x.

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22. Find the shortest distance between the lines whose vector equations are

$\begin{array}{l} \vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} = (3 - 2 t) \hat{k} a n d \\ \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} = (2 s + 1) \hat{k} \end{array}$

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