solve: def is_word_there (text, word):What it takes?a text to be searcheda word to be searched within a textwhat it does?searches if a word occurs in a textWhat it returns?returnreturns 1 if word is in the text.returns -1 if word is not in the text# your code goes in here Test Case for Question 1. Do not delete or alter the cell belowscore = dict()text =word = "amazing"try:Python is an amazing programming language.Python is also one of the primary languages in the data analysis field"""if is word_there (text, word) ==1:score['question 1'] = 'pass'score['question 1'] = 'fail'else:except:score['question 1'] = 'fail'is_word_there (text, word)

I have provided PYTHON CODE along with CODE SCREENSHOT and OUTPUT SCREENSHOT--------------------# creating a function that takes two parametersdef is_word_there(text,word): # variables declaration flag = 0 # strip the new line text.strip() # creating a list of words lst = text.split(' ') # access each word for ele in lst: # if word found then if ele == word: # update flag and break the loop flag = 1 break return flag# creating a dictionaryscore = dict()# variables declaration and initializationtext = """ Python is an amazing programming language. Python is also on of the primary language in th data analysis field"""word = "amazing"# handle exceptionstry: # if function returns 1 then if is_word_there(text, word) == 1: score['question 1'] = 'pass' # otherwise else: score['question 1'] = 'fail'except: score['question 1'] = 'fail'# print the score valueprint(score)